Cara Membuktikan Teorema Limit Fungsi Trigonometri dan Pembahasan 50+ Soal Latihan

Cara Alternatif Membuktikan Teorema Limit Fungsi Trigonometri Dan Pembahasan Soal Latihan

Calon Guru belajar matematika dari Cara Alternatif Membuktikan Teorema Limit Fungsi Trigonometri yang kita lengkapi dengan contoh soal atau pembahasan soal latihan.

Limit fungsi ini termasuk materi yang sangat penting dan sudah sering kita gunakan dalam kehidupan kita sehari-hari. Hanya saja kita tidak sadar ternyata sedang menggunakan istilah atau bagian dari limit fungsi.

Contoh sederhananya ketika kita mengukur berat badan dan hasilnya terlihat adalah $70,5\ \text{kg}$. Hasil $70,5\ \text{kg}$ ini sebenarnya belum hasil pengukuran yang sebenarnya tetapi sudah bisa mewakili hasil pengukuran, karena berat badan kita adalah mendekati $70,5\ \text{kg}$. Kata "mendekati" adalah salah satu kata kunci dalam belajar limit fungsi.

Definisi Limit Fungsi

Definisi limit fungsi dituliskan: Sebuah limit fungsi mempunyai nilai, Jika nilai $\text{Limit Kiri = Limit Kanan}$ secara simbol dituliskan $\lim\limits_{x \to a^{+}}f(x)=\lim\limits_{x \to a^{-}}f(x)=L$ Maka nilai $\lim\limits_{x \to a}f(x)=L$.

Dari definisi limit fungsi di atas, secara sederhana dapat kita tuliskan, limit fungsi adalah nilai yang dihampiri (didekati) suatu fungsi saat variabelnya mendekati suatu titik tertentu.

Dalam menyelesaikan limit fungsi baik itu limit fungsi aljabar atau limit fungsi trigonometri, langkah awalnya adalah menentukan limit kiri dan limit kanan fungsi tersebut. Akan tetapi kita akan memerlukan energi yang lebih banyak apabila untuk setiap menentukan nilai sebuah limit fungsi kita gunakan langkah-langkah tersebut.

Sehingga untuk menghemat energi, dalam menentukan nilai limit fungsi $\lim\limits_{x \to a}f(x)$, langkah pertama yang kita lakukan adalah dengan cara mensubstitusi nilai $x=a$ ke fungsi $f(x)$ (substitusi langsung).

Setelah dilakukan substitusi langsung dan diperoleh hasilnya bentuk tak tentu seperti $\dfrac{0}{0}$, $\dfrac{\infty}{\infty}$, $0 \times \infty$, $\infty - \infty$, $0^{0}$, $\infty^{0}$ atau $1^{\infty}$ maka dilakukan manipulasi aljabar dengan cara memfaktorkan, mengalikan dengan akar sekawan, atau dengan manipulasi aljabar lainnya dengan tidak melanggar aturan dalam matematika sampai nilai limit fungsi hasilnya bukan bentuk tak tentu.

Teorema Limit Fungsi Trigonometri

Limit fungsi trigonometri adalah pengembangan dari definisi limit fungsi aljabar. Menyelesaikan limit fungsi trignometri dengan menggunkan teorema ini pastinya akan mempercepat menyelesaikan masalah matematika tentang limit fungsi trigonometri. Beberapa teorema limit fungsi trigonometri yang akan kita buktikan antara lain:

$\lim\limits_{x \to 0} \dfrac{\sin x }{x} = 1 , \, \, $ atau $\lim\limits_{x \to 0} \dfrac{ x }{\sin x} = 1$
$\lim\limits_{x \to 0} \dfrac{\tan x }{x} = 1 , \, \, $ atau $\lim\limits_{x \to 0} \dfrac{ x }{\tan x} = 1$
$\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b} , \, \, $ atau $\lim\limits_{x \to 0} \dfrac{ ax }{\sin bx} = \dfrac{a}{b}$
$\lim\limits_{x \to 0} \dfrac{\tan ax }{bx} = \dfrac{a}{b} , \, \, $ atau $\lim\limits_{x \to 0} \dfrac{ ax }{\tan bx} = \dfrac{a}{b}$
$\lim\limits_{x \to 0} \dfrac{\sin ax }{\sin bx} = \dfrac{a}{b} , \, \, $ atau $\lim\limits_{x \to 0} \dfrac{\tan ax }{\tan bx} = \dfrac{a}{b}$
$\lim\limits_{x \to 0} \dfrac{\tan ax }{\sin bx} = \dfrac{a}{b} , \, \, $ atau $\lim\limits_{x \to 0} \dfrac{\sin ax }{\tan bx} = \dfrac{a}{b}$

Soal-soal dan pembahasan tentang limit fungsi yang sudah diujikan pada SBMPTN (Seleksi Bersama Masuk Perguruan Tinggi Negeri), soal SMMPTN (Seleksi Mandiri Masuk Perguruan Tinggi Negeri), atau soal UN (Ujian Nasional) dapat disimak pada catatan berikut.

Cara Pembuktian Teorema Limit Fungsi Trigonmeteri

Teorema Limit $\lim\limits_{x \to 0} \dfrac{x}{\sin x } = 1$

Alternatif Pembuktian:

Teorema $\lim\limits_{x \to 0} \dfrac{\sin x }{x} = 1$ ini menjadi teorema dasar dalam limit fungsi trigonometri yang nanti akan sangat membantu untuk membuktikan teorema-teorema limit fungsi trigonometri yang lain.

Misal pada sebuah lingkaran dengan pusat $O$ dan berjari-jari $r$ kita gambarkan $(1)$ Segitiga $OBA$, $(2)$ Juring $OCA$, dan $(3)$ Segitiga $OCD$.
Sebagai ilustrasi perhatikan gambar berikut:

Cara Alternatif Membuktikan Teorema Limit Fungsi Trigonometri Dengan Sederhana

Dari ketiga gambar di atas jika kita bandingkan berdasarkan luas, maka dapat kita tuliskan sebagai berikut:
\begin{align} \left[ OBA \right] \lt \left[ OCA \right] \lt \left[ OCD \right] \end{align}

Pernahkah luas ketiga bangun di atas sama?
Jawabnya pernah, saat sudut yang dibentuk oleh $OB\ \text{dan}\ OA$, $OA\ \text{dan}\ OC$, atau $OD\ \text{dan}\ OC$ pada gambar di atas yang dimisalkan $x$ besarnya nol derajat sehingga titik $A$ dan $D$ berimpit di $C$. Ketika ketiga titik ini berimpit maka luas ketiga bangun tersebut adalah sama, sehingga dapat juga kita tuliskan:
\begin{align} \left[ OBA \right] \leq \left[ OCA \right] \leq \left[ OCD \right] \end{align}

Dengan bantuan trigonometri, mari kita coba hitung luas ketiga bangun tersebut satu persatu:
$\begin{align}
\left[ OBA \right]\ & = \dfrac{1}{2} \cdot OB \cdot AB \\
& = \dfrac{1}{2} \cdot r \cdot \cos x \cdot r \cdot \sin x \\
& = \dfrac{1}{2} \cdot r^{2} \cdot \cos x \cdot \sin x \\
\hline \left[ OCA \right]\ & = \dfrac{x}{2 \pi} \cdot \text{Luas}\ \bigodot \\
& = \dfrac{x}{2 \pi} \cdot \pi \cdot r^{2} \\
& = \dfrac{1}{2} \cdot x \cdot r^{2} \\
\hline \left[ OCA \right]\ & = \dfrac{1}{2} \cdot OC \cdot CD \\
& = \dfrac{1}{2} \cdot r \cdot OC \cdot \tan x \\
& = \dfrac{1}{2} \cdot r \cdot r \cdot \tan x \\
& = \dfrac{1}{2} \cdot r^{2} \cdot \tan x \\
\end{align}$

Kita kembali kepada ketidaksamaan:
\begin{align} \left[ OBA \right] \leq & \left[ OCA \right] \leq \left[ OCD \right] \\
\dfrac{1}{2} \cdot r^{2} \cdot \cos x \cdot \sin x \leq & \dfrac{1}{2} \cdot x \cdot r^{2} \leq \dfrac{1}{2} \cdot r^{2} \cdot \tan x \\
\end{align}

Jika bentuk di atas, setiap ruas kita bagikan dengan $\left( \frac{1}{2}r^{2} \right)$, maka akan kita peroleh:
\begin{align} \cos x \cdot \sin x \leq x \leq \tan x \\
\cos x \cdot \sin x \leq x \leq \dfrac{\sin x}{\cos x} \\
\end{align}

Lalu bentuk di atas, setiap ruas kita bagikan dengan $\sin x$, maka akan kita peroleh:
\begin{align} \cos x \leq \dfrac{x}{\sin x} \leq \dfrac{1}{\cos x} \\
\end{align}

Berikutnya setiap ruas kita beri $\lim\limits_{x \to 0}$, sehingga dapat kita tuliskan:
\begin{align} \lim\limits_{x \to 0}\ \cos x \leq & \lim\limits_{x \to 0}\ \dfrac{x}{\sin x} \leq \lim\limits_{x \to 0}\ \dfrac{1}{\cos x} \\
\cos 0 \leq & \lim\limits_{x \to 0}\ \dfrac{x}{\sin x} \leq \dfrac{1}{\cos 0} \\
1 \leq & \lim\limits_{x \to 0}\ \dfrac{x}{\sin x} \leq 1 \\
\end{align}

Berdasarkan Teorema Apit nilai $\lim\limits_{x \to 0}\ \dfrac{x}{\sin x} \leq 1$ dan $ \lim\limits_{x \to 0}\ \dfrac{x}{\sin x} \geq 1$ dipenuhi hanya pada saat $\lim\limits_{x \to 0}\ \dfrac{x}{\sin x}=1$.

Sampai pada tahap ini kita sudah membuktikan dan mempunyai sebuah teorema limit fungsi trogonometri yaitu $\lim\limits_{x \to 0}\ \dfrac{x}{\sin x}=1$ dan dapat dikembangkan kepada beberapa teorema limit fungsi trigonometri yang lain.

Teorema Limit $\lim\limits_{x \to 0} \dfrac{\sin x }{x} = 1$

Alternatif Pembuktian:

Teorema Limit $\lim\limits_{x \to 0} \dfrac{\tan x }{x} = 1$

Alternatif Pembuktian:

Untuk membuktikan teorema ini, kita gunakan bantuan teorema yang sudah kita dapatkan sebelumnya $\lim\limits_{x \to 0} \dfrac{ \sin x }{x} = 1$.
$\begin{align}
\lim\limits_{x \to 0} \dfrac{ \tan x }{x}\ & = \lim\limits_{x \to 0} \left( \dfrac{ \frac{\sin x}{\cos x} }{x} \right) \\
& = \lim\limits_{x \to 0} \left( \dfrac{ \sin x }{\cos x} \cdot \dfrac{1}{x} \right) \\
& = \lim\limits_{x \to 0} \left( \dfrac{ \sin x }{x} \cdot \dfrac{1}{\cos x} \right) \\
& = \lim\limits_{x \to 0} \dfrac{ \sin x }{x} \cdot \lim\limits_{x \to 0} \dfrac{1}{\cos x} \\
& = 1 \cdot \dfrac{1}{\cos 0} \\
& = 1 \cdot 1 = 1 \\
& \therefore\ \text{Terbukti}
\end{align}$

Teorema Limit $\lim\limits_{x \to 0} \dfrac{x }{\tan x} = 1$

Alternatif Pembuktian:

Untuk membuktikan teorema ini, kita gunakan bantuan teorema yang sudah kita dapatkan sebelumnya $\lim\limits_{x \to 0} \dfrac{ \sin x }{x} = 1$.
$\begin{align}
\lim\limits_{x \to 0} \dfrac{ x }{\tan x}\ & = \lim\limits_{x \to 0} \left( \dfrac{ x }{ \frac{\sin x}{\cos x} } \right) \\
& = \lim\limits_{x \to 0} \left( x \cdot \dfrac{\cos x}{\sin x} \right) \\
& = \lim\limits_{x \to 0} \left( \dfrac{ x }{\sin x} \cdot \cos x \right) \\
& = \lim\limits_{x \to 0} \dfrac{ x }{\sin x} \cdot \lim\limits_{x \to 0} \cos x \\
& = 1 \cdot \cos 0 \\
& = 1 \cdot 1 = 1 \\
& \therefore\ \text{Terbukti}
\end{align}$

Teorema Limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a }{b}$

Alternatif Pembuktian:

Untuk membuktikan teorema ini, kita gunakan bantuan teorema yang sudah kita dapatkan sebelumnya $\lim\limits_{x \to 0} \dfrac{ \sin x }{x} = 1$.
$\begin{align}
\lim\limits_{x \to 0} \dfrac{\sin ax }{bx}\ & = \lim\limits_{x \to 0} \left( \dfrac{\sin ax }{bx} \cdot \dfrac{a }{a} \right) \\
& = \lim\limits_{x \to 0} \left( \dfrac{a \cdot \sin ax }{a \cdot bx} \right) \\
& = \lim\limits_{x \to 0} \left( \dfrac{a \cdot \sin ax }{b \cdot ax} \right) \\
& = \lim\limits_{x \to 0} \left( \dfrac{\sin ax }{ax} \cdot \dfrac{a}{b} \right) \\
& = \lim\limits_{x \to 0} \dfrac{\sin ax }{ax} \cdot \lim\limits_{x \to 0} \dfrac{a}{b} \\
& = 1 \cdot \dfrac{a}{b} \\
& = \dfrac{a}{b} \\
& \therefore\ \text{Terbukti}
\end{align}$

Teorema Limit $\lim\limits_{x \to 0} \dfrac{ ax }{\sin bx} = \dfrac{a }{b}$

Alternatif Pembuktian:

Untuk membuktikan teorema ini, kita gunakan bantuan teorema yang sudah kita dapatkan sebelumnya $\lim\limits_{x \to 0} \dfrac{ x }{\sin x} = 1$.
$\begin{align}
\lim\limits_{x \to 0} \dfrac{ ax }{\sin bx}\ & = \lim\limits_{x \to 0} \left( \dfrac{ ax }{\sin bx} \cdot \dfrac{b }{b} \right) \\
& = \lim\limits_{x \to 0} \left( \dfrac{b \cdot ax }{b \cdot \sin bx} \right) \\
& = \lim\limits_{x \to 0} \left( \dfrac{a \cdot bx }{b \cdot \sin bx} \right) \\
& = \lim\limits_{x \to 0} \left( \dfrac{bx }{\sin bx} \cdot \dfrac{a}{b} \right) \\
& = \lim\limits_{x \to 0} \dfrac{bx }{\sin bx} \cdot \lim\limits_{x \to 0} \dfrac{a}{b} \\
& = 1 \cdot \dfrac{a}{b} \\
& = \dfrac{a}{b} \\
& \therefore\ \text{Terbukti}
\end{align}$

Teorema Limit $\lim\limits_{x \to 0} \dfrac{\tan ax }{bx} = \dfrac{a }{b}$

Alternatif Pembuktian:

Untuk membuktikan teorema ini, kita gunakan bantuan teorema yang sudah kita dapatkan sebelumnya $\lim\limits_{x \to 0} \dfrac{ \tan x }{x} = 1$.
$\begin{align}
\lim\limits_{x \to 0} \dfrac{\tan ax }{bx}\ & = \lim\limits_{x \to 0} \left( \dfrac{\tan ax }{bx} \cdot \dfrac{a }{a} \right) \\
& = \lim\limits_{x \to 0} \left( \dfrac{a \cdot \tan ax }{a \cdot bx} \right) \\
& = \lim\limits_{x \to 0} \left( \dfrac{a \cdot \tan ax }{b \cdot ax} \right) \\
& = \lim\limits_{x \to 0} \left( \dfrac{\tan ax }{ax} \cdot \dfrac{a}{b} \right) \\
& = \lim\limits_{x \to 0} \dfrac{\tan ax }{ax} \cdot \lim\limits_{x \to 0} \dfrac{a}{b} \\
& = 1 \cdot \dfrac{a}{b} \\
& = \dfrac{a}{b} \\
& \therefore\ \text{Terbukti}
\end{align}$

Teorema Limit $\lim\limits_{x \to 0} \dfrac{ ax }{\tan bx} = \dfrac{a }{b}$

Alternatif Pembuktian:

Untuk membuktikan teorema ini, kita gunakan bantuan teorema yang sudah kita dapatkan sebelumnya $\lim\limits_{x \to 0} \dfrac{ x }{\tan x} = 1$.
$\begin{align}
\lim\limits_{x \to 0} \dfrac{ ax }{\tan bx}\ & = \lim\limits_{x \to 0} \left( \dfrac{ ax }{\tan bx} \cdot \dfrac{b }{b} \right) \\
& = \lim\limits_{x \to 0} \left( \dfrac{b \cdot ax }{b \cdot \tan bx} \right) \\
& = \lim\limits_{x \to 0} \left( \dfrac{a \cdot bx }{b \cdot \tan bx} \right) \\
& = \lim\limits_{x \to 0} \left( \dfrac{bx }{\tan bx} \cdot \dfrac{a}{b} \right) \\
& = \lim\limits_{x \to 0} \dfrac{bx }{\tan bx} \cdot \lim\limits_{x \to 0} \dfrac{a}{b} \\
& = 1 \cdot \dfrac{a}{b} \\
& = \dfrac{a}{b} \\
& \therefore\ \text{Terbukti}
\end{align}$

