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70+ Soal dan Pembahasan Matematika Dasar SMA Limit Fungsi Trigonometri

Matematika Dasar Limit Fungsi Trigonometri (*Soal Dari Berbagai Sumber)Calon Guru belajar Matematika dasar SMA lewat Soal dan Pembahasan Matematika Dasar Limit Fungsi Trigonometri. Catatan limit fungsi kita bagi dalam tiga catatan yaitu matematika dasar limit fungsi aljabar, matematika dasar limit fungsi trigonometri dan matematika dasar limit fungsi takhingga.

Penerapan Limit Fungsi Trigonometri dalam kehidupan sehari-hari mungkin tidak terlihat langsung, limit fungsi ini merupakan pengembangan dari Limit Fungsi Aljabar yang merupakan dasar dalam matematika bagaimana kita bisa belajar Limit Fungsi Tak hingga, Diferensial Fungsi (Turunan) dan sampai kepada Integral Fungsi.

Bagaimana menggunakan aturan-aturan pada limit fungsi trigonometri untuk menyelesaikan soal-soal yang berkembang bukan sesuatu yang sulit. Jika kita mengikuti step by step penjabaran pada pembahasan soal dibawah ini, maka limit fungsi trigonometri sedikit demi sedikit akan semakin kita pahami.

Limit fungsi ini termasuk materi yang sangat penting 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 $70,5\ kg$. Hasil $70,5\ kg$ ini sebenarnya belum hasil pengukuran yang paling tepat tetapi sudah bisa mewakili hasil pengukuran, karena berat badan kita adalah mendekati $70,5\ kg$. Kata "mendekati" adalah salah satu kata kunci dalam belajar limit fungsi.

Beberapa contoh soal Limit Fungsi Aljabar untuk kita diskusikan, yang kita sadur dari soal-soal SBMPTN (Seleksi Bersama Masuk Perguruan Tinggi Negeri), soal SMMPTN (Seleksi Mandiri Masuk Perguruan Tinggi Negeri), soal UN (Ujian Nasional), Soal simulasi yang dilaksanakan oleh bimbingan belajar atau soal ujian yang dilaksanakan di sekolah.

Sedikit informasi tambahan yang mungkin tidak terlalu penting, kemarin siswa baru selesai penilaian harian tentang limit dan ada beberapa siswa yang mendapat nilai sempurna, sehingga sebagai kenang-kenangan hasil pekerjaan siswa dengan hasil sempurna kita photo dan ditampilkan sebagai photo dari artikel ini.

Berdasarkan definisi limit fungsi, Jika nilai Limit Kiri = Limit Kanan=L secara simbol dituliskan $\lim\limits_{x \to a^{+}}f(x)=\lim\limits_{x \to a^{-}}f(x)=L$ maka nilai $\lim\limits_{x \to a}f(x)=L$.

Teorema dasar pada limit fungsi trigonometri


  • $\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 $\lim\limits_{x \to 0} \dfrac{\sin ax }{\tan bx} = \dfrac{a}{b}$
Limit fungsi trigonometri ini umumnya tingkat kesulitan bukan pada limit fungsi trigonometrinya tetapi lebih banyak kesulitan tentang trigonometri terkhusus Identitas Trigonometri Dasar.

Teorema Limit Fungsi Yang Sangat Penting Dalam menyelesaikan Masalah Limit Fungsi


Andaikan $n$ bilangan bulat positif, $k$ konstanta, dan $f$ dan $f$ dan $g$ adalah fungsi yang mempunyai limit di $c$. Maka berlaku:
  • $\lim\limits_{x \to c} k=k$
  • $\lim\limits_{x \to c} c=c$
  • $\lim\limits_{x \to c} kf(x)=k \cdot \lim\limits_{x \to c} f(x)$
  • $\lim\limits_{x \to c} \left( f(x)+g(x) \right) = \lim\limits_{x \to c} f(x)+\lim\limits_{x \to c} g(x)$
  • $\lim\limits_{x \to c} \left( f(x)-g(x) \right) = \lim\limits_{x \to c} f(x)-\lim\limits_{x \to c} g(x)$
  • $\lim\limits_{x \to c} \left( f(x) \cdot g(x) \right) = \lim\limits_{x \to c} f(x) \cdot \lim\limits_{x \to c} g(x)$
  • $\lim\limits_{x \to c} \left( \dfrac{f(x)}{g(x)} \right) = \dfrac{\lim\limits_{x \to c} f(x)}{\lim\limits_{x \to c} g(x)}$ dimana $\lim\limits_{x \to c} g(x) \neq 0$
  • $\lim\limits_{x \to c} \left( f(x) \right)^{n} = \left( \lim\limits_{x \to c} f(x) \right)^{n}$
  • $\lim\limits_{x \to c} \sqrt[n]{f(x)} = \sqrt[n]{ \lim\limits_{x \to c} f(x)}$ dimana $\lim\limits_{x \to c} f(x) \gt 0$ bilamana $n$ genap

Dalam menyelesaikan limit fungsi baik itu limit fungsi aljabar, trigonometri atau limit menuju tak hingga, langkah awalnya adalah substitusi langsung. Setelah dilakukan substitusi diperoleh hasilnya bentuk tak tentu seperti $\dfrac{0}{0}$, $\dfrac{\infty}{\infty}$, $0 \times \infty$, $\infty - \infty$, $0^{0}$, atau $\infty^{\infty}$ maka dilakukan manipulasi aljabar dengan cara memfaktorkan, mengalikan dengan akar sekawan, atau dengan manipulasi aljabar lainnya selama tidak menyalahi aturan-aturan dalam bermatematik.

Menyelesaikan Limit Fungsi dengan Aturan L'Hospital atau Menggunakan Turunan


Cara alternatif menyelesaikan limit fungsi adalah dengan Aturan L'Hospital atau pakai turunan fungsi. Cara ini dapat kita gunakan jika kita sudah mengenal atau belajar Turunan Fungsi, apabila belum mengenal atau belajar Fungsi Turunan maka menggunakan cara ini tidak dianjurkan.

Jika nilai $\lim\limits_{x \to a} \dfrac{f(x)}{g(x)}=\dfrac{0}{0}$ dan $\lim\limits_{x \to a} \dfrac{f^{\prime}(x)}{g^{\prime}(x)}$ ada,
maka $ \lim\limits_{x \to a} \dfrac{f(x)}{g(x)} = \lim\limits_{x \to a} \dfrac{f^{\prime} (x)}{g^{\prime} (x)}= \dfrac{f^{\prime} (a)}{g^{\prime} (a)}$

Mari kita simak beberapa Soal Limit Fungsi Trigonometri yang sudah pernah di ujikan pada Ujian Sekolah, Ujian Nasional atau Seleksi Masuk Perguruan Tinggi Negeri yang dilaksanakan secara nasional atau mandiri.

1. Soal EBATANAS Matematika SMA IPA 1998 |*Soal Lengkap

Nilai $\lim\limits_{x \to 5 } \dfrac{\left( 4x-10 \right) \sin \left( x-5 \right)}{x^{2}-25} =\cdots$
$\begin{align}
(A)\ & -3 \\
(B)\ & -1 \\
(C)\ & 1 \\
(D)\ & 2 \\
(E)\ & 4
\end{align}$
Alternatif Pembahasan:
show

Dengan menggunakan identitas trigonometri $\sin 2a = 2\ \sin a\ \cos a$ dan teorema limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$, kita coba selesaikan soal di atas seperti penjabaran berikut ini:
$\begin{align}
& \lim\limits_{x \to 5 } \dfrac{\left( 4x-10 \right) \sin \left( x-5 \right)}{x^{2}-25} \\
& = \lim\limits_{x \to 5 } \dfrac{\left( 4x-10 \right) \sin \left( x-5 \right)}{\left( x-5 \right)\left( x+5 \right)} \\
& = \dfrac{\left( 4(5)-10 \right) \left( 1 \right)}{\left( 1 \right)\left( 5+5 \right)} \\
& = \dfrac{10}{10} = 1
\end{align}$

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

2. Soal EBATANAS Matematika SMA IPA 1996 |*Soal Lengkap

Nilai $\lim\limits_{x \to 0 } \dfrac{\sin 4x + \sin 2x}{3x\ \cos x} =\cdots$
$\begin{align}
(A)\ & \dfrac{1}{4} \\
(B)\ & \dfrac{1}{2} \\
(C)\ & 1 \\
(D)\ & \dfrac{3}{2} \\
(E)\ & 2
\end{align}$
Alternatif Pembahasan:
show

Dengan menggunakan identitas trigonometri $\sin 2a = 2\ \sin a\ \cos a$ dan teorema limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$, kita coba selesaikan soal di atas seperti penjabaran berikut ini:
$\begin{align}
& \lim\limits_{x \to 0 } \dfrac{\sin 4x + \sin 2x}{3x\ \cos x} \\
& = \lim\limits_{x \to 0 } \dfrac{2\ \sin 3x\ \cos x}{3x\ \cos x} \\
& = \lim\limits_{x \to 0 } \dfrac{2\ \sin 3x }{3x } \\
& = \dfrac{2 \cdot 3 }{3 }= 2
\end{align}$

$\therefore$ Pilihan yang sesuai $(E)\ 2$

3. Soal EBATANAS Matematika SMA IPA 2001 |*Soal Lengkap

$\lim\limits_{x \to 0 } \dfrac{2x}{2\ \sin x+\sin 2x} =\cdots$
$\begin{align}
(A)\ & -\dfrac{1}{2} \\
(B)\ & -\dfrac{1}{4} \\
(C)\ & \dfrac{1}{4} \\
(D)\ & \dfrac{1}{2} \\
(E)\ & 1
\end{align}$
Alternatif Pembahasan:
show

Dengan menggunakan identitas trigonometri $\sin 2a = 2\ \sin a\ \cos a$ dan teorema limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$, kita coba selesaikan soal di atas seperti penjabaran berikut ini:
$\begin{align}
& \lim\limits_{x \to 0 } \dfrac{2x}{2\ \sin x+\sin 2x} \\
& = \lim\limits_{x \to 0 } \dfrac{2x}{2\ \sin x+2\ \sin x\ \cos x} \\
& = \lim\limits_{x \to 0 } \dfrac{2x}{2\ \sin x \left(1 + \cos x \right)} \\
& = \dfrac{2 }{2 \left(1 + \cos 0 \right)} \\
& = \dfrac{2 }{2 \left(1 + 1 \right)}=\dfrac{1 }{2}
\end{align}$

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

4. Soal EBATANAS Matematika SMA IPA 2000 |*Soal Lengkap

$\lim\limits_{x \to 0 } \dfrac{\sin 2x}{3-\sqrt{2x+9}} =\cdots$
$\begin{align}
(A)\ & 3 \\
(B)\ & 1 \\
(C)\ & 0 \\
(D)\ & -3 \\
(E)\ & -6
\end{align}$
Alternatif Pembahasan:
show

Dengan menggunakan identitas trigonometri $\sin 2a = 2\ \sin a\ \cos a$ dan teorema limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$, kita coba selesaikan soal di atas seperti penjabaran berikut ini:
$\begin{align}
& \lim\limits_{x \to 0 } \dfrac{\sin 2x}{3-\sqrt{2x+9}} \\
& = \lim\limits_{x \to 0 } \dfrac{\sin 2x}{3-\sqrt{2x+9}} \cdots \dfrac{3+\sqrt{2x+9}}{3+\sqrt{2x+9}} \\
& = \lim\limits_{x \to 0 } \dfrac{ \left( \sin 2x \right)\left( 3+\sqrt{2x+9} \right)}{9-(2x+9)} \\
& = \lim\limits_{x \to 0 } \dfrac{ \left( \sin 2x \right)\left( 3+\sqrt{2x+9} \right)}{-2x} \\
& = \dfrac{ \left( 2 \right)\left( 3+\sqrt{2(0)+9} \right)}{-2} \\
& = \dfrac{12}{-2}=-6
\end{align}$

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

5. Soal UN Matematika SMA IPA 2004 |*Soal Lengkap

$\lim\limits_{x \to -2} \dfrac{\left( x+6 \right)\ \sin \left( x+2 \right)}{x^{2}-3x-10} =\cdots$
$\begin{align}
(A)\ & -\dfrac{4}{3} \\
(B)\ & -\dfrac{4}{7} \\
(C)\ & -\dfrac{2}{5} \\
(D)\ & 0 \\
(E)\ & 1
\end{align}$
Alternatif Pembahasan:
show