Teorema Limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{\sin bx} = \dfrac{a }{b}$

Alternatif Pembuktian:

Untuk membuktikan teorema ini, kita gunakan bantuan teorema yang sudah kita dapatkan sebelumnya $\lim\limits_{x \to 0} \dfrac{ ax }{\sin bx} = \dfrac{a}{b}$ dan $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} =\dfrac{a}{b}$.
$\begin{align}
\lim\limits_{x \to 0} \dfrac{ \sin ax }{\sin bx}\ & = \lim\limits_{x \to 0} \left( \dfrac{ \sin ax }{\sin bx} \cdot \dfrac{\frac{1}{ax}}{\frac{1}{ax}} \right) \\
& = \lim\limits_{x \to 0} \left( \dfrac{ \frac{\sin ax}{ax} }{\frac{\sin bx}{ax}}\right) \\
& = \dfrac{\lim\limits_{x \to 0} \frac{\sin ax}{ax} }{\lim\limits_{x \to 0} \frac{\sin bx}{ax}} \\
& = \dfrac{1 }{ \frac{b}{a}} \\
& = \dfrac{a }{b} \\
& \therefore\ \text{Terbukti}
\end{align}$

Teorema Limit $\lim\limits_{x \to 0} \dfrac{\tan ax }{\tan bx} = \dfrac{a }{b}$

Alternatif Pembuktian:

Untuk membuktikan teorema ini, kita gunakan bantuan teorema yang sudah kita dapatkan sebelumnya $\lim\limits_{x \to 0} \dfrac{ ax }{\tan bx} = \dfrac{a}{b}$ dan $\lim\limits_{x \to 0} \dfrac{\tan ax }{bx} =\dfrac{a}{b}$ .
$\begin{align}
\lim\limits_{x \to 0} \dfrac{ \tan ax }{\tan bx}\ & = \lim\limits_{x \to 0} \left( \dfrac{ \tan ax }{\tan bx} \cdot \dfrac{\frac{1}{ax}}{\frac{1}{ax}} \right) \\
& = \lim\limits_{x \to 0} \left( \dfrac{ \frac{\tan ax}{ax} }{\frac{\tan bx}{ax}}\right) \\
& = \dfrac{\lim\limits_{x \to 0} \frac{\tan ax}{ax} }{\lim\limits_{x \to 0} \frac{\tan bx}{ax}} \\
& = \dfrac{1 }{ \frac{b}{a}} \\
& = \dfrac{a }{b} \\
& \therefore\ \text{Terbukti}
\end{align}$

Teorema Limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{\tan bx} = \dfrac{a }{b}$

Alternatif Pembuktian:

Untuk membuktikan teorema ini, kita gunakan bantuan teorema yang sudah kita dapatkan sebelumnya $\lim\limits_{x \to 0} \dfrac{ ax }{\tan bx} = \dfrac{a}{b}$ dan $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} =\dfrac{a}{b}$ .
$\begin{align}
\lim\limits_{x \to 0} \dfrac{ \sin ax }{\tan bx}\ & = \lim\limits_{x \to 0} \left( \dfrac{ \sin ax }{\tan bx} \cdot \dfrac{\frac{1}{ax}}{\frac{1}{ax}} \right) \\
& = \lim\limits_{x \to 0} \left( \dfrac{ \frac{\sin ax}{ax} }{\frac{\tan bx}{ax}}\right) \\
& = \dfrac{\lim\limits_{x \to 0} \frac{\sin ax}{ax} }{\lim\limits_{x \to 0} \frac{\tan bx}{ax}} \\
& = \dfrac{1 }{ \frac{b}{a}} \\
& = \dfrac{a }{b} \\
& \therefore\ \text{Terbukti}
\end{align}$

Teorema Limit $\lim\limits_{x \to 0} \dfrac{\tan ax }{\sin bx} = \dfrac{a }{b}$

Alternatif Pembuktian:

Untuk membuktikan teorema ini, kita gunakan bantuan teorema yang sudah kita dapatkan sebelumnya $\lim\limits_{x \to 0} \dfrac{ ax }{\tan bx} = \dfrac{a}{b}$ dan $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} =\dfrac{a}{b}$ .
$\begin{align}
\lim\limits_{x \to 0} \dfrac{ \tan ax }{\sin bx}\ & = \lim\limits_{x \to 0} \left( \dfrac{ \tan ax }{\sin bx} \cdot \dfrac{\frac{1}{ax}}{\frac{1}{ax}} \right) \\
& = \lim\limits_{x \to 0} \left( \dfrac{ \frac{\tan ax}{ax} }{\frac{\sin bx}{ax}} \right) \\
& = \dfrac{\lim\limits_{x \to 0} \frac{\tan ax}{ax} }{\lim\limits_{x \to 0} \frac{\sin bx}{ax}} \\
& = \dfrac{1 }{ \frac{b}{a}} \\
& = \dfrac{a }{b} \\
& \therefore\ \text{Terbukti}
\end{align}$

SOAL LATIHAN dan PEMBAHASAN LIMIT FUNGSI TRIGONOMETRI

Untuk menambah pengetahuan kita terkait Limit Fungsi Trigonometri mari kita lihat beberapa soal latihan berikut. Soal latihan ini kita pilih dari Modul Matematika SMA Limit Fungsi Trigonometri atau soal-soal yang ditanyakan pada media sosial.

Soal latihan limit fungsi trigonometri berikut ini, silahkan dikerjakan terlebih dahulu secara mandiri sebelum membuka buku atau sumber lain untuk melihat pembahasan soal. Setelah selesai Periksa Jawaban dan jika hasilnya belum memuaskan, pilih Ulangi Tes untuk tes ulang.

Ayo dicoba terlebih dahulu, Sebelum melihat pembahasan soal.
Tunjukkan Kemampuan Terbaikmu!

Nama Peserta :
Tanggal Tes :
Jumlah Soal :	56 soal

Petunjuk Pengerjaan Soal:
Bentuk soal pilihan ganda, pilihlah jawaban yang benar di antara pilihan jawaban yang tersedia. Apabila Kamu merasa terdapat lebih dari satu jawaban yang benar, maka pilihlah yang paling benar.

1. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \dfrac{ \sin 3x }{2x} + \lim\limits_{x \to 0} \dfrac{ 4x }{\tan 2x}=\cdots$

$(A)\ 2$

$(B)\ \frac{7}{2}$

$(C)\ 4$

$(D)\ \frac{11}{2}$

$(E)\ 6$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \dfrac{ \sin 3x }{2x} + \lim\limits_{x \to 0} \dfrac{ 4x }{\tan 2x} \\ & = \dfrac{ \sin 3(0) }{2(0)} + \dfrac{ 4(0) }{\tan 2(0)} \\ & = \dfrac{ 0 }{0} + \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

Dengan menggunakan teorema limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx}=\dfrac{a}{b}$ dan $\lim\limits_{x \to 0} \dfrac{ ax }{\tan bx} = \dfrac{a}{b}$ dapat kita peroleh:
$\begin{align}
& \lim\limits_{x \to 0} \dfrac{ \sin 3x }{2x} + \lim\limits_{x \to 0} \dfrac{ 4x }{\tan 2x} \\ & = \dfrac{ 3 }{2 } + \dfrac{ 4 }{2} \\ & = \dfrac{7}{2} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ \frac{7}{2}$

2. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \dfrac{ \sin 6x }{2 \cdot \sin 2x} + \lim\limits_{x \to 0} \dfrac{ 4 \cdot \tan 3x }{\sin 2x}=\cdots$

$(A)\ 7,5$

$(B)\ 5$

$(C)\ 6,5$

$(D)\ 8$

$(E)\ 8,5$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \dfrac{ \sin 6x }{2 \cdot \sin 2x} + \lim\limits_{x \to 0} \dfrac{ 4 \cdot \tan 3x }{\sin 2x} \\ & = \dfrac{ \sin 6(0) }{2 \cdot \sin 2(0)} + \dfrac{ 4 \cdot \tan 3(0) }{\sin 2(0)} \\ & = \dfrac{ 0 }{0} + \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

Dengan menggunakan teorema limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{\sin bx}=\dfrac{a}{b}$ dan $\lim\limits_{x \to 0} \dfrac{ \tan ax }{\sin bx} = \dfrac{a}{b}$ dapat kita peroleh:
$\begin{align}
& \dfrac{ \sin 6x }{2 \cdot \sin 2x} + \lim\limits_{x \to 0} \dfrac{ 4 \cdot \tan 3x }{\sin 2x} \\ & = \dfrac{ 6}{2 \cdot 2} + \dfrac{4 \cdot 3}{2} \\ & = \dfrac{6}{4} + 6 =7,5 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(A)\ 7,5$

3. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \dfrac{4 \cdot \sin 3x }{6x} + \lim\limits_{x \to 0} \dfrac{ 3\cdot \sin 2x }{4 \cdot \tan 6x}=\cdots$

$(A)\ 2$

$(B)\ 3,5$

$(C)\ 3$

$(D)\ 2,5$

$(E)\ 2,25$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \dfrac{4 \cdot \sin 3x }{6x} + \lim\limits_{x \to 0} \dfrac{ 3\cdot \sin 2x }{4 \cdot \tan 6x} \\ & = \dfrac{4 \cdot \sin 3(0) }{6(0)} + \dfrac{ 3\cdot \sin 2(0) }{4 \cdot \tan 6(0)} \\ & = \dfrac{ 0 }{0} + \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

Dengan menggunakan teorema limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{ bx}=\dfrac{a}{b}$ dan $\lim\limits_{x \to 0} \dfrac{ \sin ax }{\tan bx} = \dfrac{a}{b}$ dapat kita peroleh:
$\begin{align}
& \lim\limits_{x \to 0} \dfrac{4 \cdot \sin 3x }{6x} + \lim\limits_{x \to 0} \dfrac{ 3\cdot \sin 2x }{4 \cdot \tan 6x} \\ & = 4 \cdot \dfrac{3}{6} + \dfrac{3}{4} \cdot \dfrac{2}{6} \\ & = \dfrac{12}{6} + \dfrac{6}{24} = 2 + \dfrac{1}{4} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(E)\ 2,25$

4. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \dfrac{\sin^{2} 2x }{8 \cdot \tan^{2} 3x}=\cdots$

$(A)\ \frac{3}{2}$

$(B)\ \frac{1}{2}$

$(C)\ \frac{1}{18}$

$(D)\ \frac{2}{3}$

$(E)\ 2$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \dfrac{\sin^{2} 2x }{8 \cdot \tan^{2} 3x} \\ & = \dfrac{\sin^{2} 2(0) }{8 \cdot \tan^{2} 3(0)} \\ & = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

Dengan menggunakan teorema limit $\lim\limits_{x \to 0} \dfrac{ \sin ax }{\tan bx} = \dfrac{a}{b}$ dapat kita peroleh:
$\begin{align}
& \lim\limits_{x \to 0} \dfrac{\sin^{2} 2x }{8 \cdot \tan^{2} 3x} \\ & = \lim\limits_{x \to 0} \dfrac{\color{red}{\sin 2x} \cdot \color{blue}{\sin 2x} }{8 \cdot \color{red}{\tan 3x} \cdot \color{blue}{\tan 3x}} \\ & = \dfrac{1}{8} \cdot \color{red}{\dfrac{2}{3}} \cdot \color{blue}{\dfrac{2}{3}} = \dfrac{1}{18} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(C)\ \frac{1}{18}$

5. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \dfrac{6x^{2} }{ \tan^{2} 3x}=\cdots$

$(A)\ \frac{3}{2}$

$(B)\ \frac{2}{3}$

$(C)\ \frac{1}{2}$

$(D)\ 2$

$(E)\ 3$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \dfrac{6x^{2} }{ \tan^{2} 3x} \\ & = \dfrac{6(0)^{2} }{ \tan^{2} 3(0)} \\ & = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

Dengan menggunakan teorema limit $\lim\limits_{x \to 0} \dfrac{ ax }{\tan bx} = \dfrac{a}{b}$ dapat kita peroleh:
$\begin{align}
& \lim\limits_{x \to 0} \dfrac{6x^{2} }{ \tan^{2} 3x} \\ & = \lim\limits_{x \to 0} \dfrac{6 \cdot \color{red}{x} \cdot \color{blue}{x} }{\color{red}{\tan 3x} \cdot \color{blue}{\tan 3x}} \\ & = \dfrac{6}{1} \cdot \color{red}{\dfrac{1}{3}} \cdot \color{blue}{\dfrac{1}{3}} = \dfrac{2}{3} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ \frac{2}{3}$

6. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \dfrac{\sin^{3} 2x }{4x^{2} \cdot \tan x}=\cdots$

$(A)\ 2$

$(B)\ 3$

$(C)\ 1$

$(D)\ 5$

$(E)\ \frac{1}{2}$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \dfrac{\sin^{3} 2x }{4x^{2} \cdot \tan x} \\ & = \dfrac{\sin^{3} 2(0) }{4(0)^{2} \cdot \tan (0)} \\ & = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

Dengan menggunakan teorema limit $\lim\limits_{x \to 0} \dfrac{ \sin ax }{ bx} = \dfrac{a}{b}$ dan $\lim\limits_{x \to 0} \dfrac{ \sin ax }{\tan bx} = \dfrac{a}{b}$ dapat kita peroleh:
$\begin{align}
& \lim\limits_{x \to 0} \dfrac{\sin^{3} 2x }{4x^{2} \cdot \tan x} \\ & = \lim\limits_{x \to 0} \dfrac{\sin 2x \cdot \color{red}{\sin x} \cdot \color{blue}{\sin x} }{ 4x \cdot \color{red}{x} \cdot \color{blue}{\tan x} } \\ & = \dfrac{2}{4}\cdot \color{red}{\dfrac{2}{1}} \cdot \color{blue}{\dfrac{2}{1}} = 2 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(A)\ 2$

7. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \dfrac{2x \cdot \tan^{2} 3x }{\sin^{3} 6x}=\cdots$

$(A)\ \frac{1}{6}$

$(B)\ \frac{2}{3}$

$(C)\ \frac{1}{12}$

$(D)\ \frac{1}{3}$

$(E)\ \frac{3}{8}$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \dfrac{2x \cdot \tan^{2} 3x }{\sin^{3} 6x} \\ & = \dfrac{2(0) \cdot \tan^{2} 3(0) }{\sin^{3} 6(0)} \\ & = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

Dengan menggunakan teorema limit $\lim\limits_{x \to 0} \dfrac{ ax }{\sin bx} = \dfrac{a}{b}$ dan $\lim\limits_{x \to 0} \dfrac{ \tan ax }{\sin bx} = \dfrac{a}{b}$ dapat kita peroleh:
$\begin{align}
& \lim\limits_{x \to 0} \dfrac{2x \cdot \tan^{2} 3x }{\sin^{3} 6x} \\ & = \lim\limits_{x \to 0} \dfrac{2x \cdot \color{red}{\tan 3x} \cdot \color{blue}{\tan 3x} }{ \sin 6x \cdot \color{red}{\sin 6x} \cdot \color{blue}{\sin 6x} } \\ & = \dfrac{2}{6}\cdot \color{red}{\dfrac{3}{6}} \cdot \color{blue}{\dfrac{3}{6}} = \dfrac{1}{12} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(C)\ \frac{1}{12}$

8. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \dfrac{4 \cdot \sin^{3} 3x }{x \cdot \tan 3x \cdot \sin 4x}=\cdots$

$(A)\ \frac{2}{5}$

$(B)\ \frac{8}{3}$

$(C)\ \frac{1}{3}$

$(D)\ 9$

$(E)\ \frac{3}{5}$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \dfrac{4 \cdot \sin^{3} 3x }{x \cdot \tan 3x \cdot \sin 4x} \\ &= \dfrac{4 \cdot \sin^{3} 3(0) }{x \cdot \tan 3(0) \cdot \sin 4(0)} \\ & = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