Dengan menggunakan teorema limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$, kita coba selesaikan soal di atas seperti penjabaran berikut ini:
$\begin{align}
& \lim\limits_{x \to -2} \dfrac{\left( x+6 \right)\ \sin \left( x+2 \right)}{x^{2}-3x-10} \\
& = \lim\limits_{x \to -2} \dfrac{\left( x+6 \right)\ \sin \left( x+2 \right)}{\left( x+2 \right)\left( x-5 \right)} \\
& = \lim\limits_{x \to -2} \dfrac{\left( x+6 \right)\ \sin \left( x+2 \right)}{\left( x+2 \right)\left( x-5 \right)} \\
& = \dfrac{\left( -2+6 \right) (1) }{(1)\left( -2-5 \right)} \\
& = \dfrac{4}{-7}
\end{align}$

$\therefore$ Pilihan yang sesuai $(B)\ -\dfrac{4}{7}$

6. Soal UN Matematika SMA IPA 2003 |*Soal Lengkap

$\lim\limits_{x \to \frac{\pi}{4}} \dfrac{\cos 2x}{\cos x - \sin x} =\cdots$
$\begin{align}
(A)\ & -\sqrt{2} \\
(B)\ & -\dfrac{1}{2}\sqrt{2} \\
(C)\ & \dfrac{1}{2}\sqrt{2} \\
(D)\ & \sqrt{2} \\
(E)\ & 2\sqrt{2}
\end{align}$
Alternatif Pembahasan:
show

Dengan menggunakan identitas trigonometri $\cos 2a = cos^{2}\ a- sin^{2}\ a$ dan teorema limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$, kita coba selesaikan soal di atas seperti penjabaran berikut ini:
$\begin{align}
& \lim\limits_{x \to \frac{\pi}{4}} \dfrac{\cos 2x}{\cos x - \sin x} \\
& = \lim\limits_{x \to \frac{\pi}{4}} \dfrac{cos^{2}\ x - sin^{2}\ x}{\cos x - \sin x} \\
& = \lim\limits_{x \to \frac{\pi}{4}} \dfrac{\left(\cos x + \sin x \right)\left(\cos x - \sin x \right)}{\cos x - \sin x} \\
& = \lim\limits_{x \to \frac{\pi}{4}} \dfrac{\left(\cos x + \sin x \right)}{1} \\
& = \cos \frac{\pi}{4} + \sin \frac{\pi}{4} \\
& = \dfrac{1}{2}\sqrt{2} + \dfrac{1}{2}\sqrt{2} =\sqrt{2}
\end{align}$

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

7. Soal UN Matematika SMA IPA 2002 |*Soal Lengkap

$\lim\limits_{x \to \infty } \sin \dfrac{1}{x} =\cdots$
$\begin{align}
(A)\ & \infty \\
(B)\ & 0 \\
(C)\ & 1 \\
(D)\ & 2 \\
(E)\ & 3
\end{align}$
Alternatif Pembahasan:
show

$\begin{align}
\lim\limits_{x \to \infty } \sin \dfrac{1}{x} & = \sin \dfrac{1}{\infty} \\
& = \sin 0 = 0 \\
\end{align}$

$\therefore$ Pilihan yang sesuai $(B)\ 0$

8. Soal UN Matematika SMA IPA 2007 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{2x\ \sin 3x}{1-\cos 6x} =\cdots$
$\begin{align}
(A)\ & -1 \\
(B)\ & -\dfrac{1}{3} \\
(C)\ & 0 \\
(D)\ & \dfrac{1}{3} \\
(E)\ & 1
\end{align}$
Alternatif Pembahasan:
show

Dengan menggunakan identitas trigonometri $sin^{2}\ a+cos^{2}\ a=1$ dan $\cos 2a = cos^{2}\ a-sin^{2}\ a$, kita coba selesaikan soal di atas seperti penjabaran berikut ini:
$\begin{align}
& \lim\limits_{x \to 0} \dfrac{2x\ \sin 3x}{1-\cos 6x} \\
& = \lim\limits_{x \to 0} \dfrac{2x\ \sin 3x}{1-\cos 6x} \\
& = \lim\limits_{x \to 0} \dfrac{2x\ \sin 3x}{1-\cos 2(3x)} \\
& = \lim\limits_{x \to 0} \dfrac{2x\ \sin 3x}{1-\left(cos^{2}\ (3x)-sin^{2}\ (3x) \right)} \\
& = \lim\limits_{x \to 0} \dfrac{2x\ \sin 3x}{1-\left(1-sin^{2}\ (3x)-sin^{2}\ (3x) \right)} \\
& = \lim\limits_{x \to 0} \dfrac{2x\ \sin 3x}{2\ sin^{2}\ (3x) } \\
& = \lim\limits_{x \to 0} \dfrac{2x\ \sin 3x}{2\ \sin (3x) \cdot \sin (3x) } \\
& = \dfrac{2 \cdot 3}{2 \cdot 3 \cdot 3 } = \dfrac{1}{3}
\end{align}$

$\therefore$ Pilihan yang sesuai $(D)\ \dfrac{1}{3}$

9. Soal UN Matematika SMA IPA 2005 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{\tan 2x\ \cos 8x-\tan 2x}{16x^{3}} =\cdots$
$\begin{align}
(A)\ & -4 \\
(B)\ & -6 \\
(C)\ & -8 \\
(D)\ & -16 \\
(E)\ & -32
\end{align}$
Alternatif Pembahasan:
show

Dengan menggunakan identitas trigonometri $\cos 2a = 1- 2sin^{2}\ a$ dan teorema limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$, kita coba selesaikan soal di atas seperti penjabaran berikut ini:
$\begin{align}
& \lim\limits_{x \to 0} \dfrac{\tan 2x\ \cos 8x-\tan 2x}{16x^{3}} \\
& = \lim\limits_{x \to 0} \dfrac{\tan 2x \left( \cos 8x- 1 \right)}{16x^{3}} \\
& = \lim\limits_{x \to 0} \dfrac{\tan 2x \left( \cos 2(4x)- 1 \right)}{16x^{3}} \\
& = \lim\limits_{x \to 0} \dfrac{\tan 2x \left(cos^{2}(4x)-sin^{2}\ (4x)- 1 \right)}{16x^{3}} \\
& = \lim\limits_{x \to 0} \dfrac{\tan 2x \left( -2\ sin^{2}\ (4x) \right)}{16x^{3}} \\
& = \dfrac{2 \cdot (-2) \cdot 4 \cdot 4 }{16} = -4
\end{align}$

$\therefore$ Pilihan yang sesuai $(A)\ -4$


10. Soal UN Matematika SMA IPA 2016 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{\cos 4x-1}{1-\cos 2x} =\cdots$
$\begin{align}
(A)\ & -4 \\
(B)\ & -2 \\
(C)\ & -\dfrac{1}{2} \\
(D)\ & -\dfrac{1}{4} \\
(E)\ & 0
\end{align}$
Alternatif Pembahasan:
show

Dengan menggunakan identitas trigonometri $sin^{2}\ a+cos^{2}\ a=1$ dan $\cos 2a = cos^{2}\ a-sin^{2}\ a$, kita coba selesaikan soal di atas seperti penjabaran berikut ini:
$\begin{align}
& \lim\limits_{x \to 0} \dfrac{\cos 4x-1}{1-\cos 2x} \\
& = \lim\limits_{x \to 0} \dfrac{\cos 2(2x)-1}{1-\cos 2x} \\
& = \lim\limits_{x \to 0} \dfrac{cos^{2}\ (2x)-sin^{2}\ (2x)-1}{1-\cos 2x} \\
& = \lim\limits_{x \to 0} \dfrac{cos^{2}\ (2x)-\left( 1-cos^{2}\ (2x) \right)-1}{1-\cos 2x} \\
& = \lim\limits_{x \to 0} \dfrac{cos^{2}\ (2x)-1+cos^{2}\ (2x)-1}{1-\cos 2x} \\
& = \lim\limits_{x \to 0} \dfrac{2cos^{2}\ (2x)-2}{1-\cos 2x} \\
& = \lim\limits_{x \to 0} \dfrac{2 \left( cos^{2}\ (2x)-1 \right)}{1-\cos 2x} \\
& = \lim\limits_{x \to 0} \dfrac{2 \left( \cos (2x)-1 \right)\left( \cos (2x)+1 \right)}{-\left( \cos (2x)-1 \right)} \\
& = \lim\limits_{x \to 0} \dfrac{2 \left( \cos (2x)+1 \right)}{-1} \\
& = \dfrac{2 \left( \cos 0+1 \right)}{-1} = \dfrac{2 \left( 1+1 \right)}{-1}=-4 \\
\end{align}$

$\therefore$ Pilihan yang sesuai $(A)\ -4$


11. Soal UN Matematika SMA IPA 2015 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{x\ \tan 3x}{1-cos^{2}2x} =\cdots$
$\begin{align}
(A)\ & 0 \\
(B)\ & \dfrac{1}{4} \\
(C)\ & \dfrac{2}{4} \\
(D)\ & \dfrac{3}{4} \\
(E)\ & 1
\end{align}$
Alternatif Pembahasan:
show

Dengan menggunakan identitas trigonometri $sin^{2}\ a+cos^{2}\ a=1$ dan teorema limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$, kita coba selesaikan soal di atas seperti penjabaran berikut ini:
$\begin{align}
& \lim\limits_{x \to 0} \dfrac{x\ \tan 3x}{1-cos^{2}2x} \\
& = \lim\limits_{x \to 0} \dfrac{x\ \tan 3x}{sin^{2}2x} \\
& = \lim\limits_{x \to 0} \dfrac{x\ \tan 3x}{\sin 2x \cdot \sin 2x} \\
& =\lim\limits_{x \to 0} \dfrac{x}{\sin 2x } \cdot \lim\limits_{x \to 0} \dfrac{\tan 3x}{\sin 2x}\\
& = \dfrac{1}{2} \cdot \dfrac{3}{2} \\
& = \dfrac{3}{4}
\end{align}$

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

12. Soal UN Matematika SMA IPA 2014 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{1-\cos 8x}{\sin 2x\ \tan 2x} =\cdots$
$\begin{align}
(A)\ & 16 \\
(B)\ & 12 \\
(C)\ & 8 \\
(D)\ & 4 \\
(E)\ & 2
\end{align}$
Alternatif Pembahasan:
show

Dengan menggunakan identitas trigonometri $sin^{2}\ a+cos^{2}\ a=1$ dan teorema limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$, kita coba selesaikan soal di atas seperti penjabaran berikut ini:
$\begin{align}
& \lim\limits_{x \to 0} \dfrac{1-\cos 8x}{\sin 2x\ \tan 2x} \\
& = \lim\limits_{x \to 0} \dfrac{1-\cos 2(4x)}{\sin 2x\ \tan 2x} \\
& = \lim\limits_{x \to 0} \dfrac{1-\left( cos^{2}(4x)-sin^{2}(4x) \right)}{\sin 2x\ \tan 2x} \\
& = \lim\limits_{x \to 0} \dfrac{1-cos^{2}(4x)+sin^{2}(4x)}{\sin 2x\ \tan 2x} \\
& = \lim\limits_{x \to 0} \dfrac{sin^{2}(4x)+sin^{2}(4x)}{\sin 2x\ \tan 2x} \\
& = \lim\limits_{x \to 0} \dfrac{2\ sin^{2}(4x)}{\sin 2x\ \tan 2x} \\
& = \lim\limits_{x \to 0} \dfrac{2\ \sin (4x) \cdot \sin (4x)}{\sin 2x\ \tan 2x} \\
& = \dfrac{2 \cdot 4 \cdot 4}{2 \cdot 2}=8
\end{align}$

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

13. Soal UN Matematika SMA IPA 2013 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{4\ sin^{2}\ 2x}{x\ \tan 2x}=\cdots$
$\begin{align}
(A)\ & -8 \\
(B)\ & -4 \\
(C)\ & 0 \\
(D)\ & 4 \\
(E)\ & 8
\end{align}$
Alternatif Pembahasan:
show

Dengan menggunakan identitas trigonometri $sin^{2}\ a+cos^{2}\ a=1$ dan teorema limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$, kita coba selesaikan soal di atas seperti penjabaran berikut ini:
$\begin{align}
& \lim\limits_{x \to 0} \dfrac{4\ sin^{2}\ 2x}{x\ \tan 2x} \\
& = \lim\limits_{x \to 0} \dfrac{4\ \sin 2x\ \cdot \sin 2x}{x\ \tan 2x} \\
& = \dfrac{4 \cdot 2 \cdot 2}{1 \cdot 2}=8
\end{align}$