Dengan menggunakan teorema limit $\lim\limits_{x \to 0} \dfrac{ \sin ax }{ bx} = \dfrac{a}{b}$, $\lim\limits_{x \to 0} \dfrac{ \sin ax }{\sin bx} = \dfrac{a}{b}$, dan $\lim\limits_{x \to 0} \dfrac{ \sin ax }{\tan bx} = \dfrac{a}{b}$ dapat kita peroleh:
$\begin{align}
& \lim\limits_{x \to 0} \dfrac{4 \cdot \sin^{3} 3x }{x \cdot \tan 3x \cdot \sin 4x} \\ & = \lim\limits_{x \to 0} \dfrac{4 \cdot \sin 3x \cdot \color{blue}{\sin 3x} \cdot \color{blue}{\sin 3x} }{ x \cdot \color{red}{\tan 3x} \cdot \color{blue}{\sin 4x} } \\ & = 4 \cdot \dfrac{3}{1} \cdot \color{red}{\dfrac{3}{3}} \cdot \color{blue}{\dfrac{3}{4}} = 9 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(D)\ 9$

9. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \dfrac{3 \cdot \tan^{3} x }{2x \cdot \sin x}=\cdots$

$(A)\ 0$

$(B)\ \frac{1}{2}$

$(C)\ 3$

$(D)\ 4$

$(E)\ \frac{2}{3}$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \dfrac{3 \cdot \tan^{3} x }{2x \cdot \sin x} \\ & = \dfrac{3 \cdot \tan^{3} (0) }{2(0) \cdot \sin (0)} \\ & = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

Dengan menggunakan teorema limit $\lim\limits_{x \to 0} \dfrac{ ax }{\sin bx} = \dfrac{a}{b}$ dan $\lim\limits_{x \to 0} \dfrac{ \tan ax }{\sin bx} = \dfrac{a}{b}$ dapat kita peroleh:
$\begin{align}
& \lim\limits_{x \to 0} \dfrac{3 \cdot \tan^{3} x }{2x \cdot \sin x} \\ & = \lim\limits_{x \to 0} \dfrac{3 \cdot \tan x \cdot \color{red}{\tan x} \cdot \color{blue}{\tan x} }{ 2x \cdot \color{red}{\sin x} } \\ & = 3 \cdot \dfrac{1}{2} \cdot \color{red}{\dfrac{1}{1}} \cdot \color{blue}{\lim\limits_{x \to 0} \tan x } \\ & = \dfrac{3}{2} \cdot \color{blue}{ \tan 0 } \\ & = \dfrac{3}{2} \cdot 0 \\ & = 0 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(A)\ 0$

10. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \left( \dfrac{\sin 2x - \tan 3x}{\tan 6x} \right)=\cdots$

$(A)\ -\frac{1}{3}$

$(B)\ \frac{2}{3}$

$(C)\ -\frac{1}{6}$

$(D)\ 3$

$(E)\ -2$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \left( \dfrac{\sin 2x - \tan 3x}{\tan 6x} \right) \\ & = \dfrac{\sin 2(0) - \tan 3(0)}{\tan 6(0)} \\ & = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

Dengan menggunakan teorema limit $\lim\limits_{x \to 0} \dfrac{ \sin ax }{\tan bx} = \dfrac{a}{b}$ dan $\lim\limits_{x \to 0} \dfrac{ \tan ax }{\tan bx} = \dfrac{a}{b}$ dapat kita peroleh:
$\begin{align}
& \lim\limits_{x \to 0} \left( \dfrac{\sin 2x - \tan 3x}{\tan 6x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{\sin 2x}{\tan 6x} - \dfrac{\sin 2x}{\tan 6x} \right) \\ & = \lim\limits_{x \to 0} \dfrac{\sin 2x}{\tan 6x} - \lim\limits_{x \to 0} \dfrac{\sin 3x}{\tan 6x} \\ & = \dfrac{2}{6} - \dfrac{3}{6} \\ & = -\dfrac{1}{6} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(C)\ -\frac{1}{6}$

11. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \left( \dfrac{4x^{2} + \sin^{2} 2x}{\tan^{2} 4x} \right)=\cdots$

$(A)\ 3$

$(B)\ \frac{1}{2}$

$(C)\ -2$

$(D)\ -\frac{1}{2}$

$(E)\ 1$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \left( \dfrac{4x^{2} + \sin^{2} 2x}{\tan^{2} 4x} \right) \\ & = \dfrac{4(0)^{2} + \sin^{2} 2(0)}{\tan^{2} 4(0)} \\ & = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

Dengan menggunakan teorema limit $\lim\limits_{x \to 0} \dfrac{ax }{\tan bx} = \dfrac{a}{b}$ dan $\lim\limits_{x \to 0} \dfrac{ \sin ax }{\tan bx} = \dfrac{a}{b}$ dapat kita peroleh:
$\begin{align}
& \lim\limits_{x \to 0} \left( \dfrac{4x^{2} + \sin^{2} 2x}{\tan^{2} 4x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{4x^{2}}{\tan^{2} 4x} + \dfrac{\sin^{2} 2x}{\tan^{2} 4x} \right) \\ & = \lim\limits_{x \to 0} \dfrac{4x^{2}}{\tan^{2} 4x} + \lim\limits_{x \to 0} \dfrac{\sin^{2} 2x}{\tan^{2} 4x} \\ & = \lim\limits_{x \to 0} \dfrac{4x \cdot \color{red}{x}}{\tan 4x \cdot \color{red}{\tan 4x}} + \lim\limits_{x \to 0} \dfrac{\sin 2x \cdot \color{red}{\sin 2x}}{\tan 4x \cdot \color{red}{\tan 4x}} \\ & = \dfrac{4}{4} \cdot \color{red}{\dfrac{1}{4}} + \dfrac{2}{4} \cdot \color{red}{\dfrac{2}{4}} \\ & = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{2}{4} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ \frac{1}{2}$

12. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \left( \dfrac{2 \sin^{2}2x - \sin^{2} x}{x \cdot \tan x} \right)=\cdots$

$(A)\ -2$

$(B)\ -1$

$(C)\ \frac{1}{3}$

$(D)\ \frac{1}{2}$

$(E)\ 7$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \left( \dfrac{2 \sin^{2}2x - \sin^{2} x}{x \cdot \tan x} \right) \\ & = \dfrac{2 \sin^{2}2(0) - \sin^{2} (0)}{0 \cdot \tan (0)} \\ & = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

Dengan menggunakan teorema limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{ bx} = \dfrac{a}{b}$ dan $\lim\limits_{x \to 0} \dfrac{ \sin ax }{\tan bx} = \dfrac{a}{b}$ dapat kita peroleh:
$\begin{align}
& \lim\limits_{x \to 0} \left( \dfrac{2 \sin^{2}2x - \sin^{2} x}{x \cdot \tan x} \right) \\ & =\lim\limits_{x \to 0} \left( \dfrac{2 \sin^{2}2x}{x \cdot \tan x} - \dfrac{\sin^{2} x}{x \cdot \tan x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{2 \sin 2x \cdot \color{red}{\sin 2x}}{x \cdot \color{red}{\tan x}}- \dfrac{\sin x \cdot \color{red}{\sin x}}{x \cdot \color{red}{\tan x}} \right) \\ & = 2 \cdot \dfrac{2}{1} \cdot \color{red}{\dfrac{2}{1}}- \dfrac{1}{1} \cdot \color{red}{\dfrac{1}{1}} \\ & = 8-1 = 7 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(E)\ 7$

13. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \left( \dfrac{\sin 2x - 3x}{\tan 3x+\sin 2x} \right)=\cdots$

$(A)\ -\frac{1}{5}$

$(B)\ -2$

$(C)\ 3$

$(D)\ \frac{1}{2}$

$(E)\ \frac{2}{3}$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \left( \dfrac{\sin 2x - 3x}{\tan 3x+\sin 2x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{\sin 2(0) - 3(0)}{\tan 3(0)+\sin 2(0)} \right) \\ & = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(A)\ \frac{-1}{5}$

14. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \left( \dfrac{\tan^{2}4x + 3 \sin^{2}2x}{2 \sin^{2}x - 4x^{2}} \right)=\cdots$

$(A)\ -12$

$(B)\ -14$

$(C)\ 13$

$(D)\ 9$

$(E)\ -8$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \left( \dfrac{\tan^{2}4x + 3 \sin^{2}2x}{2 \sin^{2}x - 4x^{2}} \right) \\ & = \dfrac{\tan^{2}4(0) + 3 \sin^{2}2(0)}{2 \sin^{2}(0) - 4(0)^{2}} \\ & = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

Dengan menggunakan teorema limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{\sin bx} = \dfrac{a}{b}$ dan $\lim\limits_{x \to 0} \dfrac{ \tan ax }{\sin bx} = \dfrac{a}{b}$ dapat kita peroleh:
$\begin{align}
& \lim\limits_{x \to 0} \left( \dfrac{\tan^{2}4x + 3 \sin^{2}2x}{2 \sin^{2}x - 4x^{2}} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{\tan^{2}4x + 3 \sin^{2}2x}{2 \sin^{2}x - 4x^{2}} \right) \cdot \dfrac{\frac{1}{x^{2}}}{\frac{1}{x^{2}}}\\ & = \lim\limits_{x \to 0} \left( \dfrac{\frac{\tan^{2}4x}{x^{2}} + \frac{3 \sin^{2}2x}{x^{2}}}{\frac{2\sin^{2}x}{x^{2}} - \frac{4x^{2}}{x^{2}}} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{\frac{\tan 4x \cdot \color{red}{\tan 4x}}{x \cdot \color{red}{x}} + \frac{3 \sin 2x \cdot \color{red}{\sin 2x}}{x \cdot \color{red}{x}}}{\frac{2\sin x \cdot \color{red}{\sin x}}{x \cdot \color{red}{x}} - 4} \right) \\ & = \dfrac{\frac{4}{1} \cdot \color{red}{\frac{4}{1}} + \frac{3}{1} \cdot \frac{2}{1} \cdot \color{red}{\frac{2}{1}}}{2 \cdot 1 \cdot \color{red}{1}- 4} \\ & = \dfrac{16 + 12}{2- 4} \\ & = \dfrac{28}{-2} = -14 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ -14$

15. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \left( \dfrac{1-\cos 4x}{x^{2}} \right)=\cdots$

$(A)\ 2$

$(B)\ 6$

$(C)\ 8$

$(D)\ 16$

$(E)\ 4$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \left( \dfrac{1-\cos 4x}{x^{2}} \right) \\ & = \dfrac{1-\cos 4(0)}{(0)^{2}} \\ & = \dfrac{1-1}{0} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

Untuk menggunakaan teorema limit trigonometri $\lim\limits_{x \to 0} \dfrac{\sin ax }{ bx} = \dfrac{a}{b}$ ke soal di atas, kita membutuhkan kreativitas dalam mengolah bentuk trigonometri yang ada sampai kepada bentuk teorema limit.

$\cos 2x=\cos^{2}x-\sin^{2}x$
$\cos 2x=1-2\sin^{2}x$
$\cos 2x=2\cos^{2}x+1$
$\cos 4x=\cos^{2}2x-\sin^{2}2x$
$\cos 4x=1-2\sin^{2}2x$
$\cos 4x=2\cos^{2}2x+1$

$\begin{align}
& \lim\limits_{x \to 0} \left( \dfrac{1-\cos 4x}{x^{2}} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{1-\left( 1-2\sin^{2}2x \right)}{x^{2}} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{2\sin^{2}2x }{x^{2}} \right) \\ & = \lim\limits_{x \to 0} \dfrac{2 \sin 2x \cdot \color{red}{\sin 2x} }{x \cdot \color{red}{x}} \\ & = 2 \cdot \dfrac{2}{1} \cdot \color{red}{\dfrac{2}{1}} = 8 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(C)\ 8$

16. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \left( \dfrac{\cos 6x-1}{3 \tan^{2}x} \right)=\cdots$

$(A)\ 6$

$(B)\ -6$

$(C)\ 4$

$(D)\ -\frac{2}{3}$

$(E)\ 3$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \left( \dfrac{\cos 6x-1}{3 \tan^{2}x} \right) \\ & = \dfrac{\cos 6(0)-1}{3 \tan^{2}(0)} \\ & = \dfrac{1-1}{0} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

Untuk menggunakaan teorema limit trigonometri $\lim\limits_{x \to 0} \dfrac{\sin ax }{\tan bx} = \dfrac{a}{b}$ ke soal di atas, kita membutuhkan kreativitas dalam mengolah bentuk trigonometri yang ada sampai kepada bentuk teorema limit.

$\cos 2x=\cos^{2}x-\sin^{2}x$
$\cos 2x=1-2\sin^{2}x$
$\cos 2x=2\cos^{2}x+1$
$\cos 6x=\cos^{2}3x-\sin^{2}3x$
$\cos 6x=1-2\sin^{2}3x$
$\cos 6x=2\cos^{2}3x+1$

$\begin{align}
& \lim\limits_{x \to 0} \left( \dfrac{\cos 6x-1}{3 \tan^{2}x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{ \left( 1-2\sin^{2}3x \right) - 1}{3 \tan^{2}x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{-2\sin^{2}3x }{3 \tan^{2}x} \right) \\ & = \lim\limits_{x \to 0} \dfrac{-2 \sin 3x \cdot \color{red}{\sin 3x} }{3 \tan x \cdot \color{red}{\tan x}} \\ & = \dfrac{-2}{3} \cdot \dfrac{3}{1} \cdot \color{red}{\dfrac{3}{1}} = -6 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ -6$

17. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \left( \dfrac{3-3\cos 2x}{ \sin^{2}3x} \right)=\cdots$

$(A)\ -\frac{2}{3}$

$(B)\ \frac{2}{3}$

$(C)\ -3$

$(D)\ 3$

$(E)\ -2$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \left( \dfrac{3-3\cos 2x}{ \sin^{2}3x} \right) \\ & = \dfrac{3-3\cos 2(0)}{ \sin^{2}3(0)} \\ & = \dfrac{3-3}{0} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

Untuk menggunakaan teorema limit trigonometri $\lim\limits_{x \to 0} \dfrac{\sin ax }{\sin bx} = \dfrac{a}{b}$ ke soal di atas, kita membutuhkan kreativitas dalam mengolah bentuk trigonometri yang ada sampai kepada bentuk teorema limit.

$\cos 2x=\cos^{2}x-\sin^{2}x$
$\cos 2x=1-2\sin^{2}x$
$\cos 2x=2\cos^{2}x+1$

$\begin{align}
& \lim\limits_{x \to 0} \left( \dfrac{3-3\cos 2x}{ \sin^{2}3x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{3 \left( 1 - \cos 2x \right)}{ \sin^{2}3x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{3 \left( 2\sin^{2}x \right)}{ \sin^{2}3x} \right) \\ & = \lim\limits_{x \to 0} \dfrac{6 \sin x \cdot \color{red}{\sin x} }{ \sin 3x \cdot \color{red}{\sin 3x}} \\ & = 6 \cdot \dfrac{1}{3} \cdot \color{red}{\dfrac{1}{3}} = \dfrac{2}{3} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ \frac{2}{3}$

18. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \left( \dfrac{8-8\cos 4x}{4-4\cos 8x} \right)=\cdots$

$(A)\ \frac{1}{16}$

$(B)\ \frac{1}{6}$

$(C)\ \frac{1}{4}$

$(D)\ \frac{1}{2}$

$(E)\ \frac{1}{8}$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \left( \dfrac{8-8\cos 4x}{4-4\cos 8x} \right) \\ & = \dfrac{8-8\cos 4(0)}{4-4\cos 8(0)} = \dfrac{8-8 \cdot \cos 0}{4-4 \cdot \cos 0} \\ & = \dfrac{8-8}{4-4} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\cos 2x=\cos^{2}x-\sin^{2}x$
$\cos 2x=1-2\sin^{2}x$
$\cos 2x=2\cos^{2}x+1$
$\cos 4x=\cos^{2}2x-\sin^{2}2x$
$\cos 4x=1-2\sin^{2}2x$
$\cos 4x=2\cos^{2}2x+1$
$\cos 8x=\cos^{2}4x-\sin^{2}4x$
$\cos 8x=1-2\sin^{2}4x$
$\cos 8x=2\cos^{2}4x+1$

$\begin{align}
& \lim\limits_{x \to 0} \left( \dfrac{8-8\cos 4x}{4-4\cos 8x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{8 \left( 1 - \cos 4x \right)}{4 \left( 1 - \cos 8x \right)} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{8 \left( 2\sin^{2}2x \right)}{4 \left( 2\sin^{2}4x \right)} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{16 \sin 2x \cdot \color{red}{\sin 2x}}{8 \sin 4x \cdot \color{red}{\sin 4x}} \right) \\ & = \dfrac{16}{8} \cdot \dfrac{2}{4} \cdot \color{red}{\dfrac{2}{4}} = \dfrac{1}{2} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(D)\ \frac{1}{2}$

19. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \left( \dfrac{6\cos 2x-6}{2-2\cos 4x} \right)=\cdots$

$(A)\ \frac{1}{16}$

$(B)\ \frac{1}{6}$

$(C)\ -\frac{3}{4}$

$(D)\ \frac{3}{8}$

$(E)\ \frac{1}{8}$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \left( \dfrac{6\cos 2x-6}{2-2\cos 4x} \right) \\ & = \dfrac{6 \cdot \cos 2(0)-6}{2-2 \cdot \cos 4(0)} = \dfrac{6 \cdot \cos 0-6}{2-2 \cdot \cos 0} \\ & = \dfrac{6-6}{2-2} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\cos 2x=\cos^{2}x-\sin^{2}x$
$\cos 2x=1-2\sin^{2}x$
$\cos 2x=2\cos^{2}x+1$
$\cos 4x=\cos^{2}2x-\sin^{2}2x$
$\cos 4x=1-2\sin^{2}2x$
$\cos 4x=2\cos^{2}2x+1$

$\begin{align}
& \lim\limits_{x \to 0} \left( \dfrac{6\cos 2x-6}{2-2\cos 4x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{6 \left( \cos 2x-1 \right)}{2 \left( 1 - \cos 4x \right)} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{6 \left( -2\sin^{2}x \right)}{2 \left( 2\sin^{2}2x \right)} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{-12 \sin x \cdot \color{red}{\sin x}}{4 \sin 2x \cdot \color{red}{\sin 2x}} \right) \\ & = \dfrac{-12}{4} \cdot \dfrac{1}{2} \cdot \color{red}{\dfrac{1}{2}} = \dfrac{-12}{16} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(C)\ -\frac{3}{4}$

20. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \left( \dfrac{1- \cos^{2}x}{1- \cos 4x} \right)=\cdots$

$(A)\ \frac{1}{4}$

$(B)\ \frac{1}{2}$

$(C)\ \frac{2}{3}$

$(D)\ -\frac{1}{4}$

$(E)\ \frac{1}{8}$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \left( \dfrac{1- \cos^{2}x}{1- \cos 4x} \right) \\ & = \dfrac{1- \cos^{2}(0)}{1- \cos 4(0)} = \dfrac{1- \cos (0)}{1- \cos (0)} \\ & = \dfrac{1-1}{1-1} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\sin^{2}x+\cos^{2}x=1$
$\cos 2x=\cos^{2}x-\sin^{2}x$
$\cos 2x=1-2\sin^{2}x$
$\cos 2x=2\cos^{2}x+1$
$\cos 4x=\cos^{2}2x-\sin^{2}2x$
$\cos 4x=1-2\sin^{2}2x$
$\cos 4x=2\cos^{2}2x+1$

$\begin{align}
& \lim\limits_{x \to 0} \left( \dfrac{1- \cos^{2}x}{1- \cos 4x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{\sin^{2}x}{2\sin^{2}2x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{ \sin x \cdot \color{red}{\sin x}}{2 \sin 2x \cdot \color{red}{\sin 2x}} \right) \\ & = \dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \color{red}{\dfrac{1}{2}} = \dfrac{1}{8} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(E)\ \frac{1}{8}$

21. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \left( \dfrac{1- \cos 6x}{1- \cos^{2}2x} \right)=\cdots$

$(A)\ 2$

$(B)\ \frac{9}{2}$

$(C)\ 4$

$(D)\ \frac{1}{4}$

$(E)\ \frac{1}{8}$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \left( \dfrac{1- \cos 6x}{1- \cos^{2}2x} \right) \\ & = \dfrac{1- \cos 6(0)}{1- \cos^{2}2(0)} = \dfrac{1- \cos (0)}{1- \cos (0)} \\ & = \dfrac{1-1}{1-1} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\sin^{2}x+\cos^{2}x=1$
$\sin^{2}2x+\cos^{2}2x=1$
$\cos 2x=\cos^{2}x-\sin^{2}x$
$\cos 2x=1-2\sin^{2}x$
$\cos 2x=2\cos^{2}x+1$
$\cos 6x=\cos^{2}3x-\sin^{2}3x$
$\cos 6x=1-2\sin^{2}3x$
$\cos 6x=2\cos^{2}3x+1$

$\begin{align}
& \lim\limits_{x \to 0} \left( \dfrac{1- \cos 6x}{1- \cos^{2}2x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{2\sin^{2}3x}{\sin^{2}2x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{ 2 \sin 3x \cdot \color{red}{\sin 3x} }{\sin 2x \cdot \color{red}{\sin 2x}} \right) \\ & = 2 \cdot \dfrac{3}{2} \cdot \color{red}{\dfrac{3}{2}} = \dfrac{9}{2} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ \frac{9}{2}$

22. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \left( \dfrac{ \cos x -1}{2- 2\cos^{2} x} \right)=\cdots$

$(A)\ -1$

$(B)\ 1$

$(C)\ -\frac{1}{4}$

$(D)\ \frac{1}{4}$

$(E)\ -\frac{1}{2}$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \left( \dfrac{ \cos x -1}{2- 2\cos^{2} x} \right) \\ & = \dfrac{ \cos (0) -1}{2- 2\cos^{2} (0)} \\ & = \dfrac{1-1}{2-2} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\sin^{2}x+\cos^{2}x=1$
$\cos 2x=\cos^{2}x-\sin^{2}x$
$\cos 2x=1-2\sin^{2}x$
$\cos 2x=2\cos^{2}x+1$
$\cos x=\cos^{2}\frac{1}{2}x-\sin^{2}\frac{1}{2}x$
$\cos x=1-2\sin^{2}\frac{1}{2}x$
$\cos x=2\cos^{2}\frac{1}{2}x+1$

$\begin{align}
& \lim\limits_{x \to 0} \left( \dfrac{ \cos x -1}{2- 2\cos^{2} x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{ -2\sin^{2}\frac{1}{2}x }{2 \left( 1 - \cos^{2} x \right)} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{ 1-2\sin^{2}\frac{1}{2}x-1 }{2 \sin^{2}x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{ -2 \sin \frac{1}{2}x \cdot \color{red}{\sin \frac{1}{2}x} }{2 \sin x \cdot \color{red}{\sin x}} \right) \\ & = \dfrac{2}{2} \cdot \dfrac{-1}{2} \cdot \color{red}{\dfrac{1}{2}} = -\dfrac{1}{4} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(C)\ -\frac{1}{4}$

23. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \left( \dfrac{ 1- \cos 8x }{2x \cdot \sin x \cdot cos x} \right)=\cdots$

$(A)\ 8$

$(B)\ 16$

$(C)\ 2$

$(D)\ \frac{1}{2}$

$(E)\ \frac{1}{4}$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \left( \dfrac{ 1- \cos 8x }{2x \cdot \sin x \cdot cos x} \right) \\ & = \dfrac{ 1- \cos 8(0) }{2(0) \cdot \sin (0) \cdot cos (0)} \\ & = \dfrac{1-1}{0} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\sin 2x =2 \sin x \cdot \cos x$
$\cos 2x=\cos^{2}x-\sin^{2}x$
$\cos 2x=1-2\sin^{2}x$
$\cos 2x=2\cos^{2}x+1$
$\cos 8x=\cos^{2}4x-\sin^{2}4x$
$\cos 8x=1-2\sin^{2}4x$
$\cos 8x=2\cos^{2}4x+1$

$\begin{align}
& \lim\limits_{x \to 0} \left( \dfrac{ 1- \cos 8x }{2x \cdot \sin x \cdot cos x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{ 1- \left( 1-2 \sin^{2}4x \right) }{x \cdot 2 \cdot \sin x \cdot cos x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{ 2 \sin^{2}4x }{x \cdot 2 \cdot \sin x \cdot cos x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{ 2 \sin 4x \cdot \color{red}{\sin 4x} }{ x \cdot \color{red}{\sin 2x}} \right) \\ & = 2 \cdot \dfrac{4}{1} \cdot \color{red}{\dfrac{4}{2}} = 16 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ 16$

24. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \left( \dfrac{ \sin 3x - \sin 3x \cdot \cos 2x }{2x^{3}} \right)=\cdots$

$(A)\ \frac{3}{2}$

$(B)\ 3$

$(C)\ \frac{1}{2}$

$(D)\ 2$

$(E)\ \frac{1}{4}$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \left( \dfrac{ \sin 3x - \sin 3x \cdot \cos 2x }{2x^{3}} \right) \\ & = \dfrac{ \sin 3(0) - \sin 3(0) \cdot \cos 2(0) }{2(0)^{3}} \\ & = \dfrac{0-0}{0} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\cos 2x=\cos^{2}x-\sin^{2}x$
$\cos 2x=1-2\sin^{2}x$
$\cos 2x=2\cos^{2}x+1$

$\begin{align}
& \lim\limits_{x \to 0} \left( \dfrac{ \sin 3x - \sin 3x \cdot \cos 2x }{2x^{3}} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{ \sin 3x \left( 1-2 \cos 2x \right) }{2x \cdot x \cdot x \cdot x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{ \sin 3x \cdot 2\sin^{2}x}{2x \cdot x \cdot x } \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{ 2 \sin 3x \cdot \color{red}{\sin x} \cdot \color{blue}{\sin x} }{ 2x \cdot \color{red}{x} \cdot \color{blue}{x} } \right) \\ & = 2 \cdot \dfrac{3}{2} \cdot \color{red}{1} \cdot \color{blue}{1} = 3 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ 3$

25. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 3} \left( \dfrac{ \sin \left( x-3 \right)}{x^{2}-7x+12} \right)=\cdots$

$(A)\ -1$

$(B)\ 2$

$(C)\ 3$

$(D)\ -3$

$(E)\ 4$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 3} \left( \dfrac{ \sin \left( x-3 \right)}{x^{2}-7x+12} \right) \\ & = \dfrac{ \sin \left( 3-3 \right)}{3^{2}-7(3)+12} \\ & = \dfrac{\sin 0}{0} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\begin{align}
& \lim\limits_{x \to 3} \left( \dfrac{ \sin \left( x-3 \right)}{x^{2}-7x+12} \right) \\ & = \lim\limits_{x \to 3} \left( \dfrac{ \sin \left( x-3 \right)}{\left( x-3 \right)\left( x-4 \right)} \right) \\ & = \lim\limits_{x \to 3} \left( \dfrac{ \sin \left( x-3 \right)}{\left( x-3 \right)} \right) \cdot \lim\limits_{x \to 3} \left( \dfrac{ 1}{\left( x-4 \right)} \right) \\ & = 1 \cdot \dfrac{ 1}{\left( 3-4 \right)} \\ & = \dfrac{1}{-1} = -1 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(A)\ -1$

26. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 2} \left( \dfrac{ \tan \left( 3x-6 \right)}{x^{2}-4} \right)=\cdots$

$(A)\ -2$

$(B)\ 2$

$(C)\ \frac{2}{3}$

$(D)\ \frac{3}{4}$

$(E)\ \frac{3}{5}$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 2} \left( \dfrac{ \tan \left( 3x-6 \right)}{x^{2}-4} \right) \\ & = \dfrac{ \tan \left( 3(2)-6 \right)}{(2)^{2}-4} \\ & = \dfrac{\tan 0}{0} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

Untuk menggunakaan teorema limit trigonometri $\lim\limits_{x \to 0} \dfrac{\tan ax }{ bx} = \dfrac{a}{b}$ ke soal di atas, kita membutuhkan kreativitas dalam mengolah bentuk trigonometri atau bentuk suku banyak sampai kepada bentuk teorema limit.

$\begin{align}
& \lim\limits_{x \to 2} \left( \dfrac{ \tan \left( 3x-6 \right)}{x^{2}-4} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{ \tan 3\left( x-2 \right)}{\left( x-2 \right)\left( x+2 \right)} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{ \tan 3\left( x-2 \right)}{\left( x-2 \right)} \right) \cdot \lim\limits_{x \to 2} \left( \dfrac{ 1}{\left( x+2 \right)} \right) \\ & = 3 \cdot \dfrac{ 1}{\left( 2+2 \right)} \\ & = \dfrac{3}{4} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(D)\ \frac{3}{4}$

27. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 3} \left( \dfrac{ \sin \left( x^{2}-9 \right)}{x-3} \right)=\cdots$

$(A)\ 3$

$(B)\ 6$

$(C)\ 9$

$(D)\ 2$

$(E)\ -1$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 3} \left( \dfrac{ \tan \left( x^{2}-9 \right)}{x-3} \right) \\ & = \dfrac{ \tan \left( 3^{2}-9 \right)}{3-3} \\ & = \dfrac{\tan 0}{0} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\begin{align}
& \lim\limits_{x \to 3} \left( \dfrac{ \sin \left( x^{2}-9 \right)}{x-3} \right) \\ & = \lim\limits_{x \to 3} \left( \dfrac{ \sin \left( x-3 \right)\left( x+3 \right)}{ \left( x-3 \right)} \right) \\ & = \lim\limits_{x \to 3} \left( \dfrac{ \sin \left( x-3 \right)}{\left( x-3 \right)} \right) \cdot \lim\limits_{x \to 3} \left( x+3 \right) \\ & = 1 \cdot \left( 3+3 \right) \\ & = 6 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ 6$

28. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 4} \left( \dfrac{ x-4 }{\sin \left( x^{2}-5x+4 \right)} \right)=\cdots$

$(A)\ 2$

$(B)\ \frac{1}{2}$

$(C)\ \frac{1}{3}$

$(D)\ 1$

$(E)\ -\frac{1}{2}$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 4} \dfrac{ x-4 }{\sin \left( x^{2}-5x+4 \right)} \\ & = \dfrac{ 4-4 }{\sin \left( 4^{2}-5(4)+4 \right)} \\ & = \dfrac{0}{\sin 0} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\begin{align}
& \lim\limits_{x \to 4} \left( \dfrac{ x-4 }{\sin \left( x^{2}-5x+4 \right)} \right) \\ & = \lim\limits_{x \to 4} \left( \dfrac{ x-4 }{\sin \left( x-1 \right)\left( x-4 \right)} \right) \\ & = \lim\limits_{x \to 4} \dfrac{ 1}{\left( x-1 \right)} \\ & = 1 \cdot \dfrac{ 1}{\left( 4-1 \right)} \\ & = \dfrac{ 1}{3} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(C)\ \frac{1}{3}$

29. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 5} \dfrac{ \sqrt{x-1} - 2 }{\tan \left( x-5 \right)}=\cdots$

$(A)\ 4$

$(B)\ \frac{1}{4}$

$(C)\ 3$

$(D)\ \frac{1}{2}$

$(E)\ \frac{1}{3}$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 5} \dfrac{ \sqrt{x-1} - 2 }{\tan \left( x-5 \right)} \\ & = \dfrac{ \sqrt{5-1} - 2 }{\tan \left( 5-5 \right)} \\ & = \dfrac{0}{\tan 0} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

Untuk menggunakaan teorema limit trigonometri $\lim\limits_{x \to 0} \dfrac{ ax }{\tan bx} = \dfrac{a}{b}$ ke soal di atas, kita membutuhkan kreativitas dalam mengolah bentuk trigonometri atau bentuk akar sampai kepada bentuk teorema limit.