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

14. Soal UN Matematika SMA IPA 2012 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{\cos 4x-1}{x\ \tan 2x}=\cdots$
$\begin{align}
(A)\ & 4 \\
(B)\ & 2 \\
(C)\ & -1 \\
(D)\ & -2 \\
(E)\ & -4
\end{align}$
Alternatif Pembahasan:
show

Dengan menggunakan identitas trigonometri $sin^{2}\ a+cos^{2}\ a=1$ dan teorema limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$, kita coba selesaikan soal di atas seperti penjabaran berikut ini:
$\begin{align}
& \lim\limits_{x \to 0} \dfrac{\cos 4x-1}{x\ \tan 2x} \\
& = \lim\limits_{x \to 0} \dfrac{\cos 2(2x)-1}{x\ \tan 2x} \\
& = \lim\limits_{x \to 0} \dfrac{cos^{2}(2x)-sin^{2}(2x)-1}{x\ \tan 2x} \\
& = \lim\limits_{x \to 0} \dfrac{1-sin^{2}(2x)-sin^{2}(2x)-1}{x\ \tan 2x} \\
& = \lim\limits_{x \to 0} \dfrac{-2\ sin^{2}(2x)}{x\ \tan 2x} \\
& = \lim\limits_{x \to 0} \dfrac{-2\ \sin (2x) \cdot \sin (2x)}{x\ \tan 2x} \\
& = \dfrac{-2 \cdot 2 \cdot 2}{1 \cdot 2} = -4
\end{align}$

$\therefore$ Pilihan yang sesuai $(E)\ -4$

15. Soal UN Matematika SMA IPA 2011 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{1-\cos 2x}{2x\ \sin 2x}=\cdots$
$\begin{align}
(A)\ & \dfrac{1}{8} \\
(B)\ & \dfrac{1}{6} \\
(C)\ & \dfrac{1}{4} \\
(D)\ & \dfrac{1}{2} \\
(E)\ & 1
\end{align}$
Alternatif Pembahasan:
show

Dengan menggunakan identitas trigonometri $sin^{2}\ a+cos^{2}\ a=1$ dan teorema limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$, kita coba selesaikan soal di atas seperti penjabaran berikut ini:
$\begin{align}
& \lim\limits_{x \to 0} \dfrac{1-\cos 2x}{2x\ \sin 2x} \\
& = \lim\limits_{x \to 0} \dfrac{1- \left( 1-2\ sin^{2} x \right)}{2x\ \sin 2x} \\
& = \lim\limits_{x \to 0} \dfrac{2\ sin^{2} x }{2x\ \sin 2x} \\
& = \lim\limits_{x \to 0} \dfrac{2 \cdot \sin x \cdot \sin x }{2x\ \sin 2x} \\
& = \dfrac{2 \cdot 1 \cdot 1 }{2 \cdot 2}= \dfrac{1}{2}
\end{align}$

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

16. Soal UN Matematika SMA IPA 2010 |*Soal Lengkap

$\lim\limits_{x \to 0} \left( \dfrac{\sin x + \sin 5x}{6x} \right)=\cdots$
$\begin{align}
(A)\ & 2 \\
(B)\ & 1 \\
(C)\ & \dfrac{1}{2} \\
(D)\ & \dfrac{1}{3} \\
(E)\ & -1
\end{align}$
Alternatif Pembahasan:
show

Dengan menggunakan identitas trigonometri $sin^{2}\ a+cos^{2}\ a=1$ dan teorema limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$, kita coba selesaikan soal di atas seperti penjabaran berikut ini:
$\begin{align}
& \lim\limits_{x \to 0} \left( \dfrac{\sin x + \sin 5x}{6x} \right) \\
& = \lim\limits_{x \to 0} \left( \dfrac{2\ \sin (3x)\ \cos (-2x)}{6x} \right) \\
& = \lim\limits_{x \to 0} \dfrac{2\ \sin (3x)}{6x} \cdot \lim\limits_{x \to 0} \dfrac{\cos (-2x)}{1} \\
& = \dfrac{2 \cdot 3 }{6} \cdot \lim\limits_{x \to 0} \dfrac{\cos (0)}{1} \\
& = 1 \cdot 1= 1
\end{align}$

$\therefore$ Pilihan yang sesuai $(B)\ 1$

17. Soal SPMB 2006 Kode 310 |*Soal Lengkap

$\lim\limits_{x \to \frac{1}{2}\pi} \dfrac{\sin x\ tan(2x-\pi)}{2\pi-4x}=\cdots$
$\begin{align}
(A)\ & -\dfrac{1}{2} \\
(B)\ & \dfrac{1}{2} \\
(C)\ & \dfrac{1}{3} \sqrt{3} \\
(D)\ & 1 \\
(E)\ & \sqrt{3}
\end{align}$
Alternatif Pembahasan:
show

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

$\therefore$ Pilihan yang sesuai $(A)\ -\dfrac{1}{2}$

18. Soal SPMB 2006 Kode 111 |*Soal Lengkap

$\lim\limits_{x \to \frac{1}{2}\pi} \dfrac{ \left(x-\frac{1}{2} \pi \right)^{2}\ \sin x}{cos^{2}x}=\cdots$
$\begin{align}
(A)\ & -1 \\
(B)\ & -\dfrac{1}{2} \\
(C)\ & 0 \\
(D)\ & 1 \\
(E)\ & 2
\end{align}$
Alternatif Pembahasan:
show

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

$\therefore$ Pilihan yang sesuai $(D)\ 1$

19. Soal SPMB 2006 Kode 420 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{x^{2}\ \sqrt{4-x^{3}}}{\cos x-\cos 3x}=\cdots$
$\begin{align}
(A)\ & -\dfrac{3}{2} \\
(B)\ & -\dfrac{1}{2} \\
(C)\ & 0 \\
(D)\ & \dfrac{1}{2} \\
(E)\ & \dfrac{3}{2}
\end{align}$
Alternatif Pembahasan:
show

Untuk menyelesaikan soal limit trigonometri di atas, seperti kita sampaikan sebelumnya beberapa Identitas Trigonometri Dasar setidaknya dapat kita gunakan pada manipulasi aljabar;

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

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

20. Soal UM UGM 2005 Kode 611 |*Soal Lengkap

$\lim\limits_{x \to \frac{1}{4}\pi} \dfrac{ \left(x-\frac{\pi}{4} \right) tan \left(3x-\frac{3\pi}{4} \right) }{2 \left( 1-\sin 2x \right)}=\cdots$
$\begin{align}
(A)\ & 0 \\
(B)\ & -\dfrac{3}{2} \\
(C)\ & \dfrac{3}{2} \\
(D)\ & -\dfrac{3}{4} \\
(E)\ & \dfrac{3}{4}
\end{align}$
Alternatif Pembahasan:
show

Untuk menyelesaikan soal limit trigonometri di atas, seperti kita sampaikan sebelumnya beberapa Identitas Trigonometri Dasar setidaknya dapat kita gunakan pada manipulasi aljabar;

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

$\therefore$ Pilihan yang sesuai $(E)\ \dfrac{3}{4}$

21. Soal UM UGM 2005 Kode 812 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{x\ \tan 5x}{\cos 2x - \cos 7x}=\cdots$
$\begin{align}
(A)\ & \dfrac{1}{9} \\
(B)\ & -\dfrac{1}{9} \\
(C)\ & \dfrac{2}{9} \\
(D)\ & -\dfrac{2}{9} \\
(E)\ & 0
\end{align}$
Alternatif Pembahasan:
show

Untuk menyelesaikan soal limit trigonometri di atas, seperti kita sampaikan sebelumnya beberapa Identitas Trigonometri Dasar harus kita bisa gunakan pada manipulasi aljabar;

  • $\cos A +\cos B = 2cos \left( \dfrac{A+B}{2} \right)\ sin \left( \dfrac{A-B}{2} \right)$
  • $\cos A -\cos B= 2sin \left( \dfrac{A+B}{2} \right)\ sin \left( \dfrac{A-B}{2} \right)$
  • $\cos 2x -\cos 7x= -2sin \left( \dfrac{2x+7x}{2} \right)\ sin \left( \dfrac{2x-7x}{2} \right)$
    $\cos 2x -\cos 7x= -2sin \left( \dfrac{9}{2}x \right)\ sin \left( \dfrac{-5}{2}x \right)$
$\begin{align}
& \lim\limits_{x \to 0} \dfrac{x\ \tan 5x}{\cos 2x - \cos 7x} \\
& = \lim\limits_{x \to 0} \dfrac{x\ \tan 5x}{-2sin \left( \dfrac{9}{2}x \right)\ sin \left( \dfrac{-5}{2}x \right)} \\
& = \lim\limits_{x \to 0} \left( \dfrac{x}{-2sin \left( \dfrac{9}{2}x \right)} \times \dfrac{ \tan 5x}{sin \left( \dfrac{-5}{2}x \right)} \right) \\
& = \dfrac{1}{-2 \cdot \dfrac{9}{2}} \times \dfrac{5}{ \dfrac{-5}{2}} \\
& = \dfrac{1}{-9} \times -2 = \dfrac{2}{9}
\end{align}$

$\therefore$ Pilihan yang sesuai $(C)\ \dfrac{2}{9}$

22. Soal SPMB 2005 Kode 270 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{1-\cos x}{2x\ \sin 3x}=\cdots$
$\begin{align}
(A)\ & 0 \\
(B)\ & \dfrac{1}{12} \\
(C)\ & \dfrac{1}{6} \\
(D)\ & \dfrac{1}{3} \\
(E)\ & \dfrac{1}{2}
\end{align}$
Alternatif Pembahasan:
show

Untuk menyelesaikan soal limit trigonometri di atas, seperti kita sampaikan sebelumnya beberapa Identitas Trigonometri Dasar harus kita bisa gunakan pada manipulasi aljabar;

  • $\cos \left( \frac{1}{2}\pi -x \right) = \sin \left( x \right)$
  • $\cos 2x= cos^{2}x-sin^{2}x$
  • $\cos 2x= 1-2sin^{2}x$
  • $\cos x= 1-2sin^{2} \left( \frac{1}{2}x \right)$
$\begin{align}
& \lim\limits_{x \to 0} \dfrac{1-\cos x}{2x\ \sin 3x} \\
& = \lim\limits_{x \to 0} \dfrac{2sin^{2} \left( \frac{1}{2}x \right)}{2x\ \sin 3x} \\
& = \lim\limits_{x \to 0} \left( \dfrac{2 sin \left( \frac{1}{2}x \right)}{2x} \times \dfrac{ sin \left( \frac{1}{2}x \right)}{\sin 3x} \right)\\
& = \dfrac{2 \cdot \dfrac{1}{2}}{2} \times \dfrac{ \dfrac{1}{2}}{3} \\
& = \dfrac{1}{2} \times \dfrac{ \dfrac{1}{2} }{3}=\dfrac{1}{12}
\end{align}$

$\therefore$ Pilihan yang sesuai $(B)\ \dfrac{1}{12}$

23. Soal SPMB 2005 Kode 181 |*Soal Lengkap

$\lim\limits_{x \to 2} \dfrac{\tan \left( 2-\sqrt{2x} \right)}{x^{2}-2x}=\cdots$
$\begin{align}
(A)\ & \dfrac{1}{4} \\
(B)\ & \dfrac{1}{8} \\
(C)\ & 0 \\
(D)\ & -\dfrac{1}{6} \\
(E)\ & -\dfrac{1}{4}
\end{align}$
Alternatif Pembahasan:
show

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

$\therefore$ Pilihan yang sesuai $(E)\ -\dfrac{1}{4}$

24. Soal SPMB 2005 Kode 780 |*Soal Lengkap

$\lim\limits_{x \to 1} \dfrac{ \left( x^{2}+x-2 \right) \sin (x-1)}{x^{2}-2x+1}=\cdots$
$\begin{align}
(A)\ & 4 \\
(B)\ & 3 \\
(C)\ & 0 \\
(D)\ & -\dfrac{1}{4} \\
(E)\ & -\dfrac{1}{2}
\end{align}$
Alternatif Pembahasan:
show

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

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

25. Soal SPMB 2005 Kode 370 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{-x^{2}}{1-\cos x}=\cdots$
$\begin{align}
(A)\ & -2 \\
(B)\ & -1 \\
(C)\ & 0 \\
(D)\ & 1 \\
(E)\ & 2
\end{align}$
Alternatif Pembahasan:
show

Untuk menyelesaikan soal limit trigonometri di atas, seperti kita sampaikan sebelumnya beberapa Identitas Trigonometri Dasar harus kita bisa gunakan pada manipulasi aljabar;