$\begin{align}
& \lim\limits_{x \to 5} \dfrac{ \sqrt{x-1} - 2 }{\tan \left( x-5 \right)} \\ & = \lim\limits_{x \to 5} \dfrac{ \sqrt{x-1} - 2 }{\tan \left( x-5 \right)} \cdot \dfrac{ \sqrt{x-1} + 2 }{ \sqrt{x-1} + 2 } \\ & = \lim\limits_{x \to 5} \dfrac{ x-1 - 4 }{\tan \left( x-5 \right) \left( \sqrt{x-1} + 2 \right) } \\ & = \lim\limits_{x \to 5} \dfrac{ x-5 }{\tan \left( x-5 \right) \left( \sqrt{x-1} + 2 \right) } \\ & = \lim\limits_{x \to 5} \dfrac{ x-5 }{\tan \left( x-5 \right) } \cdot \lim\limits_{x \to 5} \dfrac{ 1 }{ \left( \sqrt{x-1} + 2 \right) } \\ & = 1 \cdot \dfrac{ 1}{\sqrt{5-1} + 2} \\ & = \dfrac{ 1}{4} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ \frac{ 1}{4}$

30. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \dfrac{ 1 - \sqrt{\cos 2x}}{x^{2}}=\cdots$

$(A)\ 2$

$(B)\ -2$

$(C)\ 4$

$(D)\ -4$

$(E)\ 1$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \dfrac{ 1 - \sqrt{\cos 2x}}{x^{2}} \\ & = \dfrac{ 1 - \sqrt{\cos 0}}{0} \\ & = \dfrac{1- \sqrt{1}}{0} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\begin{align}
& \lim\limits_{x \to 0} \dfrac{ 1 - \sqrt{\cos 2x}}{x^{2}} \\ & = \lim\limits_{x \to 0} \dfrac{ 1 - \sqrt{\cos 2x}}{x^{2}} \cdot \dfrac{ 1+\sqrt{\cos 2x}}{ 1+\sqrt{\cos 2x} } \\ & = \lim\limits_{x \to 0} \dfrac{ 1 - \cos 2x }{x^{2} \left( 1+\sqrt{\cos 2x} \right)} \\ & = \lim\limits_{x \to 0} \dfrac{ 1 - \left( 1- 2 \sin^{2}x \right)}{x^{2}} \cdot \lim\limits_{x \to 0} \dfrac{ 1}{ 1+\sqrt{\cos 2x} } \\ & = \lim\limits_{x \to 0} \dfrac{ 2 \sin^{2}x }{x^{2}} \cdot \lim\limits_{x \to 0} \dfrac{ 1}{ 1+\sqrt{\cos 2x} } \\ & = 2 \cdot \dfrac{ 1}{1+ \sqrt{\cos 0}} = 2 \cdot \dfrac{ 1}{1+1} \\ & = 1 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(E)\ 1$

31. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to a} \left( \dfrac{ \left( x-a \right)}{3x-3a+ \tan \left( x-a \right)} \right)=\cdots$

$(A)\ 1$

$(B)\ \frac{1}{4}$

$(C)\ \frac{1}{2}$

$(D)\ \frac{1}{3}$

$(E)\ 2$

Alternatif Pembahasan:

$\begin{align}
& \lim\limits_{x \to a} \left( \dfrac{ \left( x-a \right)}{3x-3a+ \tan \left( x-a \right)} \right) \\ & = \dfrac{ \left( a-a \right)}{3(a)-3a+ \tan \left( a-a \right)} \\ & = \dfrac{0}{0 \tan 0} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\begin{align}
& \lim\limits_{x \to a} \left( \dfrac{ \left( x-a \right)}{3x-3a+ \tan \left( x-a \right)} \right) \\ & = \lim\limits_{x \to a} \left( \dfrac{ \left( x-a \right)}{3\left( x-a \right)+ \tan \left( x-a \right)} \cdot \dfrac{\frac{1}{x-a}}{\frac{1}{x-a}} \right) \\ & = \lim\limits_{x \to a} \left( \dfrac{ \frac{x-a}{x-a}}{\frac{3\left( x-a \right)+ \tan \left( x-a \right)}{x-a}} \right) \\ & = \lim\limits_{x \to a} \left( \dfrac{1}{3+ \frac{\tan \left( x-a \right)}{x-a}} \right) \\ & = \dfrac{1}{3+ 1} \\ & = \dfrac{1}{4} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ \frac{1}{4}$

32. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 2} \left( \dfrac{ x^{2}-x-2 }{\tan \left( 2x-4 \right) + \sin \left( x-2 \right)} \right)=\cdots$

$(A)\ \frac{3}{2}$

$(B)\ 3$

$(C)\ -\frac{1}{2}$

$(D)\ -\frac{1}{4}$

$(E)\ 1$

Alternatif Pembahasan:

$\begin{align}
& \lim\limits_{x \to 2} \left( \dfrac{ x^{2}-x-2 }{\tan \left( 2x-4 \right) + \sin \left( x-2 \right)} \right) \\ & = \dfrac{ (2)^{2}-(2)-2 }{\tan \left( 2(2)-4 \right) + \sin \left( (2)-2 \right)} \\ & = \dfrac{0}{\tan 0 + \sin 0} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\begin{align}
& \lim\limits_{x \to 2} \left( \dfrac{ x^{2}-x-2 }{\tan \left( 2x-4 \right) + \sin \left( x-2 \right)} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{ (x-2)(x+1) }{\tan 2\left( x-2 \right) + \sin \left( x-2 \right)} \cdot \dfrac{\frac{1}{x-2}}{\frac{1}{x-2}} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{ \frac{(x-2)(x+1)}{x-2} }{\frac{\tan 2\left( x-2 \right)}{x-2} + \frac{\sin \left( x-2 \right)}{x-2}} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{ (x-2)(x+1) }{\tan 2\left( x-2 \right) + \sin \left( x-2 \right)} \cdot \dfrac{\frac{1}{x-2}}{\frac{1}{x-2}} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{ x+1 }{\frac{\tan 2\left( x-2 \right)}{x-2} + \frac{\sin \left( x-2 \right)}{x-2}} \right) \\ & = \dfrac{ (2)+1 }{ 2 + 1 } \\ & = 1 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(E)\ 1$

33. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \left( \dfrac{\cos 4x -1}{\cos 5x - \cos 3x} \right)=\cdots$

$(A)\ -1$

$(B)\ \frac{1}{2}$

$(C)\ 1$

$(D)\ -\frac{1}{2}$

$(E)\ \frac{2}{3}$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \left( \dfrac{\cos 4x -1}{\cos 5x - \cos 3x} \right) \\ & = \dfrac{\cos 4(0) -1}{\cos 5(0) - \cos 3(0)} \\ & = \dfrac{1-1}{1-1} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\cos 2x=1-2\sin^{2}x$
$\cos 4x=1-2\sin^{2}2x$
$\cos A + \cos B= 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$
$\cos A - \cos B= -2 \sin \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)$
$\cos 5x - \cos 3x= -2 \sin 4x \sin x$

$\begin{align}
& \lim\limits_{x \to 0} \left( \dfrac{\cos 4x -1}{\cos 5x - \cos 3x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{1-2\sin^{2}2x -1}{-2 \sin 4x \sin x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{-2\sin^{2}2x}{-2 \sin 4x \sin x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{ -2 \sin 2x \cdot \color{red}{\sin 2x}}{-2 \sin 4x \cdot \color{red}{\sin x}} \right) \\ & = \dfrac{-2}{-2} \cdot \dfrac{2}{4} \cdot \color{red}{\dfrac{2}{1}} = 1 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(C)\ 1$

34. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \left( \dfrac{4 - 4 \cos^{2} 5x}{\cos 3x - \cos 2x} \right)=\cdots$

$(A)\ -20$

$(B)\ -40$

$(C)\ 24$

$(D)\ -24$

$(E)\ 40$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \left( \dfrac{4 - 4 \cos^{2} 5x}{\cos 3x - \cos 2x} \right) \\ & = \dfrac{4 - 4 \cos^{2} 5(0)}{\cos 3(0) - \cos 2(0)} \\ & = \dfrac{4-4}{1-1} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\sin^{2} x + \cos^{2} x=1$
$\sin^{2} 5x + \cos^{2} 5x=1$
$\cos A + \cos B= 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$
$\cos A - \cos B= -2 \sin \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)$
$\cos 3x - \cos 2x= -2 \sin \frac{5}{2}x \sin \frac{1}{2}x$

$\begin{align}
& \lim\limits_{x \to 0} \left( \dfrac{4 - 4 \cos^{2} 5x}{\cos 3x - \cos 2x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{4 \left(1 - \cos^{2} 5x \right)}{-2 \sin \frac{5}{2}x \sin \frac{1}{2}x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{4 \left( \sin^{2} 5x \right)}{-2 \sin \frac{5}{2}x \sin \frac{1}{2}x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{ 4 \sin 5x \cdot \color{red}{\sin 5x}}{-2 \sin \frac{5}{2}x \cdot \color{red}{\sin \frac{1}{2}x}} \right) \\ & = \dfrac{4}{-2} \cdot \dfrac{5}{\frac{5}{2}} \cdot \color{red}{\dfrac{5}{\frac{1}{2}}} = -40 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ -40$

35. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{1 - \tan x}{\sin x - \cos x} \right)=\cdots$

$(A)\ -\frac{1}{2}\sqrt{2}$

$(B)\ \frac{1}{2}\sqrt{2}$

$(C)\ -\sqrt{2}$

$(D)\ \sqrt{2}$

$(E)\ -1$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{1 - \tan x}{\sin x - \cos x} \right) \\ & = \dfrac{1 - \tan \frac{\pi}{4}}{\sin \frac{\pi}{4} - \cos \frac{\pi}{4}} \\ & = \dfrac{1-1}{\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\begin{align}
& \lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{1 - \tan x}{\sin x - \cos x} \right) \\ & = \lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{\frac{\cos x}{\cos x} - \frac{\sin x}{\cos x}}{\sin x - \cos x} \right) \\ & = \lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{\frac{\color{red}{\cos x - \sin x}}{\cos x}}{- \left( \color{red}{\cos x - \sin x} \right)} \right) \\ & = \lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{\frac{1}{\cos x}}{- 1} \right) \\ & = \lim\limits_{x \to \frac{\pi}{4}} \dfrac{-1}{\cos x} \\ & = \dfrac{-1}{\cos \frac{\pi}{4}} \\ & = \dfrac{-1}{\frac{1}{2}\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}} \\ & = -\sqrt{2} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(C)\ -\sqrt{2}$

36. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{\cot x - \tan x}{\cos x - \sin x} \right)=\cdots$

$(A)\ -1$

$(B)\ 2$

$(C)\ \infty $

$(D)\ -2$

$(E)\ 2\sqrt{2}$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{\cot x - \tan x}{\cos x - \sin x} \right) \\ & = \dfrac{\cot \frac{\pi}{4} - \tan \frac{\pi}{4}}{\cos \frac{\pi}{4} - \sin \frac{\pi}{4}} \\ & = \dfrac{1-1}{\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\begin{align}
& \lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{\cot x - \tan x}{\cos x - \sin x} \right) \\ & = \lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{\frac{\cos x}{\sin x} - \frac{\sin x}{\cos x}}{\cos x - \sin x} \right) \\ & = \lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{\frac{\cos^{2} x - \sin^{2} x}{\sin x \cos x}}{ \cos x - \sin x} \right) \\ & = \lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{\left( \cos x + \sin x \right)\left( \cos x - \sin x \right)}{ \sin x \cos x} \cdot \dfrac{1}{ \cos x - \sin x} \right) \\ & = \lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{ \cos x + \sin x }{ \sin x \cos x} \right) \\ & = \dfrac{ \cos \frac{\pi}{4} + \sin \frac{\pi}{4} }{ \sin \frac{\pi}{4} \cdot \cos \frac{\pi}{4}} \\ & = \dfrac{ \frac{1}{2}\sqrt{2} + \frac{1}{2}\sqrt{2} }{ \frac{1}{2}\sqrt{2} \cdot \frac{1}{2}\sqrt{2}} \\ & = \dfrac{ \sqrt{2} }{ \frac{1}{4} \left(2 \right) } \\ & = 2\sqrt{2} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(E)\ 2\sqrt{2}$

37. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to \frac{\pi}{2}} \left( \dfrac{1 + \cos 2x}{\cos x} \right)=\cdots$

$(A)\ 2$

$(B)\ 1$

$(C)\ \infty $

$(D)\ 0$

$(E)\ -2$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to \frac{\pi}{2}} \left( \dfrac{1 + \cos 2x}{\cos x} \right) \\ & = \dfrac{1 + \cos 2\left( \frac{\pi}{2} \right)}{\cos \frac{\pi}{2}} = = \dfrac{1 + \cos \pi}{\cos \frac{\pi}{2}} \\ & = \dfrac{1-1}{0} = \dfrac{0}{0} = \text{tak tentu} \end{align}$

$\begin{align}
& \lim\limits_{x \to \frac{\pi}{2}} \left( \dfrac{1 + \cos 2x}{\cos x} \right) \\ & = \lim\limits_{x \to \frac{\pi}{2}} \left( \dfrac{1 + 1 - 2 \sin^{2} x}{\cos x} \right) \\ & = \lim\limits_{x \to \frac{\pi}{2}} \left( \dfrac{2 - 2\sin^{2} x}{\cos x} \right) \\ & = \lim\limits_{x \to \frac{\pi}{2}} \dfrac{2 \left( 1 - \sin^{2} x \right)}{\cos x} \\ & = \lim\limits_{x \to \frac{\pi}{2}} \left( \dfrac{2 \cos^{2} x}{\cos x} \right) \\ & = \lim\limits_{x \to \frac{\pi}{2}} 2 \cos x \\ & = 2 \cos \frac{\pi}{2} \\ & = 2 \cdot 0 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(D)\ 0$

38. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{ \tan x - 1}{ \cos 2x} \right)=\cdots$

$(A)\ -2$

$(B)\ 2$

$(C)\ -\frac{1}{2}$

$(D)\ \frac{1}{2}$

$(E)\ -1$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{ \tan x - 1}{ \cos 2x} \right) \\ & = \dfrac{ \tan \frac{\pi}{4} - 1}{ \cos 2\left( \frac{\pi}{4} \right)} \\ & = \dfrac{1-1}{\cos \frac{\pi}{2}} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\begin{align}
& \lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{ \tan x - 1}{ \cos 2x} \right) \\ &= \lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{\frac{\sin x}{\cos x} - \frac{\cos x}{\cos x}}{\cos^{2}x-\sin^{2}x} \right) \\ &= \lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{\frac{\sin x - \cos x}{\cos x}}{\left(\cos x - \sin x \right)\left(\cos x + \sin x \right)} \right) \\ &= \lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{- \left( \cos x - \sin x \right)}{\cos x} \cdot \dfrac{1}{\left(\cos x - \sin x \right)\left(\cos x + \sin x \right)} \right) \\ &= \lim\limits_{x \to \frac{\pi}{4}} \dfrac{-1}{\left(\cos x \right)\left(\cos x + \sin x \right)} \\ &= \dfrac{-1}{\left(\cos \frac{\pi}{4} \right)\left(\cos \frac{\pi}{4} + \sin \frac{\pi}{4} \right)} \\ &= \dfrac{-1}{\left(\frac{1}{2}\sqrt{2} \right)\left(\frac{1}{2}\sqrt{2} + \frac{1}{2}\sqrt{2} \right)} \\ &= \dfrac{-1}{\left(\frac{1}{2}\sqrt{2} \right)\left( \sqrt{2} \right)} = -1 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(E)\ -1$

39. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to \frac{\pi}{2}} \left( \sec^{2} x - \sec x \cdot \tan x \right)=\cdots$

$(A)\ -\frac{1}{2}\sqrt{3}$

$(B)\ \frac{1}{2}\sqrt{3}$

$(C)\ -\frac{1}{2}$

$(D)\ \frac{1}{2}$

$(E)\ 2$

Alternatif Pembahasan:

Untuk menggunakaan teorema limit trigonometri ke soal di atas, kita membutuhkan kreativitas dalam mengolah bentuk trigonometri yang ada sampai kepada bentuk teorema limit.

$\begin{align}
& \lim\limits_{x \to \frac{\pi}{2}} \left( \sec^{2} x - \sec x \cdot \tan x \right) \\ &= \lim\limits_{x \to \frac{\pi}{2}} \left( \dfrac{1}{\cos^{2} x} - \dfrac{1}{\cos x} \cdot \dfrac{\sin x}{\cos x} \right) \\ &= \lim\limits_{x \to \frac{\pi}{2}} \left( \dfrac{1}{\cos^{2} x} - \dfrac{\sin x}{\cos^{2} x} \right) \\ &= \lim\limits_{x \to \frac{\pi}{2}} \left( \dfrac{1-\sin x}{\cos^{2} x} \right) \\ &= \lim\limits_{x \to \frac{\pi}{2}} \left( \dfrac{1-\sin x}{1-\sin^{2} x} \right) \\ &= \lim\limits_{x \to \frac{\pi}{2}} \left( \dfrac{1-\sin x}{\left( 1-\sin x\right)\left( 1+\sin x\right)} \right) \\ &= \lim\limits_{x \to \frac{\pi}{2}} \left( \dfrac{ 1}{ 1+\sin x } \right) \\ &= \dfrac{1 }{ 1+\sin \frac{\pi}{2} } \\ &= \dfrac{1 }{ 1+1 } = \dfrac{1 }{ 2 } \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(D)\ \frac{1 }{ 2 }$

40. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \left( \dfrac{ \left( x^{2}-7x+12 \right) \tan 2x }{5x^{2}+4x} \right)=\cdots$

$(A)\ -7$

$(B)\ -6$

$(C)\ -3$

$(D)\ 3$

$(E)\ 6$

Alternatif Pembahasan:

$\begin{align}
& \lim\limits_{x \to 0} \left( \dfrac{ \left( x^{2}-7x+12 \right) \tan 2x }{5x^{2}+4x} \right) \\ & = \dfrac{ \left( (0)^{2}-7(0)+12 \right) \tan 2(0) }{5(0)^{2}+4(0)} \\ & = \dfrac{(0) \tan 0}{0} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\begin{align}
& \lim\limits_{x \to 0} \left( \dfrac{ \left( x^{2}-7x+12 \right) \tan 2x }{5x^{2}+4x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{ (x-4)(x-3) \tan 2x }{\left( 5x+4 \right) x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{ (x-4)(x-3) }{ 5x+4 } \cdot \dfrac{ \tan 2x }{x} \right) \\ & = \dfrac{ (0-4)(0-3) }{ 5(0)+4 } \cdot 2 \\ & = \dfrac{ 12 }{ 4 } \cdot 2 = 6 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(E)\ 6$

41. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \left( \dfrac{1 - \cos 4x \cdot \sin^{2}x - \cos^{2}x }{x^{4}} \right)=\cdots$