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

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

26. Soal SPMB 2005 Kode 772 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{-x+ \tan x}{x}=\cdots$
$\begin{align}
(A)\ & -2 \\
(B)\ & -1 \\
(C)\ & 0 \\
(D)\ & 1 \\
(E)\ & 2
\end{align}$
Alternatif Pembahasan:
show

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

$\therefore$ Pilihan yang sesuai $(C)\ 0$

27. Soal SPMB 2005 Kode 470 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{3\ \sin \frac{1}{2}x}{\tan \frac{1}{3}x}=\cdots$
$\begin{align}
(A)\ & 0 \\
(B)\ & 3\dfrac{1}{3} \\
(C)\ & 4\dfrac{1}{2} \\
(D)\ & 6 \\
(E)\ & 9
\end{align}$
Alternatif Pembahasan:
show

$\begin{align}
& \lim\limits_{x \to 0} \dfrac{3\ \sin \frac{1}{2}x}{\tan \frac{1}{3}x} \\
& = \dfrac{3 \cdot \frac{1}{2}}{\frac{1}{3}} \\
& = 3 \cdot \dfrac{3}{2} = \dfrac{9}{2}
\end{align}$

$\therefore$ Pilihan yang sesuai $(C)\ 0$

28. Soal UM UGM 2004 Kode 322 |*Soal Lengkap

$ \lim\limits_{x \to 1} \dfrac{\tan (x-1)\ \sin \left(1-\sqrt{x} \right)}{x^{2}-2x+1}=\cdots$
$\begin{align}
(A)\ & -1 \\
(B)\ & -\dfrac{1}{2} \\
(C)\ & 0 \\
(D)\ & \dfrac{1}{2} \\
(E)\ & 1
\end{align}$
Alternatif Pembahasan:
show

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

$\therefore$ Pilihan yang sesuai $(B)\ -\dfrac{1}{2}$

29. Soal UM UGM 2004 Kode 121 |*Soal Lengkap

$\underset{a \to 0}{lim} \dfrac{1}{a} \left( \dfrac{sin^{3}2a}{\cos 2a}+\sin 2a\ \cos 2a \right)$ sama dengan
$\begin{align}
(A)\ & 0 \\
(B)\ & \dfrac{1}{2} \\
(C)\ & 1 \\
(D)\ & 2 \\
(E)\ & \infty
\end{align}$
Alternatif Pembahasan:
show

$\begin{align}
& \underset{a \to 0}{lim} \dfrac{1}{a} \left( \dfrac{sin^{3}2a}{\cos 2a}+\sin 2a\ \cos 2a \right) \\
& = \underset{a \to 0}{lim} \left( \dfrac{sin^{3}2a}{a \cdot \cos 2a}+\dfrac{\sin 2a}{a}\ \cdot \cos 2a \right) \\
& = \underset{a \to 0}{lim} \left( \dfrac{\sin 2a}{a} \cdot \dfrac{\sin 2a}{\cos 2a} \cdot \dfrac{\sin 2a}{1}+\dfrac{\sin 2a}{a}\ \cdot \cos 2a \right) \\
& = 2 \cdot 0 \cdot 0 + 2 \cdot 1 \\
& = 2
\end{align}$

$\therefore$ Pilihan yang sesuai $(D)\ 2$

30. Soal UM STIS 2011 |*Soal Lengkap

Jika $\lim\limits_{x \to 0} \dfrac{x^{a}\ sin^{4}x}{sin^{6}x}=1$, maka nilai $a$ yang memenuhi adalah...
$\begin{align}
(A)\ & 1 \\
(B)\ & 2 \\
(C)\ & 3 \\
(D)\ & 4 \\
(E)\ & 5
\end{align}$
Alternatif Pembahasan:
show

Catatan calon guru yang mungkin kita perlukan tentang Limit Trigonometri yaitu $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$ atau $\lim\limits_{x \to 0} \dfrac{ ax }{\sin bx} = \dfrac{a}{b}$.
$\begin{align}
\lim\limits_{x \to 0} \dfrac{x^{a}\ sin^{4}x}{sin^{6}x} & =1 \\
\lim\limits_{x \to 0} \dfrac{x^{a}\ sin^{4}x}{sin^{2}x \cdot sin^{4}x} & =1 \\
\lim\limits_{x \to 0} \dfrac{x^{a} }{sin^{2}x} & =1
\end{align}$
Agar nilai limit fungsi di atas benar adalah $1$, maka nilai $a=2$

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


31. Soal UM STIS 2011 |*Soal Lengkap

Nilai dari $\lim\limits_{x \to \frac{1}{4}\pi} \dfrac{1-2\ \sin x\ \cos x}{\sin x-\cos x}$ adalah...
$\begin{align}
(A)\ & \dfrac{1}{2} \\
(B)\ & \dfrac{1}{2} \sqrt{2} \\
(C)\ & 1 \\
(D)\ & 0 \\
(E)\ & -1
\end{align}$
Alternatif Pembahasan:
show

Catatan calon guru yang mungkin kita perlukan tentang Limit Trigonometri yaitu $sin^{2}x+cos^{2}x=1$.

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

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

32. Soal UTBK-SBMPTN 2019 |*Soal Lengkap

Nilai $\lim\limits_{x \to 0} \dfrac{cot\ 2x - csc\ 2x}{\cos 3x\ \tan \frac{1}{3}x } =\cdots$
$\begin{align}
(A)\ & 3 \\
(B)\ & 2 \\
(C)\ & 0 \\
(D)\ & -2 \\
(E)\ & -3
\end{align}$
Alternatif Pembahasan:
show

Catatan calon guru tentang limit fungsi trigonometri yang mungkin kita butuhkan adalah:

  • $\lim\limits_{x \to 0} \dfrac{\tan ax }{bx} = \dfrac{a}{b}$
  • $\lim\limits_{x \to 0} \dfrac{\sin ax }{\sin bx} = \dfrac{a}{b}$
  • $\lim\limits_{x \to 0} \dfrac{\tan ax }{\sin bx} = \dfrac{a}{b}$
$ \begin{align}
& \lim\limits_{x \to 0} \dfrac{cot\ 2x - csc\ 2x}{\cos 3x\ \tan \frac{1}{3}x } \\
& = \lim\limits_{x \to 0} \dfrac{\dfrac{\cos 2x}{\sin 2x} - \frac{1}{\sin 2x}}{\cos 3x\ \tan \frac{1}{3}x } \\
& = \lim\limits_{x \to 0} \dfrac{\dfrac{\cos 2x-1}{\sin 2x}}{\cos 3x\ \tan \frac{1}{3}x } \\
& = \lim\limits_{x \to 0} \dfrac{ \cos 2x-1}{\cos 3x\ \tan \frac{1}{3}x\ \sin 2x } \\
& = \lim\limits_{x \to 0} \dfrac{ 1-sin^{2} x-1}{\cos 3x\ \tan \frac{1}{3}x\ \sin 2x } \\
& = \lim\limits_{x \to 0} \dfrac{ -2sin^{2} x }{\cos 3x\ \tan \frac{1}{3}x\ \sin 2x } \\
& = \lim\limits_{x \to 0} \dfrac{ -2\ \sin x\ \sin x }{\cos 3x\ \tan \frac{1}{3}x\ \sin 2x } \\
& = \dfrac{ -2\ \cdot 1 \cdot 1 }{\cos 0\ \cdot \frac{1}{3}\ \cdot 2 } \\
& = \dfrac{ -2 }{ \frac{2}{3} } =-3
\end{align} $

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

33. Soal SNMPTN 2010 Kode 546 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{\sqrt{4x}}{\sqrt{\sin 2x}}=\cdots$
$\begin{align}
(A)\ & \sqrt{2} \\
(B)\ & 1 \\
(C)\ & \dfrac{1}{2} \\
(D)\ & \dfrac{1}{4} \\
(E)\ & 0
\end{align}$
Alternatif Pembahasan:
show

Dengan menggunakan teorema limit $\lim\limits_{x \to c} \sqrt[n]{f(x)} = \sqrt[n]{ \lim\limits_{x \to c} f(x)}$ dan $\lim\limits_{x \to 0} \dfrac{ ax }{\sin bx} = \dfrac{a}{b}$ kita coba selesaikan soal di atas seperti penjabaran berikut ini;
$\begin{align}
\lim\limits_{x \to 0} \dfrac{\sqrt{4x}}{\sqrt{\sin 2x}} & = \lim\limits_{x \to 0} \sqrt{ \dfrac{4x}{ \sin 2x} } \\
& = \sqrt{ \lim\limits_{x \to 0} \dfrac{4x}{ \sin 2x} } \\
& = \sqrt{ \dfrac{4 }{2} } \\
& = \sqrt{ 2 } \\
\end{align}$

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

34. Soal SNMPTN 2012 Kode 132 |*Soal Lengkap

Nilai $\lim\limits_{x \to 0} \dfrac{1-cos^{2}x}{x^{2}\ cot \left( x + \frac{\pi}{3} \right)}=\cdots$
$\begin{align}
(A)\ & -1 \\
(B)\ & 0 \\
(C)\ & 1 \\
(D)\ & \dfrac{\sqrt{2}}{2} \\
(E)\ & \sqrt{3}
\end{align}$
Alternatif Pembahasan:
show

Dengan menggunakan Indentitas Trigonometri, teorema limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$ atau manipulasi aljabar, kita coba selesaikan soal di atas seperti penjabaran berikut ini;
$\begin{align} & \lim\limits_{x \to 0} \dfrac{1-cos^{2}x}{x^{2}\ cot \left( x + \dfrac{\pi}{3} \right)} \\
& = \lim\limits_{x \to 0} \dfrac{sin^{2}x}{x^{2}\ cot \left( x + \dfrac{\pi}{3} \right)} \\
& = \lim\limits_{x \to 0} \left( \dfrac{sin^{2}x}{x^{2}} \cdot \dfrac{1}{cot \left( x + \dfrac{\pi}{3} \right)} \right) \\
& = \lim\limits_{x \to 0} \dfrac{sin^{2}x}{x^{2}} \cdot \lim\limits_{x \to 0} \dfrac{1}{cot \left( x + \dfrac{\pi}{3} \right)} \\
& = 1 \cdot 1 \cdot \dfrac{1}{cot \left( 0 + \dfrac{\pi}{3} \right)} \\
& = \dfrac{1}{cot\ 60^{\circ} } \\
& = \tan 60^{\circ} = \sqrt{3} \\
\end{align}$

$\therefore$ Pilihan yang sesuai $(E)\ \sqrt{3}$

35. Soal SNMPTN 2012 Kode 833 |*Soal Lengkap

Nilai $\lim\limits_{x \to 0} \dfrac{1-cos^{2}2x}{x^{2}\ tan \left( x + \dfrac{\pi}{4} \right)}=\cdots$
$\begin{align}
(A)\ & -2 \\
(B)\ & 0 \\
(C)\ & \sqrt{2} \\
(D)\ & \sqrt{3} \\
(E)\ & 4
\end{align}$
Alternatif Pembahasan:
show

Dengan menggunakan Indentitas Trigonometri, teorema limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$ atau manipulasi aljabar, kita coba selesaikan soal di atas seperti penjabaran berikut ini;
$\begin{align}
& \lim\limits_{x \to 0} \dfrac{1-cos^{2}2x}{x^{2}\ tan \left( x + \dfrac{\pi}{4} \right)} \\
& = \lim\limits_{x \to 0} \dfrac{sin^{2}2x}{x^{2}\ tan \left( x + \dfrac{\pi}{4} \right)} \\
& = \lim\limits_{x \to 0} \left( \dfrac{sin^{2}2x}{x^{2}} \cdot \dfrac{1}{tan \left( x + \dfrac{\pi}{4} \right)} \right) \\
& = \lim\limits_{x \to 0} \dfrac{sin^{2}2x}{x^{2}} \cdot \lim\limits_{x \to 0} \dfrac{1}{tan \left( x + \dfrac{\pi}{4} \right)} \\
& = 2 \cdot 2 \cdot \dfrac{1}{tan \left( 0 + \dfrac{\pi}{4} \right)} \\
& = \dfrac{4}{\tan 45^{\circ} } \\
& = \dfrac{4}{1}=4
\end{align}$

$\therefore$ Pilihan yang sesuai $(E)\ 4$

36. Soal SBMPTN 2013 Kode 138 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{\cos x\ - \cos 3x}{x^{2}\ \sqrt{4-x}} =\cdots$
$\begin{align}
(A)\ & -2 \\
(B)\ & -\dfrac{1}{2} \\
(C)\ & \dfrac{1}{2} \\
(D)\ & 1 \\
(E)\ & 2
\end{align}$
Alternatif Pembahasan:
show

Dengan menggunakan Indentitas Trigonometri, teorema limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$ atau manipulasi aljabar, kita coba selesaikan soal di atas seperti penjabaran berikut ini;
$\begin{align}
& \lim\limits_{x \to 0} \dfrac{\cos x\ - \cos 3x}{x^{2}\ \sqrt{4-x}} \\
&= \lim\limits_{x \to 0} \dfrac{-2\ \sin \left( \dfrac{x+3x}{2}\right) \cdot \sin \left( \dfrac{x-3x}{2}\right) }{x^{2}\ \sqrt{4-x}} \\
&= \lim\limits_{x \to 0} \dfrac{-2\ \sin \left( 2x \right) \cdot \sin \left( -x \right) }{x^{2}\ \sqrt{4-x}} \\
&= \lim\limits_{x \to 0} \dfrac{ 2\ \sin \left( 2x \right) \cdot \sin \left( x \right) }{x \cdot x\ \sqrt{4-x}} \\
&= \dfrac{ 2\ \cdot 2 \cdot 1}{ \sqrt{4-0}} \\
&= \dfrac{4}{ 2} = 2
\end{align}$

$\therefore$ Pilihan yang sesuai $(E)\ 2$

37. Soal SIMAK UI 2013 Kode 134 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{\cos x\ \sin x - \tan x}{x^{2}\ \sin x} =\cdots$
$\begin{align}
(A)\ & -1 \\
(B)\ & -\dfrac{1}{2} \\
(C)\ & 0 \\
(D)\ & \dfrac{1}{2} \\
(E)\ & 1
\end{align}$
Alternatif Pembahasan:
show

Dengan menggunakan Indentitas Trigonometri, teorema limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$ atau manipulasi aljabar, kita coba selesaikan soal di atas seperti penjabaran berikut ini;
$\begin{align}
& \lim\limits_{x \to 0} \dfrac{\cos x\ \cdot \sin x - \tan x}{x^{2}\ \sin x} \\
&= \lim\limits_{x \to 0} \dfrac{\cos x\ \cdot \cos x\ \cdot \tan x - \tan x}{x^{2}\ \sin x} \\
&= \lim\limits_{x \to 0} \dfrac{\tan x \left( cos^{2} x - 1 \right)}{x^{2}\ \sin x} \\
&= \lim\limits_{x \to 0} \dfrac{\tan x \left( - sin^{2} x \right)}{x^{2}\ \sin x} \\
&= \lim\limits_{x \to 0} \dfrac{\tan x }{\sin x} \cdot \lim\limits_{x \to 0} \dfrac{ \left( - sin^{2} x \right)}{x^{2}} \\
&= 1 \cdot ( -1 ) \\
&= -1
\end{align}$

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

38. Soal SIMAK UI 2013 Kode 134 |*Soal Lengkap

Nilai dari $\lim\limits_{x \to 0} \dfrac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^{3}} =\cdots$
$\begin{align}
(A)\ & -1 \\
(B)\ & -\dfrac{1}{4} \\
(C)\ & 0 \\
(D)\ & \dfrac{1}{4} \\
(E)\ & 1
\end{align}$
Alternatif Pembahasan:
show

Dengan menggunakan Indentitas Trigonometri, teorema limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$ atau manipulasi aljabar, kita coba selesaikan soal di atas seperti penjabaran berikut ini;
$\begin{align}
& \lim\limits_{x \to 0} \dfrac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^{3}} \\
&= \lim\limits_{x \to 0} \left( \dfrac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^{3}} \cdot \dfrac{\sqrt{1+\tan x}+\sqrt{1+\sin x}}{\sqrt{1+\tan x}+\sqrt{1+\sin x}} \right) \\
&= \lim\limits_{x \to 0} \left( \dfrac{ 1+\tan x- 1-\sin x }{x^{3}} \cdot \dfrac{1}{\sqrt{1+\tan x}+\sqrt{1+\sin x}} \right) \\
&= \lim\limits_{x \to 0} \left( \dfrac{ \tan x -\sin x }{x^{3}} \cdot \dfrac{1}{\sqrt{1+\tan x}+\sqrt{1+\sin x}} \right) \\
\hline
\tan x -\sin x & = \dfrac{\sin x}{\cos x} -\sin x \\
& = \dfrac{\sin x-\sin x\ \cos x}{\cos x} \\
& = \dfrac{\sin x \left( 1- \cos x \right)}{\cos x} \\
& = \dfrac{\sin x \left( 2\ sin^{2} \frac{1}{2}x \right)}{\cos x} \\
\hline
&= \lim\limits_{x \to 0} \left( \dfrac{ \dfrac{\sin x \left( 2\ sin^{2} \frac{1}{2}x \right)}{\cos x} }{x^{3}} \cdot \dfrac{1}{\sqrt{1+\tan x}+\sqrt{1+\sin x}} \right) \\
&= \lim\limits_{x \to 0} \left( \dfrac{\sin x \left( 2\ sin^{2} \frac{1}{2}x \right)}{\cos x \cdot x^{3}} \cdot \dfrac{1}{\sqrt{1+\tan x}+\sqrt{1+\sin x}} \right) \\
&= \lim\limits_{x \to 0} \left( \dfrac{\sin x \left( 2\ sin^{2} \frac{1}{2}x \right)}{ x^{3}} \cdot \dfrac{1}{\cos x } \cdot \dfrac{1}{\sqrt{1+\tan x}+\sqrt{1+\sin x}} \right) \\
&= 2\ \cdot \dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{1}{\cos 0 } \cdot \dfrac{1}{\sqrt{1+\tan 0}+\sqrt{1+\sin 0}} \\
&= \dfrac{1}{2} \cdot \dfrac{1}{1} \cdot \dfrac{1}{\sqrt{1}+\sqrt{1}} \\
&= \dfrac{1}{4}
\end{align}$

$\therefore$ Pilihan yang sesuai $(D)\ \dfrac{1}{4}$

39. Soal UM UGM 2013 Kode 261 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{1-cos^{3}x}{x\ \tan x}=\cdots$
$\begin{align}
(A)\ & 0 \\
(B)\ & \dfrac{1}{2} \\
(C)\ & \dfrac{3}{4} \\
(D)\ & \dfrac{3}{2} \\
(E)\ & 3
\end{align}$
Alternatif Pembahasan:
show

Dengan menggunakan Sifat Bilangan Berpangkat $a^{3}-b^{3}=\left( a-b \right)\left( a^{2}+ab+b^{2} \right)$, teorema limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$ atau manipulasi aljabar, kita coba selesaikan soal di atas seperti penjabaran berikut ini;
$\begin{align}
& \lim\limits_{x \to 0} \dfrac{1-cos^{3}x}{x\ \tan x} \\
& = \lim\limits_{x \to 0} \dfrac{\left( 1-\cos x \right)\left( 1+\cos x+cos^{2} x \right)}{x\ \tan x} \\
& = \lim\limits_{x \to 0} \dfrac{ 2sin^{2} \left( \frac{1}{2}x \right)\left( 1+\cos x+cos^{2} x \right)}{x\ \tan x} \\
& = \lim\limits_{x \to 0} \left( \dfrac{ 2sin^{2} \left( \frac{1}{2}x \right)}{x\ \tan x} \cdot \left( 1+\cos x+cos^{2} x \right) \right) \\
& = 2 \cdot \dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \left( 1+\cos 0+cos^{2} 0 \right) \\
& = \dfrac{1}{2} \cdot \left( 1+1+1 \right) \\
& = \dfrac{3}{2}
\end{align}$

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

40. Soal Latihan Matematika TryOut Masuk PTN (*Soal Request)

$\lim\limits_{x \to 0} \dfrac{x}{2\ csc\ x\ \left( 1-\sqrt{\cos x}\right) } =\cdots$
$\begin{align}
(A)\ & -2 \\
(B)\ & -1 \\
(C)\ & 0 \\
(D)\ & 1 \\
(E)\ & 2
\end{align}$
Alternatif Pembahasan:
show

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

$\therefore$ Pilihan yang sesuai $(E)\ 2$

41. Soal SNMPTN 2008 Kode 201 |*Soal Lengkap

$\lim\limits_{x \to \frac{1}{4}\pi} \dfrac{ 1-2 \sin x\ \cos x}{\cos x - \sin x}=\cdots$
$\begin{align}
(A)\ & \dfrac{1}{2} \\
(B)\ & \dfrac{1}{2}\sqrt{2} \\
(C)\ & 1 \\
(D)\ & 0 \\
(E)\ & -1
\end{align}$
Alternatif Pembahasan:
show

Untuk menyelesaikan soal limit trigonometri di atas, seperti kita sampaikan sebelumnya beberapa Identitas Trigonometri Dasar setidaknya dapat kita gunakan pada manipulasi aljabar;
$\begin{align}
& \lim\limits_{x \to \frac{1}{4}\pi} \dfrac{ 1-2 \sin x\ \cos x}{\sin x - \cos x} \\
& = \lim\limits_{x \to \frac{1}{4}\pi} \dfrac{ sin^{2}x+cos^{2}x-2 \sin x\ \cos x}{\sin x - \cos x} \\
& = \lim\limits_{x \to \frac{1}{4}\pi} \dfrac{ \left(\sin x-\cos x \right)^{2}}{\sin x - \cos x} \\
& = \lim\limits_{x \to \frac{1}{4}\pi} \dfrac{ \left(\sin x-\cos x \right) }{1} \\
& = \sin \frac{1}{4}\pi-\cos \frac{1}{4}\pi \\
& = \dfrac{1}{2}\sqrt{2}-\dfrac{1}{2}\sqrt{2}=0
\end{align}$

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

42. Soal UM UNDIP 2010 Kode 101 |*Soal Lengkap

$\lim\limits_{x \to -1} \dfrac{sin(1-x^{2})\ cos (1-x^{2})}{x^{2}-1}=\cdots$
$\begin{align}
(A)\ & 1 \\
(B)\ & -1 \\
(C)\ & 2 \\
(D)\ & -2 \\
(E)\ & 0
\end{align}$
Alternatif Pembahasan:
show

Untuk menyelesaikan soal di atas kita coba dengan memisalkan $1-x^{2}=m$, karena $x \to -1$ maka $m \to 0$.
$\begin{align}
& \lim\limits_{x \to -1} \dfrac{sin(1-x^{2})\ cos (1-x^{2})}{x^{2}-1} \\
& = \lim\limits_{m \to 0} \dfrac{\sin m\ \cos m}{-m} \\
& = \lim\limits_{m \to 0} \cos m \cdot \lim\limits_{m \to 0} \dfrac{\sin m}{-m} \\
& = \cos 0 \cdot -1 \\
& = 1 \cdot -1 =-1
\end{align}$

$\therefore$ Pilihan yang sesuai $(B)\ -1$

43. Soal SIMAK UI 2012 Kode 221 |*Soal Lengkap

$\lim\limits_{x \to 1} \dfrac{\sin 2(x-1)}{(x^{2}-2x+1)\ cot\ \dfrac{1}{2}(x-1)}=\cdots$
$\begin{align}
(A)\ & \dfrac{1}{4} \\
(B)\ & \dfrac{1}{2} \\
(C)\ & 1 \\
(D)\ & 2 \\ \\
(E)\ & 4
\end{align}$
Alternatif Pembahasan:
show

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

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

44. Soal SIMAK UI 2009 Kode 941 |*Soal Lengkap

$\lim\limits_{x \to \frac{\pi}{2}} \dfrac{\pi (\pi-2x)\ tan \left ( x-\dfrac{\pi}{2} \right)}{2(x-\pi)\ cos^{2}x }=\cdots$
$\begin{align}
(A)\ & -2 \\
(B)\ & -1 \\
(C)\ & -\dfrac{1}{2} \\
(D)\ & 1 \\
(E)\ & 2
\end{align}$
Alternatif Pembahasan:
show

Untuk menyelesaikan soal limit trigonometri di atas, seperti kita sampaikan sebelumnya beberapa Identitas Trigonometri Dasar setidaknya bisa kita gunakan pada manipulasi aljabar;

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

$\therefore$ Pilihan yang sesuai $(E)\ 2$

45. Soal SPMB 2006 Kode 510 |*Soal Lengkap

$\lim\limits_{x \to 4} \dfrac{\sin \left( 4-2\sqrt{x} \right)}{4-x}=\cdots$
$\begin{align}
(A)\ & -\dfrac{1}{6} \\
(B)\ & -\dfrac{1}{2} \\
(C)\ & 0 \\
(D)\ & \dfrac{1}{4} \\
(E)\ & \dfrac{1}{2}
\end{align}$
Alternatif Pembahasan:
show