$(A)\ 8$

$(B)\ 4$

$(C)\ 2$

$(D)\ 0$

$(E)\ -2$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \left( \dfrac{1 - \cos 4x \cdot \sin^{2}x - \cos^{2}x }{x^{4}} \right) \\ & = \dfrac{1 - \cos 4(0) \cdot \sin^{2}(0) - \cos^{2}(0) }{(0)^{4}} \\ & = \dfrac{1-0-1}{0} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\begin{align}
& \lim\limits_{x \to 0} \left( \dfrac{1 - \cos 4x \cdot \sin^{2}x - \cos^{2}x }{x^{4}} \right) \\ &= \lim\limits_{x \to 0} \left( \dfrac{1 - \cos^{2}x - \cos 4x \cdot \sin^{2}x }{x^{4}} \right) \\ &= \lim\limits_{x \to 0} \left( \dfrac{ \sin^{2}x - \cos 4x \cdot \sin^{2}x }{x^{4}} \right) \\ &= \lim\limits_{x \to 0} \left( \dfrac{ \sin^{2}x \left( 1 - \cos 4x \right) }{x^{4}} \right) \\ &= \lim\limits_{x \to 0} \left( \dfrac{ \sin^{2}x \left( 1 - \left( 1 - 2\sin^{2} 2x \right) \right) }{x^{4}} \right) \\ &= \lim\limits_{x \to 0} \left( \dfrac{ \sin^{2}x \cdot 2\sin^{2} 2x }{x^{4}} \right) \\ &= \lim\limits_{x \to 0} \left( \dfrac{ \sin^{2}x \cdot \color{red}{ 2\sin^{2} 2x} }{x^{2} \cdot \color{red}{x^{2}}} \right) \\ &= \lim\limits_{x \to 0} \left( \dfrac{ \sin x \cdot \sin x \cdot \color{red}{ 2\sin 2x \sin 2x} }{x \cdot x \cdot \color{red}{x \cdot x }} \right) \\ &= 1 \cdot 1 \cdot \color{red}{2 \cdot 2 \cdot 2}=8 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(A)\ 8$

42. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \left( \dfrac{ \left( x^{2}-1 \right) \sin 6x }{x^{3}+3x^{2}+2x} \right)=\cdots$

$(A)\ 6$

$(B)\ 3$

$(C)\ 0$

$(D)\ -1$

$(E)\ -3$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \left( \dfrac{ \left( x^{2}-1 \right) \sin 6x }{x^{3}+3x^{2}+2x} \right) \\ & = \dfrac{ \left( (0)^{2}-1 \right) \sin 6(0) }{(0)^{3}+3(0)^{2}+2(0)} \\ & = \dfrac{-1 \cdot \sin 0}{0} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\begin{align}
& \lim\limits_{x \to 0} \left( \dfrac{ \left( x^{2}-1 \right) \sin 6x }{x^{3}+3x^{2}+2x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{ \left( x -1 \right)\left( x +1 \right) \sin 6x }{\left( x+1 \right)\left( x+2 \right)\left( x \right)} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{ \left( x -1 \right) \sin 6x }{ \left( x+2 \right)\left( x \right)} \right) \\ & = \lim\limits_{x \to 0} \dfrac{ \left( x -1 \right) }{ \left( x+2 \right) } \cdot \lim\limits_{x \to 0} \dfrac{ \sin 6x }{x} \\ & = \dfrac{ 0 -1 }{ 0+2 } \cdot 6 = -3 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(E)\ -3$

43. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to \frac{\pi}{3}} \left( \dfrac{\sin x - \sin \frac{\pi}{3}}{\cos x - \cos \frac{\pi}{3}} \right)=\cdots$

$(A)\ -\frac{1}{3}\sqrt{3}$

$(B)\ \frac{1}{3}\sqrt{3}$

$(C)\ \sqrt{3}$

$(D)\ -\sqrt{3}$

$(E)\ -\frac{1}{2}\sqrt{3}$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to \frac{\pi}{3}} \left( \dfrac{\sin x - \sin \frac{\pi}{3}}{\cos x - \cos \frac{\pi}{3}} \right) \\ & = \dfrac{\sin \frac{\pi}{3} - \sin \frac{\pi}{3}}{\cos \frac{\pi}{3} - \cos \frac{\pi}{3}} \\ & = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

Untuk menggunakaan teorema limit trigonometri ke soal di atas, kita membutuhkan kreativitas dalam mengolah bentuk trigonometri yang ada sampai kepada bentuk teorema limit.

$\sin A + \sin B= 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$
$\sin A - \sin B= 2 \cos \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)$
$\sin x - \sin 60= 2 \cos \left( \frac{x+60}{2} \right) \sin \left( \frac{x-60}{2} \right)$
$\cos A + \cos B= 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$
$\cos A - \cos B= -2 \sin \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)$
$\cos x - \cos 60= -2 \sin \frac{x+60}{2} \sin \frac{x-60}{2}$

$\begin{align}
& \lim\limits_{x \to \frac{\pi}{3}} \left( \dfrac{\sin x - \sin \frac{\pi}{3}}{\cos x - \cos \frac{\pi}{3}} \right) \\ & = \lim\limits_{x \to 60} \left( \dfrac{\sin x - \sin 60}{\cos x - \cos 60} \right) \\ & = \lim\limits_{x \to 60} \left( \dfrac{2 \cos \frac{x+60}{2} \sin \frac{x-60}{2} }{-2 \sin \frac{x+60}{2} \sin \frac{x-60}{2}} \right) \\ & = \lim\limits_{x \to 60} \left( -\dfrac{ \cos \frac{x+60}{2} }{ \sin \frac{x+60}{2} } \right) \\ & = -\dfrac{ \cos \frac{60+60}{2} }{ \sin \frac{60+60}{2} } \\ & = -\dfrac{ \cos 60 }{ \sin 60 } \\ & = - \dfrac{ \frac{1}{2} }{ \frac{1}{2}\sqrt{3} } = -\dfrac{1}{3}\sqrt{3} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(A)\ -\frac{1}{3}\sqrt{3}$

44. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to y} \left( \dfrac{\tan x - \tan y}{ \left( 1-\frac{x}{y} \right)\left( 1+\tan x\ \tan y \right)} \right)=\cdots$

$(A)\ -1$

$(B)\ 0$

$(C)\ 1$

$(D)\ y$

$(E)\ -y$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to y} \left( \dfrac{\tan x - \tan y}{ \left( 1-\frac{x}{y} \right)\left( 1+\tan x\ \tan y \right)} \right) \\ & = \dfrac{\tan y - \tan y}{ \left( 1-\frac{y}{y} \right)\left( 1+\tan y\ \tan y \right)} \\ & = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\tan \left( A + B \right)= \dfrac{\tan A + \tan B}{ 1-\tan x\ \tan y }$
$\tan \left( A - B \right)= \dfrac{\tan x - \tan y}{ 1+\tan x\ \tan y }$

$\begin{align}
& \lim\limits_{x \to y} \left( \dfrac{\tan x - \tan y}{ \left( 1-\frac{x}{y} \right)\left( 1+\tan x\ \tan y \right)} \right) \\ & = \lim\limits_{x \to y} \left( \dfrac{1}{1-\frac{x}{y}} \cdot \dfrac{\tan x - \tan y}{ 1+\tan x\ \tan y } \right) \\ & = \lim\limits_{x \to y} \left( \dfrac{1}{\frac{y}{y}-\frac{x}{y}} \cdot \tan \left( x - y \right) \right) \\ & = \lim\limits_{x \to y} \left( \dfrac{y}{y-x} \cdot \tan \left( x - y \right) \right)\\ & = \lim\limits_{x \to y} \left( -\dfrac{y}{x-y} \cdot \tan \left( x - y \right) \right)\\ & = \lim\limits_{x \to y} \left( -\dfrac{y\ \tan \left( x - y \right) }{x-y} \right) \\ & = -y \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(E)\ -y$

45. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \left( \dfrac{ x^{2} \cdot \tan 9x }{2 \sin 3x - \sin 6x } \right)=\cdots$

$(A)\ -2$

$(B)\ -\frac{1}{3}$

$(C)\ \frac{1}{3}$

$(D)\ 2$

$(E)\ 5$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \left( \dfrac{ x^{2} \cdot \tan 9x }{2 \sin 3x - \sin 6x } \right) \\ & = \dfrac{ (0)^{2} \cdot \tan 9(0) }{2 \sin 3(0) - \sin 6(0) } \\ & = \dfrac{0}{0-0} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

Untuk menggunakaan teorema limit trigonometri $\lim\limits_{x \to 0} \dfrac{\tan ax }{\sin bx} = \dfrac{a}{b}$ ke soal di atas, kita membutuhkan kreativitas dalam mengolah bentuk trigonometri yang ada sampai kepada bentuk teorema limit.

$\sin 2x=2 \sin x\ \cos x$
$\sin 6x=2 \sin 3x\ \cos 3x$
$\cos 2x=\cos^{2}x-\sin^{2}x$
$\cos 2x=1-2\sin^{2}x$
$\cos 3x=1-2\sin^{2}\frac{3}{2}x$
$\cos 4x=1-2\sin^{2}2x$

$\begin{align}
& \lim\limits_{x \to 0} \left( \dfrac{ x^{2} \cdot \tan 9x }{2 \sin 3x - \sin 6x } \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{ x^{2} \cdot \tan 9x }{2 \sin 3x - 2 \sin 3x\ \cos 3x } \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{ x^{2} \cdot \tan 9x }{2 \sin 3x \left( 1 - \cos 3x \right) } \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{ x^{2} \cdot \tan 9x }{2 \sin 3x \left( 1 - 1-2\sin^{2}\frac{3}{2}x \right) } \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{ x^{2} \cdot \tan 9x }{2 \sin 3x \left( 2\sin^{2}\frac{3}{2}x \right) } \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{ x^{2} \cdot \tan 9x }{4\ \sin^{2}\frac{3}{2}x \cdot \sin 3x } \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{ x^{2} \cdot \color{red}{\tan 9x} }{ 4\ \sin^{2}\frac{3}{2}x \cdot \color{red}{\sin 3x} } \right) \\ & = \dfrac{1}{4} \cdot \dfrac{2}{3} \cdot \dfrac{2}{3} \cdot \color{red}{\dfrac{9}{3}} = \dfrac{1}{3} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(C)\ \frac{1}{3}$

46. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \left( \dfrac{ 2x-x\sqrt{2+\sqrt{2+{\sqrt{2+2\cos 4x}}}} }{\tan x - \sin x} \right)=\cdots$

$(A)\ -3$

$(B)\ \frac{1}{2}$

$(C)\ 2$

$(D)\ 3$

$(E)\ 5$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \left( \dfrac{ 2x-x\sqrt{2+\sqrt{2+{\sqrt{2+2\cos 4x}}}} }{\tan x - \sin x} \right) \\ & = \dfrac{ 2(0)-(0)\sqrt{2+\sqrt{2+{\sqrt{2+2\cos 4(0)}}}} }{\tan (0) - \sin (0)} \\ & = \dfrac{0-0}{0-0} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\cos 2x=\cos^{2}x-\sin^{2}x$
$\cos 2x=12\sin^{2}x$
$\cos 4x=1-2\sin^{2}2x$
$\cos x=1-2\sin^{2}\frac{1}{2}x$

$\begin{align}
& \lim\limits_{x \to 0} \left( \dfrac{ 2x-x\sqrt{2+\sqrt{2+{\sqrt{2+2\cos 4x}}}} }{\tan x - \sin x} \right) \\ &= \lim\limits_{x \to 0} \left( \dfrac{ 2x-x\sqrt{2+\sqrt{2+{\sqrt{2+2\left(2 \cos^{2}2x-1 \right)}}}} }{\tan x - \sin x} \right) \\ &=\lim\limits_{x \to 0} \left( \dfrac{ 2x-x\sqrt{2+\sqrt{2+{\sqrt{2+4 \cos^{2}2x-2 }}}} }{\color{red}{\tan x - \sin x}} \right) \\ &=\lim\limits_{x \to 0} \left( \dfrac{ 2x-x\sqrt{2+\sqrt{2+{\sqrt{4 \cos^{2}2x}}}} }{\color{red}{\frac{\sin x}{\cos x} - \sin x}} \right) \\ &=\lim\limits_{x \to 0} \left( \dfrac{ 2x-x\sqrt{2+\sqrt{2+2 \cos 2x}} }{\color{red}{\frac{\sin x- \sin x\ \cos x}{\cos x}}} \right) \\ &=\lim\limits_{x \to 0} \left( \dfrac{ 2x-x\sqrt{2+\sqrt{2+2 \left(2 \cos^{2}x-1 \right)}} }{\color{red}{\frac{\sin x \left( 1 - \cos x \right)}{\cos x}}} \right) \\ &=\lim\limits_{x \to 0} \left( \dfrac{ 2x-x\sqrt{2+\sqrt{2+4 \cos^{2}x-2}} }{\color{red}{ \sin x \left( 1 - \cos x \right)}} \cdot \color{red}{\dfrac{\cos x}{1}} \right) \\ &=\lim\limits_{x \to 0} \left( \dfrac{ 2x-x\sqrt{2+\sqrt{4 \cos^{2}x}} }{\color{red}{ \sin x \left( 1 - \cos x \right)}} \cdot \color{red}{ \cos x } \right) \\ &=\lim\limits_{x \to 0} \left( \dfrac{ 2x-x\sqrt{2+2 \cos x } }{\color{red}{ \sin x \left( 1 - \cos x \right)}} \cdot \color{red}{\cos x} \right) \\ &=\lim\limits_{x \to 0} \left( \dfrac{ 2x-x\sqrt{2+2 \left(2 \cos^{2} \frac{1}{2}x-1 \right) } }{\color{red}{ \sin x \left( 1 - \cos x \right)}} \cdot \color{red}{\cos x} \right) \\ &=\lim\limits_{x \to 0} \left( \dfrac{ 2x-x\sqrt{2+4 \cos^{2} \frac{1}{2}x-2 } }{\color{red}{ \sin x \left( 1 - \cos x \right)}} \cdot \color{red}{\cos x} \right) \\ &=\lim\limits_{x \to 0} \left( \dfrac{ 2x-x\sqrt{4 \cos^{2} \frac{1}{2}x} }{\color{red}{ \sin x \left( 1 - \cos x \right)}} \cdot \color{red}{\cos x} \right) \\ &=\lim\limits_{x \to 0} \left( \dfrac{ 2x-2x \cos \frac{1}{2}x }{\color{red}{ \sin x \left( 1 - \cos x \right)}} \cdot \color{red}{\cos x} \right) \\ &=\lim\limits_{x \to 0} \left( \dfrac{ 2x \left( 1- \cos \frac{1}{2}x \right) }{\color{red}{ \sin x \left( 1 - \cos x \right)}} \cdot \color{red}{\cos x} \right) \\ &=\lim\limits_{x \to 0} \left( \dfrac{ 2x \left( 2 \sin^{2} \frac{1}{4}x \right) }{\color{red}{ \sin x \left( 2 \sin^{2} \frac{1}{2}x \right)}} \cdot \color{red}{\cos x} \right) \\ &= \dfrac{2}{1} \cdot \dfrac{ 2}{2} \cdot \dfrac{ \frac{1}{4}}{\frac{1}{2}} \cdot \dfrac{ \frac{1}{4}}{\frac{1}{2}} \cdot \cos 0 = \dfrac{1}{2} \\ \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ \frac{1}{2}$

47. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \left( \dfrac{ 1-\sin x - \cos x}{1 +\sin x - \cos x} \right)=\cdots$

$(A)\ -3$

$(B)\ -2$

$(C)\ -1$

$(D)\ 3$

$(E)\ 5$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \left( \dfrac{ 1-\sin x - \cos x}{1 +\sin x - \cos x} \right) \\ & = \dfrac{ 1-\sin (0) - \cos (0)}{1 +\sin (0) - \cos (0)} \\ & = \dfrac{1+0-1}{1+0-1} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\cos 2x=\cos^{2}x-\sin^{2}x$
$\cos 2x=12\sin^{2}x$
$\cos 4x=1-2\sin^{2}2x$
$\cos x=1-2\sin^{2}\frac{1}{2}x$