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

$\therefore$ Pilihan yang sesuai $(E)\ \dfrac{1}{2}$

46. Soal SPMB 2006 Kode 720 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{\sin \left( 3x-\pi \right)}{\sqrt[3]{8+x}\ \tan 2x}=\cdots$
$\begin{align}
(A)\ & -\dfrac{3}{2} \\
(B)\ & -\dfrac{3}{4} \\
(C)\ & -\dfrac{1}{4} \\
(D)\ & \dfrac{1}{4} \\
(E)\ & \dfrac{3}{4}
\end{align}$
Alternatif Pembahasan:
show

$\begin{align}
& \lim\limits_{x \to 0} \dfrac{\sin \left( 3x-\pi \right)}{\sqrt[3]{8+x}\ \tan 2x} \\
& = \lim\limits_{x \to 0} \dfrac{-\sin \left( \pi-3x \right)}{\sqrt[3]{8+x}\ \tan 2x} \\
& = \lim\limits_{x \to 0} \dfrac{-\sin 3x}{\sqrt[3]{8+x}\ \tan 2x} \\
& = \lim\limits_{x \to 0} \left( \dfrac{1}{\sqrt[3]{8+x}} \times \dfrac{-\sin 3x}{\tan 2x} \right) \\
& = \dfrac{1}{\sqrt[3]{8+0}} \times \dfrac{- 3 }{ 2 } \\
& = \dfrac{1}{2} \times \dfrac{-3}{2} = -\dfrac{3}{4}
\end{align}$

$\therefore$ Pilihan yang sesuai $(B)\ -\dfrac{3}{4}$

47. Soal UM UGM 2006 Kode 381 |*Soal Lengkap

$\lim\limits_{x \to 0} \left( \dfrac{1}{x}-\dfrac{1}{x\ \cos x} \right)=\cdots$
$\begin{align}
(A)\ & -1 \\
(B)\ & -\dfrac{1}{2} \\
(C)\ & 0 \\
(D)\ & \dfrac{1}{2} \\
(E)\ & 1
\end{align}$
Alternatif Pembahasan:
show

Untuk menyelesaikan soal limit trigonometri di atas, seperti kita sampaikan sebelumnya beberapa Identitas Trigonometri Dasar setidaknya dapat kita gunakan pada manipulasi aljabar;

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

$\therefore$ Pilihan yang sesuai $(C)\ 0$

48. Soal SPMB 2006 Kode 121 |*Soal Lengkap

$\lim\limits_{x \to 5} \dfrac{2x^{3}-20x^{2}+50x}{sin^{2}(x-5)cos(2x-10)}=\cdots$
$\begin{align}
(A)\ & 0 \\
(B)\ & 1 \\
(C)\ & 5 \\
(D)\ & 10 \\
(E)\ & \infty
\end{align}$
Alternatif Pembahasan:
show

$\begin{align}
& \lim\limits_{x \to 5} \dfrac{2x^{3}-20x^{2}+50x}{sin^{2}(x-5)cos(2x-10)} \\
& = \lim\limits_{x \to 5} \dfrac{2x \left( x^{2}-10x +25 \right) }{sin^{2}(x-5)cos (2x-10)} \\
& = \lim\limits_{x \to 5} \dfrac{2x \left( x-5 \right)\left( x-5 \right) }{sin^{2}(x-5)cos(2x-10)} \\
& = \lim\limits_{x \to 5} \left( \dfrac{\left( x-5 \right)\left( x-5 \right) }{sin^{2}(x-5)} \times \dfrac{2x}{cos(2x-10)} \right)\\
& = 1 \times \dfrac{2(5)}{cos(2(5)-10)} \\
& = 1 \times \dfrac{10}{cos(0)} = 10
\end{align}$

$\therefore$ Pilihan yang sesuai $(D)\ 10$


49. Soal UM UNDIP 2010 Kode 101 |*Soal Lengkap

$\lim\limits_{x \to y} \dfrac{\sin x - \sin y}{x-y}=\cdots$
$\begin{align}
(A)\ & \sin x \\
(B)\ & \sin y \\
(C)\ & 0 \\
(D)\ & \cos x \\
(E)\ & \cos y
\end{align}$
Alternatif Pembahasan:
show

Untuk menyelesaikan bentuk ini, kita gunakan sedikit identitas trigonometri yaitu $\sin x - \sin y$ adalah $2\ \cos \dfrac{1}{2}(x+y)\ \sin \dfrac{1}{2}(x-y)$.

$\begin{align}
& \lim\limits_{x \to y} \dfrac{\sin x - \sin y}{x-y} \\
& = \lim\limits_{x \to y} \dfrac{2\ \cos \dfrac{1}{2}(x+y)\ \sin \dfrac{1}{2}(x-y)}{x-y} \\
& = \lim\limits_{x \to y} 2\ \cos \dfrac{1}{2}(x+y) \times \lim\limits_{x \to y} \dfrac{\sin \dfrac{1}{2}(x-y)}{x-y} \\
& = 2\ \cos \dfrac{1}{2}(y+y) \times \dfrac{1}{2} \\
& = \cos \dfrac{1}{2}(2y) \\
& = \cos y
\end{align}$

$\therefore$ Pilihan yang sesuai $(E)\ \cos y$

50. Soal SBMPTN 2013 Kode 338 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{x\ \tan x}{x\ \sin x - \cos x +1}=\cdots$
$\begin{align}
(A)\ & 2 \\
(B)\ & \dfrac{3}{2} \\
(C)\ & 1 \\
(D)\ & \dfrac{2}{3} \\
(E)\ & -1
\end{align}$
Alternatif Pembahasan:
show

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

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

51. Soal SBMPTN 2013 Kode 131 |*Soal Lengkap

$\lim\limits_{x \to 0} \sqrt{\dfrac{x\ \tan x}{sin^{2} x - \cos 2x +1}}=\cdots$
$\begin{align}
(A)\ & 3 \\
(B)\ & \sqrt{3} \\
(C)\ & \dfrac{\sqrt{3}}{3} \\
(D)\ & \dfrac{1}{3} \\
(E)\ & \dfrac{\sqrt{3}}{2}
\end{align}$
Alternatif Pembahasan:
show

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

$\therefore$ Pilihan yang sesuai $(C)\ \dfrac{\sqrt{3}}{3}$

52. Soal SBMPTN 2013 Kode 132 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{x\ \tan x}{sin^{2} x - \cos 2x +1}=\cdots$
$\begin{align}
(A)\ & 1 \\
(B)\ & \dfrac{1}{3} \\
(C)\ & \dfrac{2}{3} \\
(D)\ & -\dfrac{1}{2} \\
(E)\ & -1
\end{align}$
Alternatif Pembahasan:
show

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

$\therefore$ Pilihan yang sesuai $(B)\ \dfrac{1}{3}$

53. Soal SBMPTN 2017 Kode 106 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{sec\ x+\cos x-2}{x^{2}\ sin^{2}x}=\cdots$
$\begin{align}
(A)\ & -\dfrac{1}{8} \\
(B)\ & -\dfrac{1}{4} \\
(C)\ & 0 \\
(D)\ & \dfrac{1}{4} \\
(E)\ & \dfrac{1}{8}
\end{align}$
Alternatif Pembahasan:
show

Identitas trigonometri yg mungkin diperlukan:
$\cos 4x=cos^{2}2x-sin^{2}2x$
$\cos 2x=cos^{2}x-sin^{2}x$
$\cos x=cos^{2} \dfrac{1}{2}x-sin^{2}\dfrac{1}{2}x$
$1=cos^{2} \dfrac{1}{2}x+sin^{2}\dfrac{1}{2}x$
$\cos x - 1=-2sin^{2}\dfrac{1}{2}x$

Kita kembali ke soal;
$\begin{align}
& \lim\limits_{x \to 0} \dfrac{sec\ x+\cos x-2}{x^{2}\ sin^{2}x}\\
= & \lim\limits_{x \to 0} \dfrac{\dfrac{1}{\cos x}+\dfrac{cos^{2}x}{\cos x}-\dfrac{2\ \cos x}{\cos x}}{x^{2}\ sin^{2}x}\\
= & \lim\limits_{x \to 0} \dfrac{cos^{2}-2\ \cos x+1}{x^{2}\ sin^{2}x\ \cos x}\\
= & \lim\limits_{x \to 0} \dfrac{\left (\cos x-1 \right )^{2}}{x^{2}\ sin^{2}x\ \cos x}\\
= & \lim\limits_{x \to 0} \dfrac{\left (-2sin^{2}(\dfrac{1}{2}x) \right )^{2}}{x^{2}\ sin^{2}x\ \cos x}\\
= & \lim\limits_{x \to 0} \dfrac{4\ sin^{2}(\dfrac{1}{2}x)\ sin^{2}(\dfrac{1}{2}x)}{x^{2}\ sin^{2}x\ \cos x}\\
= & \lim\limits_{x \to 0} 4\ \cdot \dfrac{sin^{2}(\dfrac{1}{2}x)}{x^{2}} \cdot \dfrac{sin^{2}(\dfrac{1}{2}x)}{sin^{2}x} \cdot \dfrac{1}{\cos x}\\
= & 4\ \cdot \dfrac{1}{4} \cdot \dfrac{1}{4} \cdot \dfrac{1}{1} = \dfrac{1}{4}
\end{align}$

$\therefore$ Pilihan yang sesuai $(D)\ \dfrac{1}{4}$

54. Soal SBMPTN 2016 Kode 255 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{x^{3}}{\sqrt{1+\sin x}-\sqrt{1+\tan x}}=\cdots$
$\begin{align}
(A)\ & -4 \\
(B)\ & -2 \\
(C)\ & 0 \\
(D)\ & 2 \\
(E)\ & 4
\end{align}$
Alternatif Pembahasan:
show

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

$\therefore$ Pilihan yang sesuai $(A)\ -4$

55. Soal UM STIS 2017 |*Soal Lengkap

$\lim\limits_{x \to 2} \dfrac{\left( x^{2}-5x-6\right)\ \sin 2(x-2) }{\left( x^{2}-x-2\right)} \cdots$
$\begin{align}
(A)\ & -8 \\
(B)\ & -5 \\
(C)\ & -2 \\
(D)\ & \dfrac{3}{4} \\
(E)\ & 5
\end{align}$
Alternatif Pembahasan:
show

$\begin{align}
& \lim\limits_{x \to 2} \dfrac{\left( x^{2}-5x-6\right)\ \sin 2(x-2) }{\left( x^{2}-x-2 \right)} \\
& = \lim\limits_{x \to 2} \dfrac{(x+1)(x-6)\ \sin 2(x-2) }{(x-2)(x+1)} \\
& = \lim\limits_{x \to 2} \dfrac{(x-6)\ \sin 2(x-2) }{(x-2)} \\
& = \lim\limits_{x \to 2} (x-6) \cdot \lim\limits_{x \to 2} \dfrac{\sin 2(x-2) }{(x-2)} \\
& = (2-6) \cdot 2 \\
& = -4 \cdot 2 =-8 \\
\end{align}$

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

56. Soal SBMPTN 2018 Kode 423 |*Soal Lengkap

$\lim\limits_{x \to 2} \dfrac{\sin \left( 2x-4 \right) }{2- \sqrt{6-x}} =\cdots$
$\begin{align}
(A)\ & -8 \\
(B)\ & -2 \\
(C)\ & 0 \\
(D)\ & 2 \\
(E)\ & 8
\end{align}$
Alternatif Pembahasan:
show

$\begin{align}
& \lim\limits_{x \to 2} \dfrac{\sin \left( 2x-4 \right) }{2- \sqrt{6-x}} \\
& = \lim\limits_{x \to 2} \dfrac{\sin \left( 2x-4 \right) }{2- \sqrt{6-x}} \cdot \dfrac{2+ \sqrt{6-x}}{2+ \sqrt{6-x}} \\
& = \lim\limits_{x \to 2} \dfrac{\sin \left( 2x-4 \right) \left( 2+ \sqrt{6-x} \right) }{4- \left( 6-x \right)} \\
& = \lim\limits_{x \to 2} \dfrac{\sin \left( 2x-4 \right) \left( 2+ \sqrt{6-x} \right) }{4- 6+x } \\
& = \lim\limits_{x \to 2} \dfrac{\sin 2\left( x-2 \right) \left( 2+ \sqrt{6-x} \right) }{x-2 } \\
& = \lim\limits_{x \to 2} \dfrac{\sin 2\left( x-2 \right)}{x-2 } \cdot \lim\limits_{x \to 2} \left( 2+ \sqrt{6-x} \right) \\
& = 2 \cdot \left( 2+ \sqrt{6-2} \right) \\
& = 2 \cdot ( 2+ 2)=8 \\
\end{align}$

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

57. Soal UM UGM 2017 Kode 713 |*Soal Lengkap

$\lim\limits_{x \to -4} \dfrac{1-cos(x+4)}{x^{2}+8x+16}=\cdots$
$\begin{align}
(A)\ & -2 \\
(B)\ & -\dfrac{1}{2} \\
(C)\ & \dfrac{1}{3} \\
(D)\ & \dfrac{1}{2} \\
(E)\ & 2
\end{align}$
Alternatif Pembahasan:
show

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

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

58. Soal UM UGM 2019 Kode 624 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{ \sqrt{1-\cos 4x^{2}}}{1-\cos 2x}=\cdots$

$\begin{align} (A)\ & 1 \\ (B)\ & \sqrt{2} \\
(C)\ & 2 \\ (D)\ & 2\sqrt{2} \\ (E)\ & 4 \end{align}$

Alternatif Pembahasan:
show

Untuk menyelesaikan soal limit trigonometri di atas, seperti kita sampaikan sebelumnya beberapa Identitas Trigonometri Dasar setidaknya dapat kita gunakan pada manipulasi aljabar;

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

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

59. Soal UM UGM 2016 Kode 582 |*Soal Lengkap

$\lim\limits_{x \to 3} \dfrac{\left( x+6 \right)\ \tan (2x-6) }{\left( x^{2}-x-6\right)} =\cdots$
maka $b=\cdots$

$\begin{align} (A)\ & -\dfrac{18}{5} \\ (B)\ & -\dfrac{9}{5} \\ (C)\ & \dfrac{9}{5} \\ (D)\ & \dfrac{18}{5} \\ (E)\ & \dfrac{27}{5} \end{align}$

Alternatif Pembahasan:
show

$\begin{align}
& \lim\limits_{x \to 3} \dfrac{\left( x+6 \right)\ \tan (2x-6) }{\left( x^{2}-x-6\right)} \\ & = \lim\limits_{x \to 3} \dfrac{\left( x+6 \right)\ \tan 2( x-3) }{\left( x-3 \right) \left( x+2 \right)} \\ & = \lim\limits_{x \to 3} \dfrac{\left( x+6 \right) }{ \left( x+2 \right)} \cdot \lim\limits_{x \to 3} \dfrac{\tan 2( x-3) }{\left( x-3 \right)} \\ & = \dfrac{\left( 3+6 \right) }{ \left( 3+2 \right)} \cdot \dfrac{ 2 }{1} \\ & = \dfrac{\left( 18 \right) }{ 5} \end{align}$

$\therefore$ Pilihan yang sesuai $(D)\ \dfrac{18}{5}$

60. Soal UM UGM 2015 Kode 632 |*Soal Lengkap

Jika $b,c \neq 0$ dan
$\lim\limits_{x \to a} \dfrac{\left( x-a \right)\ \tan b\left( a-x \right) }{\cos c\left( x-a \right)-1}=d$
maka $b=\cdots$

$\begin{align} (A)\ & 2c^{2}d \\ (B)\ & c^{2}d \\ (C)\ & \dfrac{1}{2} c^{2}d \\ (D)\ & -\dfrac{1}{2} c^{2}d \\ (E)\ & -c^{2}d \end{align}$

Alternatif Pembahasan:
show

Untuk menyelesaikan soal limit trigonometri di atas, seperti kita sampaikan sebelumnya beberapa Identitas Trigonometri Dasar setidaknya dapat kita gunakan pada manipulasi aljabar;

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

$\therefore$ Pilihan yang sesuai $(C)\ \dfrac{1}{2}c^{2}d$


61. Soal UM UNDIP 2014 Kode 141 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{x^{2}+sin^{2}(2x)}{x^{2}\ cos (2x)}=\cdots$

$\begin{align} (A)\ & 5 \\ (B)\ & 4 \\ (C)\ & 3 \\ (D)\ & 0 \\ (E)\ & -2 \end{align}$

Alternatif Pembahasan:
show

Untuk menyelesaikan soal limit trigonometri di atas, seperti kita sampaikan sebelumnya beberapa Identitas Trigonometri Dasar setidaknya dapat kita gunakan pada manipulasi aljabar;

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


$\therefore$ Pilihan yang sesuai $(A)\ 5$

62. Soal UM UNDIP 2011 Kode 112 |*Soal Lengkap

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

$\begin{align} (A)\ & -4 \\ (B)\ & -\dfrac{3}{4} \\ (C)\ & \dfrac{1}{2} \\ (D)\ & 3 \\ (E)\ & 6 \end{align}$

Alternatif Pembahasan:
show

Untuk menyelesaikan soal limit trigonometri di atas, seperti kita sampaikan sebelumnya beberapa Identitas Trigonometri Dasar setidaknya dapat kita gunakan pada manipulasi aljabar;

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


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

63. Soal UM UNDIP 2013 Kode 132 |*Soal Lengkap

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

$\begin{align} (A)\ & -\dfrac{1}{3} \\ (B)\ & -\dfrac{2}{3} \\ (C)\ & \dfrac{1}{3} \\ (D)\ & \dfrac{2}{3} \\ (E)\ & 1 \end{align}$

Alternatif Pembahasan:
show

Dengan menggunakan teorema limit trigonometri dan catatan Pak Anang bentuk Limit Berputar kita akan selesaikan limit di atas. Limit Berputar istilah yang mungkin dibuat untuk mempermudah kita dalam memahami bentuknya, yaitu:

  • $\lim\limits_{x \to 0} \dfrac{x - \sin x}{x^{3}}=\dfrac{1}{6}$
  • $\lim\limits_{x \to 0} \dfrac{(2x) - \sin (2x)}{(2x)^{3}}=\dfrac{1}{6}$
  • $\lim\limits_{x \to 0} \dfrac{(3x) - \sin (3x)}{(3x)^{3}}=\dfrac{1}{6}$

Kita terapkan pada soal di atas, menjadi seperti berikut ini:
$\begin{align} & \lim\limits_{x \to 0} \left[ csc^{2} \left(2x \right) - \dfrac{1}{4x^{2}} \right] \\ & = \lim\limits_{x \to 0} \left[ \dfrac{1}{sin^{2}(2x)} - \dfrac{1}{4x^{2}} \right] \\ & = \lim\limits_{x \to 0} \left[ \dfrac{4x^{2}-sin^{2}(2x)}{4x^{2} sin^{2}(2x)} \right] \\ & = \lim\limits_{x \to 0} \left[ \dfrac{\left( (2x)-sin(2x) \right)\left( (2x)+sin(2x) \right) }{(2x)^{2} sin^{2}(2x)} \right] \cdot \dfrac{(2x)^{2}}{(2x)^{2}} \\ & = \lim\limits_{x \to 0} \left[ \dfrac{\left( (2x)-sin(2x) \right)\left( (2x)+sin(2x) \right) \cdot (2x)^{2}}{(2x)^{4} sin^{2}(2x)} \right] \\ & = \lim\limits_{x \to 0} \left[ \dfrac{\left( (2x)-sin(2x) \right)}{(2x)^{3}} \cdot \dfrac{ \left( 2x+sin(2x) \right)}{(2x)} \cdot \dfrac{ (2x)^{2}}{sin^{2}(2x)} \right] \\ & = \dfrac{1}{6} \cdot (1+1) \cdot 1 \\ & = \dfrac{1}{3} \end{align}$


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

64. Soal Simulasi UTBK-SBMPTN TKA SAINTEK 2021

Nilai dari $ \lim\limits_{x \to 0} \dfrac{\sqrt{x} - \sqrt{\sin x}}{x^{2}\sqrt{x}}$ adalah...

$\begin{align} (A)\ & 12 \\ (B)\ & 6 \\ (C)\ & \dfrac{1}{3} \\ (D)\ & \dfrac{1}{12} \\ (E)\ & \dfrac{2}{3} \end{align}$

Alternatif Pembahasan:
show

Dengan menggunakan teorema limit trigonometri dan catatan Pak Anang bentuk Limit Berputar kita akan selesaikan limit di atas. Limit Berputar istilah yang mungkin dibuat untuk mempermudah kita dalam memahami bentuknya, yaitu:

  • $\lim\limits_{x \to 0} \dfrac{x - \sin x}{x^{3}}=\dfrac{1}{6}$
  • $\lim\limits_{x \to 0} \dfrac{(2x) - \sin (2x)}{(2x)^{3}}=\dfrac{1}{6}$
  • $\lim\limits_{x \to 0} \dfrac{(3x) - \sin (3x)}{(3x)^{3}}=\dfrac{1}{6}$

Kita terapkan pada soal di atas, menjadi seperti berikut ini:
$\begin{align} & \lim\limits_{x \to 0} \dfrac{\sqrt{x} - \sqrt{\sin x}}{x^{2}\sqrt{x}} \\ & = \lim\limits_{x \to 0} \dfrac{\sqrt{x} - \sqrt{\sin x}}{x^{2}\sqrt{x}} \times \dfrac{\sqrt{x} + \sqrt{\sin x}}{\sqrt{x} + \sqrt{\sin x}} \\ & = \lim\limits_{x \to 0} \dfrac{x - \sin x}{\left( x^{2}\sqrt{x} \right) \left(\sqrt{x} + \sqrt{\sin x} \right)} \\ & = \lim\limits_{x \to 0} \dfrac{x - \sin x}{\left( x^{2}\sqrt{x} \right) \sqrt{x} \left( 1 + \sqrt{\frac{\sin x}{x}} \right)} \\ & = \lim\limits_{x \to 0} \dfrac{x - \sin x}{\left( x^{3} \right) \left( 1 + \sqrt{\frac{\sin x}{x}} \right)} \\ & = \lim\limits_{x \to 0} \dfrac{x - \sin x}{x^{3}} \cdot \lim\limits_{x \to 0} \dfrac{1}{ 1 + \sqrt{\frac{\sin x}{x}}} \\ & = \dfrac{1}{6} \cdot \dfrac{1}{1+ \sqrt{1}} \\ & = \dfrac{1}{6} \cdot \dfrac{1}{2}=\dfrac{1}{12} \end{align}$


$\therefore$ Pilihan yang sesuai $(D)\ \dfrac{1}{12}$

65. Soal UM UGM 2019 Kode 624 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{\sqrt{1-\cos 4x^{2}}}{1-\cos 2x}=\cdots$

$\begin{align} (A)\ & 1 \\ (B)\ & \sqrt{2} \\ (C)\ & 2 \\ (D)\ & 3\sqrt{2} \\ (E)\ & 4 \end{align}$

Alternatif Pembahasan:
show

Dengan menggunakan beberapa identitas trigonometri antara lain $2sin^{2}\ a=1-\cos 2a$ dan teorema limit $\lim\limits_{x \to 0} \dfrac{\sin ax }{bx} = \dfrac{a}{b}$, kita coba selesaikan soal di atas seperti penjabaran berikut ini:

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

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

66. Soal SBMPTN 2016 Kode 249 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{\sqrt{x^{2}+1}-1}{\sqrt{3x^{5}+4 \sin^{4} x}}=\cdots$

$\begin{align} (A)\ & 0 \\ (B)\ & \dfrac{1}{4} \\ (C)\ & \dfrac{1}{\sqrt{7}} \\ (D)\ & \dfrac{1}{2} \\ (E)\ & \dfrac{1}{\sqrt{3}} \end{align}$

Alternatif Pembahasan:
show

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


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

67. Soal SBMPTN 2017 Kode 124 |*Soal Lengkap

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

$\begin{align} (A)\ & 8 \\ (B)\ & 7 \\ (C)\ & 6 \\ (D)\ & 5 \\ (E)\ & 2 \end{align}$

Alternatif Pembahasan:
show

$\begin{align} & \lim\limits_{x \to 0} \dfrac{4x + 3x \cos 2x}{\sin x\ \cos x} \\ & = \lim\limits_{x \to 0} \dfrac{4x + 3x \cos 2x}{\sin x\ \cos x} \cdot \dfrac{\frac{1}{x}}{\frac{1}{x}} \\ & = \lim\limits_{x \to 0} \dfrac{\frac{4x}{x} + \frac{3x \cos 2x}{x}}{\frac{\sin x\ \cos x}{x}} \\ & = \lim\limits_{x \to 0} \dfrac{ 4 + 3 \cos 2x }{\frac{\sin x}{x} \cdot \cos x} \\ & = \dfrac{ 4 + 3 \cos 2(0) }{1 \cdot \cos 0} \\ & = \dfrac{ 4 + 3}{1 \cdot 1} =7 \end{align}$