$\begin{align}
& \lim\limits_{x \to 0} \left( \dfrac{ 1-\sin x - \cos x}{1 +\sin x - \cos x} \right) \\ &=\lim\limits_{x \to 0} \left( \dfrac{ 1-\cos x - \sin x}{1-\cos x +\sin x } \right) \\ &=\lim\limits_{x \to 0} \left( \dfrac{ 1- \left( 1-2\sin^{2}\frac{1}{2}x \right) - \sin x}{1- \left( 1-2\sin^{2}\frac{1}{2}x \right) +\sin x } \right) \\ &=\lim\limits_{x \to 0} \left( \dfrac{ 2\sin^{2}\frac{1}{2}x - \sin x}{ 2\sin^{2}\frac{1}{2}x +\sin x } \cdot \dfrac{\frac{1}{x}}{\frac{1}{x}} \right)\\ &=\lim\limits_{x \to 0} \left( \dfrac{ \frac{2\sin^{2}\frac{1}{2}x}{x} - \frac{\sin x}{x}}{\frac{ 2\sin^{2}\frac{1}{2}x}{x} +\frac{\sin x}{x} } \right) \\ &= \dfrac{ \lim\limits_{x \to 0} \frac{2\sin^{2}\frac{1}{2}x}{x} - \lim\limits_{x \to 0}\frac{\sin x}{x}}{\lim\limits_{x \to 0}\frac{ 2\sin^{2}\frac{1}{2}x}{x} +\lim\limits_{x \to 0}\frac{\sin x}{x} } \\ &= \dfrac{ \lim\limits_{x \to 0} \frac{2 \sin \frac{1}{2}(0)}{1} \cdot \lim\limits_{x \to 0} \frac{ \sin \frac{1}{2}x}{x} - \lim\limits_{x \to 0}\frac{\sin x}{x}}{\lim\limits_{x \to 0} \frac{2 \sin \frac{1}{2}(0)}{1} \cdot \lim\limits_{x \to 0} \frac{ \sin \frac{1}{2}x}{x} + \lim\limits_{x \to 0}\frac{\sin x}{x}} \\ &= \dfrac{ 2 \cdot \sin 0 \cdot \frac{1}{2} - 1}{2 \cdot \sin 0 \cdot \frac{1}{2} + 1} \\ &= \dfrac{ 0 - 1}{0 + 1} =-1 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(C)\ -1$

48. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{ \sin x - \cos x}{ x - \frac{\pi}{4}} \right)=\cdots$

$(A)\ -\sqrt{3}$

$(B)\ -\sqrt{2}$

$(C)\ \sqrt{2}$

$(D)\ \sqrt{3}$

$(E)\ 2\sqrt{2}$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{ \sin x - \cos x}{ x - \frac{\pi}{4}} \right) \\ & = \dfrac{ \sin \frac{\pi}{4} - \cos \frac{\pi}{4}}{ \frac{\pi}{4} - \frac{\pi}{4}} \\ & = \dfrac{ \frac{1}{2}\sqrt{2} - \frac{1}{2}\sqrt{2} }{ 0} = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\sin \left( 90-x \right)=\cos x $
$\cos \left( 90-x \right)=\sin x $
$\sin A + \sin B= 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$
$\sin A - \sin B= 2 \cos \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)$

$\begin{align}
& \lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{ \sin x - \cos x}{ x - \frac{\pi}{4}} \right) \\ &=\lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{ \sin x-\sin \left( 90-x \right)}{ x - \frac{\pi}{4}} \right) \\ & =\lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{ \sin x-\sin \left( 90-x \right)}{ x - 45 } \right) \\ & =\lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{ 2 \cos \left( \frac{x+90-x}{2} \right) \sin \left( \frac{x-90+x}{2} \right)}{ x - 45} \right) \\ & =\lim\limits_{x \to \frac{\pi}{4}} \left( \dfrac{ 2 \cos 45\ \sin \left( x-45 \right)}{ x - 45} \right) \\ & =\lim\limits_{x \to \frac{\pi}{4}} \left( 2 \cos 45 \right) \cdot \lim\limits_{x \to \frac{\pi}{4}} \dfrac{ \sin \left( x-45 \right)}{ x - 45} \\ & = 2 \cdot \dfrac{1}{2}\sqrt{2} = \sqrt{2} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(C)\ \sqrt{2}$

49. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to a} \left( \dfrac{\tan x - \tan a}{\sin x - \sin a} \right)=\cdots$

$(A)\ \sec^{3}a$

$(B)\ \cos^{3}a$

$(C)\ 2 \cdot \sin^{3}a$

$(D)\ 2 \cdot \sec^{2}a$

$(E)\ 2 \cdot \cos^{2}a$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to a} \left( \dfrac{\tan x - \tan a}{\sin x - \sin a} \right) \\ & = \dfrac{\tan a - \tan a}{\sin a - \sin a} \\ & = \dfrac{ 0 }{0} = \text{tak tentu} \end{align}$

$\sin^{2} A + \cos^{2} A= 1$
$1 + \cot^{2} A= \csc^{2} A$
$\tan^{2} A + 1= \sec^{2} A$
$\tan \left( A + B \right)= \dfrac{\tan A + \tan B}{ 1-\tan A\ \tan B }$
$\tan \left( A - B \right)= \dfrac{\tan A - \tan B}{ 1+\tan A\ \tan B }$

$\begin{align}
& \lim\limits_{x \to a} \left( \dfrac{\tan x - \tan a}{\sin x - \sin a} \right) \\ & = \lim\limits_{x \to a} \left( \dfrac{ \tan (x-a) \left( 1-\tan x \cdot \tan a \right)}{2 \cos \left( \frac{x+a}{2} \right) \cdot \sin \left( \frac{x-a}{2} \right) } \right) \\ & = \lim\limits_{x \to a} \left( \dfrac{ \tan (x-a) \left( 1+\tan x \cdot \tan a \right)}{2 \cos \left( \frac{x+a}{2} \right) \cdot \sin \left( \frac{x-a}{2} \right) } \right) \\ & = \lim\limits_{x \to a} \left( \dfrac{ \tan (x-a) \left( 1+\tan x \cdot \tan a \right)}{2\ \sin \left( \frac{x-a}{2} \right) \cdot \cos \left( \frac{x+a}{2} \right) } \right) \\ & = \lim\limits_{x \to a} \left( \dfrac{ \tan (x-a) }{2\ \sin \frac{1}{2}\left( x-a \right) } \right) \cdot \lim\limits_{x \to a} \left( \dfrac{ 1+\tan x \cdot \tan a }{\cos \frac{1}{2}\left( x+a \right) } \right) \\ & = \dfrac{ 1 }{2\ \cdot \frac{1}{2} } \cdot \left( \dfrac{ 1+\tan a \cdot \tan a }{\cos \frac{1}{2}\left( a+a \right) } \right) \\ & = \dfrac{ 1+\tan^{2} a }{\cos a} \\ & = \dfrac{ \sec^{2} a }{\cos a} = \sec^{3} a \\ \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(A)\ \sec^{3} a$

50. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to \frac{\pi}{2}} \left( 1 - \tan x \right)\left( 1 - \tan \frac{1}{2}x \right)=\cdots$

$(A)\ -3$

$(B)\ -2$

$(C)\ -1$

$(D)\ 3$

$(E)\ 5$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to \frac{\pi}{2}} \left( 1 - \tan x \right)\left( 1 - \tan \frac{1}{2}x \right) \\ & = \left( 1 - \tan \frac{\pi}{2} \right)\left( 1 - \tan \frac{\pi}{4} \right) \\ & = \left( 1 - \infty \right)\left( 1 - 1 \right) \\ & = \left( \infty \right)\left( 0 \right) = \text{tak tentu} \end{align}$

$\sin^{2} A + \cos^{2} A= 1$
$1 + \cot^{2} A= \csc^{2} A$
$\tan^{2} A + 1= \sec^{2} A$
$\cos 2x=\cos^{2}x-\sin^{2}x$
$\cos x=\cos^{2} \frac{1}{2}x-\sin^{2}\frac{1}{2}x$

$\begin{align}
& \lim\limits_{x \to \frac{\pi}{2}} \left( 1 - \tan x \right)\left( 1 - \tan \frac{1}{2}x \right) \\ & = \lim\limits_{x \to 90} \left( \dfrac{\cos x}{\cos x} - \dfrac{\sin x}{\cos x} \right)\left( \dfrac{\cos \frac{1}{2}x}{\cos \frac{1}{2}x} - \dfrac{\sin \frac{1}{2}x}{\cos \frac{1}{2}x} \right) \\ & = \lim\limits_{x \to 90} \left( \dfrac{\cos x-\sin x}{\cos x} \right)\left( \dfrac{\cos \frac{1}{2}x-\sin \frac{1}{2}x}{\cos \frac{1}{2}x} \right) \\ & = \lim\limits_{x \to 90} \left( \dfrac{\cos x-\sin x}{\cos^{2}\frac{1}{2}x-\sin^{2}\frac{1}{2}x} \right)\left( \dfrac{\cos \frac{1}{2}x-\sin \frac{1}{2}x}{\cos \frac{1}{2}x} \right) \\ & = \lim\limits_{x \to 90} \left( \dfrac{\cos x-\sin x}{\color{red}{\left( \cos \frac{1}{2}x-\sin \frac{1}{2}x \right)}\left( \cos \frac{1}{2}x + \sin \frac{1}{2}x \right)} \right)\left( \dfrac{ \color{red}{\cos \frac{1}{2}x-\sin \frac{1}{2}x}}{\cos \frac{1}{2}x} \right) \\ & = \lim\limits_{x \to 90} \left( \dfrac{\cos x-\sin x}{\left( \cos \frac{1}{2}x \right)\left( \cos \frac{1}{2}x + \sin \frac{1}{2}x \right)} \right) \\ & = \dfrac{\cos 90-\sin 90}{\left( \cos 45 \right)\left( \cos 45 + \sin 45 \right)} \\ & = \dfrac{0-1}{\left( \frac{1}{2}\sqrt{2} \right)\left( \frac{1}{2}\sqrt{2} + \frac{1}{2}\sqrt{2} \right)} \\ & = \dfrac{-1}{1}=-1 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(C)\ -1$

51. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to 0} \dfrac{\sin 4x - \tan 4x}{\sin 2x - \tan 2x}=\cdots$

$(A)\ -4$

$(B)\ -2$

$(C)\ 3$

$(D)\ 4$

$(E)\ 8$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \dfrac{\sin 4x - \tan 4x}{\sin 2x - \tan 2x} \\ & = \dfrac{\sin (0) - \tan (0)}{\sin (0) - \tan (0)} \\ & = \dfrac{0 - 0}{0 - 0} =\dfrac{0}{0} = \text{tak tentu} \end{align}$

$\cos 2x=\cos^{2}x-\sin^{2}x$
$\cos 2x=1-2\sin^{2}x$
$\cos 4x=\cos^{2}2x-\sin^{2}2x$
$\cos 4x=1-2\sin^{2}2x$

$\begin{align}
& \lim\limits_{x \to 0} \dfrac{\sin 4x - \tan 4x}{\sin 2x - \tan 2x} \\ & = \lim\limits_{x \to 0} \dfrac{\sin 4x - \frac{\sin 4x}{\cos 4x}}{\sin 2x - \frac{\sin 2x}{\cos 2x}} \\ & = \lim\limits_{x \to 0} \left( \dfrac{\frac{\sin 4x \cos 4x-\sin 4x}{\cos 4x}}{ \frac{\sin 2x \cos 2x-\sin 2x}{\cos 2x}} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{\sin 4x \cos 4x-\sin 4x}{\cos 4x} \cdot \dfrac{\cos 2x}{\sin 2x \cos 2x-\sin 2x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{\sin 4x \left( \cos 4x- 1 \right)}{\cos 4x} \cdot \dfrac{\cos 2x}{\sin 2x \left( \cos 2x- 1 \right)} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{\sin 4x \left( \cos 4x- 1 \right)}{\sin 2x \left( \cos 2x- 1 \right)} \cdot \dfrac{\cos 2x}{\cos 4x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{\sin 4x \left( 2\sin^{2}2x \right)}{\sin 2x \left( 2\sin^{2}x \right)} \cdot \dfrac{\cos 2x}{\cos 4x} \right) \\ & = \dfrac{4}{2} \cdot \dfrac{2}{2} \cdot \dfrac{2}{1} \cdot \dfrac{2}{1} \cdot \dfrac{\cos 0}{\cos 0} \\ & = 8 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(E)\ 8$

52. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{p \to \frac{\pi}{2}} \dfrac{\sin 6p}{p - \frac{\pi}{2}}=\cdots$

$(A)\ 6$

$(B)\ 3$

$(C)\ 2$

$(D)\ -2$

$(E)\ -6$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to \frac{\pi}{2}} \dfrac{\sin 6p}{p - \frac{\pi}{2}} \\ & = \dfrac{\sin \left( 6 \cdot \frac{\pi}{2} \right) }{\frac{\pi}{2} - \frac{\pi}{2}} \\ & = \dfrac{\sin 3\pi}{0} =\dfrac{0}{0} = \text{tak tentu} \end{align}$

Untuk menggunakaan teorema limit trigonometri $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$ ke soal di atas, kita membutuhkan kreativitas dalam mengolah bentuk trigonometri yang ada sampai kepada bentuk teorema limit.

$\sin \left( \pi -x \right)= \sin x$
$\sin \left( 2 \pi + \pi -x \right)= \sin x$
$\sin \left( 4 \pi + \pi -x \right)= \sin x$

$\begin{align}
& \lim\limits_{x \to \frac{\pi}{2}} \dfrac{\sin 6p}{p - \frac{\pi}{2}} \\ & = \lim\limits_{p \to \frac{\pi}{2}} \dfrac{\sin 6p}{p - \frac{\pi}{2}} \\ & = \lim\limits_{p \to \frac{\pi}{2}} \dfrac{\sin (\pi-6p)}{-\left(\frac{\pi}{2}-p \right)} \\ & = \lim\limits_{p \to \frac{\pi}{2}} \dfrac{\sin (3\pi-6p)}{-\left(\frac{\pi}{2}-p \right)} \\ & = \lim\limits_{p \to \frac{\pi}{2}} \dfrac{\sin 6(\frac{\pi}{2}- p)}{-\left(\frac{\pi}{2}-p \right)} \\ & = \lim\limits_{p \to \frac{\pi}{2}} \dfrac{\sin 6(\frac{\pi}{2}- p)}{-\left(\frac{\pi}{2}-p \right)} \\ & =-6 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(E)\ -6$

53. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{\theta \to \frac{\pi}{4}} \dfrac{1 - \tan \theta}{1 - \sqrt{2} \sin \theta} =\cdots$

$(A)\ 3$

$(B)\ 2$

$(C)\ 1$

$(D)\ 0$

$(E)\ -1$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{\theta \to \frac{\pi}{4}} \dfrac{1 - \tan \theta}{1 - \sqrt{2} \sin \theta} \\ & = \dfrac{1 - \tan \frac{\pi}{4}}{1 - \sqrt{2} \sin \frac{\pi}{4}} \\ & = \dfrac{1 - \tan 45^{\circ}}{1 - \sqrt{2} \sin 45^{\circ}} \\ & = \dfrac{1 - 1}{1 - 1} =\dfrac{0}{0} = \text{tak tentu} \end{align}$

Untuk menggunakaan teorema limit trigonometri $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$ ke soal di atas, kita membutuhkan kreativitas dalam mengolah bentuk trigonometri yang ada sampai kepada bentuk teorema limit.

$\sin^{2} A + \cos^{2} A= 1$
$1 + \cot^{2} A= \csc^{2} A$
$\tan^{2} A + 1= \sec^{2} A$
$\cos 2x=\cos^{2}x-\sin^{2}x$
$\cos 2x=1 -2 \sin^{2}x$

$\begin{align}
& \lim\limits_{\theta \to \frac{\pi}{4}} \dfrac{1 - \tan \theta}{1 - \sqrt{2} \sin \theta} \\ & = \lim\limits_{\theta \to \frac{\pi}{4}} \dfrac{1 - \tan \theta}{1 - \sqrt{2} \sin \theta} \cdot \dfrac{ 1 + \sqrt{2} \sin \theta }{1 + \sqrt{2} \sin \theta} \\ & = \lim\limits_{\theta \to \frac{\pi}{4}} \dfrac{ \left( 1 - \tan \theta \right) \left( 1 + \sqrt{2} \sin \theta \right) }{1 - 2 \sin^{2} \theta} \\ & = \lim\limits_{\theta \to \frac{\pi}{4}} \dfrac{ \left( 1 - \frac{\sin \theta}{\cos \theta} \right) \left( 1 + \sqrt{2} \sin \theta \right) }{\sin^{2} \theta + \cos^{2} \theta - 2 \sin^{2} \theta} \\ & = \lim\limits_{\theta \to \frac{\pi}{4}} \dfrac{ \left( \frac{\cos \theta}{\cos \theta}-\frac{\sin \theta}{\cos \theta} \right) \left( 1 + \sqrt{2} \sin \theta \right) }{ \cos^{2} \theta - \sin^{2} \theta} \\ & = \lim\limits_{\theta \to \frac{\pi}{4}} \dfrac{ \left( \frac{\cos \theta - \sin \theta}{\cos \theta} \right) \left( 1 + \sqrt{2} \sin \theta \right) }{ \cos^{2} \theta - \sin^{2} \theta} \\ & = \lim\limits_{\theta \to \frac{\pi}{4}} \dfrac{ \color{red}{ \left( \cos \theta - \sin \theta \right) } \left( 1 + \sqrt{2} \sin \theta \right) }{ \left( \cos \theta \right) \color{red}{ \left( \cos \theta - \sin \theta \right) } \left( \cos \theta + \sin \theta \right) } \\ & = \lim\limits_{\theta \to \frac{\pi}{4}} \dfrac{ \left( 1 + \sqrt{2} \sin \theta \right) }{ \left( \cos \theta \right) \left( \cos \theta + \sin \theta \right) } \\ & = \dfrac{ \left( 1 + \sqrt{2} \sin 45^{\circ} \right) }{ \left( \cos \sin 45^{\circ} \right) \left( \cos 45^{\circ} + \sin 45^{\circ} \right) } \\ & = \dfrac{ \left( 1 + \sqrt{2} \cdot \frac{1}{2} \sqrt{2} \right) }{ \left( \frac{1}{2} \sqrt{2} \right) \left( \frac{1}{2} \sqrt{2} + \frac{1}{2} \sqrt{2} \right)} \\ & = \dfrac{ 1+1 }{ \left( \frac{1}{2} \sqrt{2} \right) \left( \sqrt{2} \right)} \\ & = \dfrac{ 2 }{ 1} =2 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ 2$

54. Soal Latihan Limit Fungsi Trigonometri

Jika $f \left( x \right) = \lim\limits_{x \to 0} \left( \dfrac{1 - \cos x \sqrt{\cos 2x} }{x^{2}} \right)$ maka $f \left( x \right) =\cdots$

$(A)\ 1$

$(B)\ 0$

$(C)\ \frac{1}{2}$

$(D)\ \frac{3}{2}$

$(E)\ -1$

Alternatif Pembahasan:

$\begin{align} f \left( x \right) & = \lim\limits_{x \to 0} \left( \dfrac{1 - \cos x \sqrt{\cos 2x} }{x^{2}} \right) \\ f \left( x \right) & = \dfrac{1 - \cos 0 \sqrt{\cos 2(0)} }{(0)^{2}} \\ f \left( x \right) & = \dfrac{1 - 1 \sqrt{1} }{0} =\dfrac{0}{0} = \text{tak tentu} \end{align}$

Untuk menggunakaan teorema limit trigonometri $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$ ke soal di atas, kita membutuhkan kreativitas dalam mengolah bentuk trigonometri yang ada sampai kepada bentuk teorema limit.