$\therefore$ Pilihan yang sesuai $(B)\ 7$

68. Soal SBMPTN 2018 Kode 408 |*Soal Lengkap

$\lim\limits_{x \to 3} \dfrac{\sin \left( 2x-6 \right) }{\sqrt{4-x}-1} =\cdots$

$\begin{align} (A)\ & 4 \\ (B)\ & 2 \\ (C)\ & 0 \\ (D)\ & -2 \\ (E)\ & -4 \end{align}$

Alternatif Pembahasan:
show

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


$\therefore$ Pilihan yang sesuai $(E)\ -4$

69. Soal SBMPTN 2018 Kode 419 |*Soal Lengkap

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

$\begin{align} (A)\ & -2\sqrt{\pi} \\ (B)\ & -\sqrt{\pi} \\ (C)\ & 0 \\ (D)\ & \sqrt{\pi} \\ (E)\ & 2\sqrt{\pi} \end{align}$

Alternatif Pembahasan:
show

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


$\therefore$ Pilihan yang sesuai $(D)\ \sqrt{\pi}$

70. Soal SBMPTN 2018 Kode 422 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{\sin (x)}{\sqrt{\pi + \tan \left( x \right)}-\sqrt{\pi - \tan \left( x \right)}} =\cdots$

$\begin{align} (A)\ & -2\sqrt{\pi} \\ (B)\ & -\sqrt{\pi} \\ (C)\ & 0 \\ (D)\ & \sqrt{\pi} \\ (E)\ & 2\sqrt{\pi} \end{align}$

Alternatif Pembahasan:
show

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


$\therefore$ Pilihan yang sesuai $(D)\ \sqrt{\pi}$


71. Soal SPMB 2002 (Regional I) |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{x^{2}+\sin x\ \tan x}{1-\cos 2x}=\cdots$

$\begin{align} (A)\ & 0 \\ (B)\ & \dfrac{1}{2} \\ (C)\ & 1 \\ (D)\ & 2 \\ (E)\ & 4 \end{align}$

Alternatif Pembahasan:
show

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


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

72. Soal SPMB 2003 (Regional I) |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{1-\cos^{2} x - \cos x\ \sin^{2} x}{x^{4}}=\cdots$

$\begin{align} (A)\ & 0 \\ (B)\ & \dfrac{1}{4} \\ (C)\ & \dfrac{1}{2} \\ (D)\ & 1 \\ (E)\ & -1 \end{align}$

Alternatif Pembahasan:
show

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


$\therefore$ Pilihan yang sesuai $(C)\ \dfrac{1}{2}$

73. Soal UMPTN 1995 (Rayon B) |*Soal Lengkap

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

$\begin{align} (A)\ & -3 \\ (B)\ & -2 \\ (C)\ & 2 \\ (D)\ & 3 \\ (E)\ & 5 \end{align}$

Alternatif Pembahasan:
show

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


$\therefore$ Pilihan yang sesuai $(A)\ -3$

74. Soal UMPTN 1995 (Rayon C) |*Soal Lengkap

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

$\begin{align} (A)\ & 0 \\ (B)\ & \dfrac{1}{4} \\ (C)\ & \dfrac{1}{2} \\ (D)\ & 2 \\ (E)\ & 4 \end{align}$

Alternatif Pembahasan:
show

Sedikit catatan tentang identitas trigonometri untuk menyelesaikan limit trigonometri di atas yaitu:

  • $\cos 2a = \cos^{2}a-\sin^{2}a$
  • $\cos 2a = 1-2\sin^{2}a$
  • $\cos a = 1-2\sin^{2} \frac{1}{2}a$
  • $\cos f(x) = 1-2\sin^{2} \frac{1}{2}f(x)$
  • $\cos (x+a) = 1-2\sin^{2} \frac{1}{2}(x+a)$

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


$\therefore$ Pilihan yang sesuai $(C)\ \dfrac{1}{2}$

75. Soal SIMAK UI 2009 Kode 954 |*Soal Lengkap

$\lim\limits_{\theta \to \frac{\pi}{2}} \dfrac{\sec^{2} \theta}{\sec^{2} 5\theta}=\cdots$

$\begin{align} (A)\ & 15 \\ (B)\ & 25 \\ (C)\ & 30 \\ (D)\ & 35 \\ (E)\ & 40 \end{align}$

Alternatif Pembahasan:
show

Untuk menyelesaikan soal limit trigonometri di atas, seperti kita sampaikan sebelumnya beberapa Identitas Trigonometri Dasar kita coba dengan memisalkan $\theta=\frac{\pi}{2}+x$.

Untuk $\theta=\frac{\pi}{2}+x$ dan $\theta \to \frac{\pi}{2}$ maka $x \to 0$ sehingga soal limit trigonometri di atas dapat kita tuliskan menjadi seperti berikut ini:

$\begin{align}
& \lim\limits_{\theta \to \frac{\pi}{2}} \dfrac{\sec^{2} \theta}{\sec^{2} 5\theta} \\ & = \lim\limits_{x \to 0} \dfrac{\sec^{2} \left( \frac{\pi}{2}+x \right)}{\sec^{2} 5\left( \frac{\pi}{2}+x \right)} \\ & = \lim\limits_{x \to 0} \dfrac{\csc^{2} x}{\csc^{2} 5x} \\ & = \lim\limits_{x \to 0} \dfrac{\frac{1}{\sin^{2} x}}{\frac{1}{\sin^{2} 5x}} \\ & = \lim\limits_{x \to 0} \dfrac{\sin^{2} 5x}{\sin^{2} x} \\ & = \lim\limits_{x \to 0} \dfrac{\sin 5x \cdot \sin 5x}{\sin x \cdot \sin x} \\ & = \dfrac{5 \cdot 5 }{1 \cdot 1} = 25 \end{align}$

$\therefore$ Pilihan yang sesuai $(D)\ 25$

76. Soal SIMAK UI 2009 Kode 964 |*Soal Lengkap

$\lim\limits_{x \to 0} \left( cosec\ x - \dfrac{1}{x} \right)=\cdots$

$\begin{align} (A)\ & \infty \\ (B)\ & 2 \\ (C)\ & 0 \\ (D)\ & -2 \\ (E)\ & -\infty \end{align}$

Alternatif Pembahasan:
show

Untuk menyelesaikan limit fungsi di atas kita gunakan Aturan L'Hospital atau pakai turunan fungsi. Jika pembaca mempunyai ide untuk menyelesaikan soal ini tanpa menggunakan Aturan L'Hospital, dapat menuliskan pada kotak komentar.

$\begin{align}
& \lim\limits_{x \to 0} \left( \csc x - \dfrac{1}{x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{1}{\sin x} - \dfrac{1}{x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{x-\sin x}{x\ \sin x} \right) \\ \hline & Aturan\ L'Hospital \\ \hline & = \lim\limits_{x \to 0} \left( \dfrac{1-\cos x}{1 \cdot \sin x+x \cdot \cos x} \right) \\ & = \lim\limits_{x \to 0} \left( \dfrac{1-\cos x}{\sin x+x \cdot \cos x} \right) \\ \hline & Aturan\ L'Hospital \\ \hline & = \lim\limits_{x \to 0} \dfrac{\sin x}{2 \cos x-x \sin x} \\ & = \dfrac{\sin 0}{2 \cos 0-0 \sin 0} \\ & = \dfrac{0}{2 \cdot 1 -0}= \dfrac{0}{2-0}= 0 \end{align}$

$\therefore$ Pilihan yang sesuai $(C)\ 0$

77. Soal SIMAK UI 2010 Kode 507 |*Soal Lengkap

Jika diketahui $\lim\limits_{x \to 0} \dfrac{ax \sin x + b}{\cos x -1}=1$, maka nilai $a$ dan $b$ yang memenuhi adalah...

$\begin{align} (A)\ & a=-\dfrac{1}{2}, b=0 \\ (B)\ & a=1, b=1 \\ (C)\ & a=\dfrac{1}{2}, b=1 \\ (D)\ & a=1, b=-1 \\ (E)\ & a=1, b=0 \end{align}$

Alternatif Pembahasan:
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Nilai $\lim\limits_{x \to 0} \dfrac{ax \sin x + b}{\cos x -1}=1$ sehingga jika kita substitusi langsung nilai $x=0$ maka nilai $ax \sin x + b$ harus $0$, karena jika $ax \sin x + b$ tidak nol maka nilai limit adalah $\infty$.


Karena nilai $ax \sin x + b$ untuk $x=0$ adalah $0$ maka dapat kita tuliskan:
$\begin{align}
ax \sin x + b &= 0 \\ a(0) \sin 0 + b &= 0 \\ b &= 0 \end{align}$


Dari hasil di atas kita peroleh:
$\begin{align}
\lim\limits_{x \to 0} \dfrac{ax \sin x + b}{\cos x -1} &= 1 \\ \lim\limits_{x \to 0} \dfrac{ax \sin x}{\cos x -1} &= 1 \\ \lim\limits_{x \to 0} \dfrac{ax \sin x}{-2 \sin^{2} \frac{1}{2}x} &= 1 \\ \lim\limits_{x \to 0} \left( \dfrac{ax}{-2 \sin \frac{1}{2}x} \cdot \dfrac{\sin x}{\sin \frac{1}{2}x} \right)&= 1 \\ \dfrac{a}{-2 \cdot \frac{1}{2}} \cdot \dfrac{1}{\frac{1}{2}} &= 1 \\ \dfrac{a}{-1} \cdot 2 &= 1 \\ a &= -\dfrac{1}{2} \end{align}$


$\therefore$ Pilihan yang sesuai $(A)\ a=-\dfrac{1}{2}, b=0$

78. Soal UMB 2011 Kode 252 |*Soal Lengkap

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

$\begin{align} (A)\ & 0 \\ (B)\ & \dfrac{1}{3} \\ (C)\ & \frac{1}{2} \\ (D)\ & 1 \\ (E)\ & 2\frac{1}{2} \end{align}$

Alternatif Pembahasan:
show

$\begin{align} & \lim\limits_{x \to 0} \dfrac{x^{2} + \sin^{2} 3x}{2 \tan \left(2x^{2} \right)} \\ & = \lim\limits_{x \to 0} \dfrac{x^{2} + \sin^{2} 3x}{2 \tan \left(2x^{2} \right)} \cdot \dfrac{\frac{1}{x^{2}}}{\frac{1}{x^{2}}} \\ & = \lim\limits_{x \to 0} \dfrac{\frac{x^{2}}{x^{2}}+ \frac{\sin^{2} 3x}{x^{2}}}{\frac{2 \tan \left(2x^{2} \right)}{x^{2}}} \\ & = \dfrac{1+ \frac{3 \cdot 3}{1}}{\frac{2 \cdot 2 }{1}} \\ & = \dfrac{10}{4}=2\frac{1}{2} \end{align}$


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

79. Soal SIMAK UI 2011 Kode 511 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{\tan a - \tan b}{1+\left(1-\frac{a}{b} \right) \tan a\ \tan b - \frac{a}{b}}=\cdots$

$\begin{align} (A)\ & \dfrac{1}{b} \\ (B)\ & b \\ (C)\ & -b \\ (D)\ & \dfrac{-1}{b} \\ (E)\ & 1 \end{align}$

Alternatif Pembahasan:
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Sedikit catatan tentang identitas trigonometri yang mungkin sangat membantu dalam menyelesaikan soal di atas yaitu $\tan \left( a-b \right)=\dfrac{\tan a - \tan b}{1+ \tan a \cdot \tan b}$.


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


$\therefore$ Pilihan yang sesuai $(C)\ -b$

80. Soal SBMPTN 2013 Kode 332 |*Soal Lengkap

$\lim\limits_{x \to 0} \dfrac{\sin^{2}x - \cos x +1}{x \tan x}=\cdots$

$\begin{align} (A)\ & \dfrac{3}{2} \\ (B)\ & \dfrac{1}{2} \\ (C)\ & -\dfrac{1}{2} \\ (D)\ & -1 \\ (E)\ & -2 \end{align}$

Alternatif Pembahasan:
show

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


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

Jika engkau tidak sanggup menahan lelahnya belajar, Maka engkau harus menanggung pahitnya kebodohan ___pythagoras

Beberapa pembahasan soal Matematika Dasar Limit Fungsi Trigonometri di atas adalah coretan kreatif siswa pada:

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