$\left( a+b \right)\left( a - b \right)=a^{2}-b^{2} $
$\left( a+ \sqrt{b} \right)\left( a - \sqrt{b} \right)=a^{2}-b $
$\sin^{2} A + \cos^{2} A= 1$
$\cos 2x=\cos^{2}x-\sin^{2}x$
$\cos 2x=1 -2 \sin^{2}x$
$\cos 2x= 2 \cos^{2}x-1$

$\begin{align}
f \left( x \right) &= \lim\limits_{x \to 0} \dfrac{1 - \cos x \sqrt{\cos 2x} }{x^{2}} \\ &= \lim\limits_{x \to 0} \left( \dfrac{1 - \cos x \sqrt{\cos 2x} }{x^{2}} \right) \cdot \left( \dfrac{ 1 + \cos x \sqrt{\cos 2x} }{1 + \cos x \sqrt{\cos 2x}} \right) \\ &= \lim\limits_{x \to 0} \dfrac{1 - \cos^{2} x \cdot \cos 2x }{x^{2} \left( 1 + \cos x \sqrt{\cos 2x} \right)} \\ &= \lim\limits_{x \to 0} \dfrac{1 - \cos^{2} x \cdot \left( 1 - 2 \sin^{2} x \right) }{x^{2} \left( 1 + \cos x \sqrt{\cos 2x} \right)} \\ &= \lim\limits_{x \to 0} \dfrac{1 - \cos^{2} x - 2 \cos^{2} x \cdot \sin^{2} x }{x^{2} \left( 1 + \cos x \sqrt{\cos 2x} \right)} \\ &= \lim\limits_{x \to 0} \dfrac{ \sin^{2} x + 2 \sin^{2} x \cdot \cos^{2} x }{x^{2} \left( 1 + \cos x \sqrt{\cos 2x} \right)} \\ &= \lim\limits_{x \to 0} \dfrac{ \sin^{2} x \left( 1 + 2 \cos^{2} x \right) }{x^{2} \left( 1 + \cos x \sqrt{\cos 2x} \right)} \\ &= \lim\limits_{x \to 0} \dfrac{ \sin^{2} x }{x^{2}} \cdot \dfrac{1 + 2 \cos^{2} x}{1 + \cos x \sqrt{\cos 2x}} \\ &= \dfrac{ 1^{2} }{1^{2}} \cdot \dfrac{1 + 2 \cos^{2} 0}{1 + \cos 0 \sqrt{\cos 2(0)}} \\ &= 1 \cdot \dfrac{1 + 2 (1)}{1 + 1 \cdot (1) } = \dfrac{3}{1 + 1 \cdot (1) } =\dfrac{3}{2} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(D)\ \dfrac{3}{2}$

55. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to \frac{\pi}{2}} \dfrac{\cot x - \cos x}{\left( \pi - 2x \right)^{3}}=\cdots$

$(A)\ \frac{1}{16}$

$(B)\ \frac{1}{8}$

$(C)\ \frac{1}{4}$

$(D)\ \frac{\pi}{2}$

$(E)\ \frac{\pi}{4}$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to \frac{\pi}{2}} \dfrac{\cot x - \cos x}{\left( \pi - 2x \right)^{3}} \\ & = \dfrac{\cot \frac{\pi}{2} - \cos \frac{\pi}{2}}{\left( \pi - 2 \cdot \frac{\pi}{2} \right)^{3}} \\ & = \dfrac{\cot 90^{\circ} - \cos 90^{\circ}}{\left( \pi - \pi \right)^{3}} \\ & = \dfrac{0}{0}\ = \text{tak tentu} \end{align}$

Untuk menggunakaan teorema limit trigonometri $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$ ke soal di atas, kita membutuhkan kreativitas dalam mengolah bentuk trigonometri yang ada sampai kepada bentuk teorema limit.

$\sin \left( \frac{\pi}{2} -x \right)= \cos x$
$\cos \left( \frac{\pi}{2} -x \right)= \sin x$
$\sin \left( \pi -x \right)= \sin x$
$\sin \left( 2 \pi + \pi -x \right)= \sin x$
$\sin \left( 4 \pi + \pi -x \right)= \sin x$

$\begin{align}
& \lim\limits_{x \to \frac{\pi}{2}} \dfrac{\cot x - \cos x}{\left( \pi - 2x \right)^{3}} \\ & = \lim\limits_{x \to \frac{\pi}{2}} \dfrac{\frac{\cos x}{\sin x} - \cos x}{\left( \pi - 2x \right)^{3}} \\ & = \lim\limits_{x \to \frac{\pi}{2}} \dfrac{\frac{\cos x-\sin x \cos x}{\sin x} }{\left( \pi - 2x \right)^{3}} \\ & = \lim\limits_{x \to \frac{\pi}{2}} \dfrac{ \cos x-\sin x \cos x }{\left( \pi - 2x \right)^{3} \cdot \sin x} \\ & = \lim\limits_{x \to \frac{\pi}{2}} \dfrac{ \cos x \left( 1-\sin x \right) }{\left( \pi - 2x \right)^{3} \cdot \sin x} \cdot \dfrac{ 1+\sin x }{1+\sin x} \\ & = \lim\limits_{x \to \frac{\pi}{2}} \dfrac{ \cos x \left( 1-\sin^{2} x \right) }{\left( \pi - 2x \right)^{3} \cdot \sin x \left( 1+\sin x \right)} \\ & = \lim\limits_{x \to \frac{\pi}{2}} \dfrac{ \cos x \left( \cos^{2} x \right) }{\left( \pi - 2x \right)^{3} \cdot \sin x \left( 1+\sin x \right)} \\ & = \lim\limits_{x \to \frac{\pi}{2}} \dfrac{ \cos^{3} x }{\left( \pi - x \right)^{3} \cdot \sin x \left( 1+\sin x \right)} \\ & = \lim\limits_{x \to \frac{\pi}{2}} \dfrac{ \sin^{3} \left( \frac{\pi}{2} - x \right) }{2^{3}\left( \frac{\pi}{2} - x \right)^{3} \cdot \sin x \left( 1+\sin x \right)} \\ & = \lim\limits_{x \to \frac{\pi}{2}} \dfrac{ \sin^{3} \left( \frac{\pi}{2} - x \right) }{2^{3} \left( \frac{\pi}{2} - x \right)^{3}} \cdot \dfrac{1}{\sin x \left( 1+\sin x \right)} \\ & = \dfrac{ 1^{3} }{2^{3} \cdot 1^{3}} \cdot \dfrac{1}{\sin \frac{\pi}{2} \left( 1+\sin \frac{\pi}{2} \right)} \\ & = \dfrac{ 1 }{8 \cdot 1} \cdot \dfrac{1}{1 \left( 1+1 \right)} \\ & = \dfrac{ 1 }{8} \cdot \dfrac{1}{2} = \dfrac{1}{16} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(A)\ \dfrac{1}{16}$

56. Soal Latihan Limit Fungsi Trigonometri

Nilai dari $\lim\limits_{x \to \frac{\pi}{4}} \dfrac{1 - \cot^{3} x}{2-\cot x - \cot^{3} x}=\cdots$

$(A)\ \frac{11}{4}$

$(B)\ \frac{3}{4}$

$(C)\ \frac{1}{2}$

$(D)\ \frac{5}{4}$

$(E)\ \frac{1}{8}$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to \frac{\pi}{4}} \dfrac{1 - \cot^{3} x}{2-\cot x - \cot^{3} x} \\ & = \dfrac{1 - \cot^{3} \frac{\pi}{4}}{2-\cot \frac{\pi}{4} - \cot^{3} \frac{\pi}{4}} \\ & = \dfrac{1 - \cot^{3} 45^{\circ}}{2-\cot 45^{\circ} - \cot^{3} 45^{\circ}} \\ & = \dfrac{1 - 1}{2-1 - 1} = \dfrac{0}{0} = \text{tak tentu} \end{align}$

Untuk menggunakaan teorema limit trigonometri $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$ ke soal di atas, kita membutuhkan kreativitas dalam mengolah bentuk trigonometri yang ada sampai kepada bentuk teorema limit.

$a^{2}-b^{2}= \left( a-b \right) \left( a+b \right) $
$a^{3}-b^{3}= \left( a-b \right) \left( a^{2}+ab+b^{2} \right) $

$\begin{align}
& \lim\limits_{x \to \frac{\pi}{4}} \dfrac{1 - \cot^{3} x}{2-\cot x - \cot^{3} x} \cdot \dfrac{-1}{-1} \\ & = \lim\limits_{x \to \frac{\pi}{4}} \dfrac{\cot^{3} x-1}{ \cot^{3} x- \cot x - 2} \\ & = \lim\limits_{x \to \frac{\pi}{4}} \dfrac{\left( \cot x-1 \right) \left( \cot^{2} x + \cot x+1 \right)}{ \left( \cot x-1 \right) \left( \cot^{2} x + \cot x+2 \right)} \\ & = \lim\limits_{x \to \frac{\pi}{4}} \dfrac{ \cot^{2} x + \cot x+1 }{ \cot^{2} x + \cot x+2 } \\ & = \dfrac{ \cot^{2} 45^{\circ} + \cot 45^{\circ}+1 }{ \cot^{2} 45^{\circ} + \cot 45^{\circ}+2 } \\ & = \dfrac{ 1^{2} + 1+1 }{ 1^{2} + 1+2 } \\ & = \dfrac{ 3 }{4} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ \frac{3}{4}$

57. Soal Latihan Limit Fungsi Trigonometri

Nilai dari Nilai dari $\lim\limits_{x \to 0} \dfrac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\tan x} =\cdots$

$(A)\ 1$

$(B)\ 0$

$(C)\ -1$

$(D)\ -2$

$(E)\ -3$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \dfrac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\tan x} \\ & = \dfrac{\sqrt{1+\sin 0}-\sqrt{1-\sin 0}}{\tan 0} \\ & = \dfrac{\sqrt{1+ 0}-\sqrt{1+ 0}}{0} \\ & = \dfrac{1 - 1}{0} = \dfrac{0}{0} = \text{tak tentu} \end{align}$

$a^{2}-b^{2}= \left( a-b \right) \left( a+b \right) $
$a^{3}-b^{3}= \left( a-b \right) \left( a^{2}+ab+b^{2} \right) $

$\begin{align}
& \lim\limits_{x \to 0} \dfrac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\tan x} \\ &= \lim\limits_{x \to 0} \left( \dfrac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\tan x} \cdot \dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}} \right) \\ &= \lim\limits_{x \to 0} \left( \dfrac{ 1+\sin x- 1+\sin x }{\tan x} \cdot \dfrac{1}{\sqrt{1+\sin x}+\sqrt{1-\sin x}} \right) \\ &= \lim\limits_{x \to 0} \left( \dfrac{ 2 \sin x }{\tan x} \cdot \dfrac{1}{\sqrt{1+\sin x}+\sqrt{1-\sin x}} \right) \\ & = 2 \cdot \dfrac{1}{\sqrt{1+\sin 0}+\sqrt{1-\sin 0}} \\ &= 2 \cdot \dfrac{1}{\sqrt{1+ 0}+\sqrt{1- 0}} \\ &= \dfrac{2}{2}=1 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(A)\ 1$

58. Soal Latihan Limit Fungsi Trigonometri

Nilai dari Nilai dari $\lim\limits_{x \to 0} \dfrac{x \tan 2x - 2x \tan x}{\left( 1-\cos 2x \right)^{2}} =\cdots$

$(A)\ 2$

$(B)\ \frac{1}{2}$

$(C)\ 0$

$(D)\ -\frac{1}{2}$

$(E)\ -2$

Alternatif Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \dfrac{x \tan 2x - 2x \tan x}{\left( 1-\cos 2x \right)^{2}} \\ & = \dfrac{0 \cdot \tan 2(0) - 2(0) \cdot \tan 0}{\left( 1-\cos 2(0) \right)^{2}} \\ & = \dfrac{0 - 0}{\left( 1- 1 \right)^{2}} \\ & = \dfrac{0}{0} = \text{tak tentu} \end{align}$

$\left( a+b \right)\left( a - b \right)=a^{2}-b^{2} $
$\left( a+ \sqrt{b} \right)\left( a - \sqrt{b} \right)=a^{2}-b $
$\sin^{2} A + \cos^{2} A= 1$
$\cos 2x=\cos^{2}x-\sin^{2}x$
$\cos 2x=1 -2 \sin^{2}x$
$\cos 2x= 2 \cos^{2}x-1$

$\begin{align}
& \lim\limits_{x \to 0} \dfrac{x \tan 2x - 2x \tan x}{\left( 1-\cos 2x \right)^{2}} \\ &=\lim\limits_{x \to 0} \dfrac{x \cdot \frac{2 \tan x}{1-tan^{2}x} - 2x \tan x}{\left( 1- 1 +2 \sin^{2}x \right)^{2}} \\ &=\lim\limits_{x \to 0} \dfrac{ \frac{2x \tan x}{1-tan^{2}x} - \frac{2x \tan x-2x \tan^{3} x}{1-\tan^{2}x}}{\left( 2 \sin^{2}x \right)^{2}} \\ &=\lim\limits_{x \to 0} \dfrac{ \frac{2x \tan x - 2x \tan x+2x \tan^{3} x}{1-\tan^{2}x}}{\ 4 \sin^{4}x } \\ &=\lim\limits_{x \to 0} \dfrac{ \frac{2x \tan^{3} x}{1-tan^{2}x}}{4 \sin^{4}x } \\ &=\lim\limits_{x \to 0} \dfrac{ 2x \tan^{3} x}{\left( 1-\tan^{2}x \right)\left( 4\sin^{4}x \right) } \\ &=\lim\limits_{x \to 0} \dfrac{1}{\left( 1-\tan^{2}x \right)} \cdot \dfrac{ 2x \tan^{3} x}{\left( 4\sin^{4}x \right) } \\ &= \dfrac{1}{\left( 1-\tan^{2}0 \right)} \cdot \dfrac{ 2}{4} \\ &= \dfrac{1}{\left( 1- 0 \right)} \cdot \dfrac{ 2}{4} = \dfrac{1}{2} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ \frac{1}{2}$

Catatan Cara Membuktikan Teorema Limit Fungsi Trigonometri dan Pembahasan 50+ Soal Latihan di atas sifatnya "dokumen hidup" yang senantiasa diperbaiki atau diperbaharui sesuai dengan dinamika kebutuhan dan perubahan zaman. Catatan tambahan dari Anda untuk admin diharapkan dapat meningkatkan kualitas catatan ini 🙏 CMIIW.

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Cara Membuktikan Teorema Limit Fungsi Trigonometri dan Pembahasan 50+ Soal Latihan

Definisi Limit Fungsi

Teorema Limit Fungsi Trigonometri

Cara Pembuktian Teorema Limit Fungsi Trigonmeteri

SOAL LATIHAN dan PEMBAHASAN LIMIT FUNGSI TRIGONOMETRI

1. Soal Latihan Limit Fungsi Trigonometri

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