Pembahasan Contoh Soal Materi Limit Fungsi Trigonometri #4
Berikut ini mimin sajikan beberapa contoh soal dan pembahasan pada materi limit fungsi trigonometri. Selamat membaca, sobat. Semoga bermanfaat.
Contoh soal 1
Nilai dari $\lim\limits_{x \to \frac{\pi}{4}} \dfrac{2 \sin x+\cos 2x}{\cos x}$ adalah ...
Jawab:
$\begin{aligned} &\lim\limits_{x \to \frac{\pi}{4}} \dfrac{2 \sin x+\cos 2x}{\cos x} \\&= \dfrac{2 \sin \frac{\pi}{4}+\cos (\frac{\pi}{2})}{\cos \frac{\pi}{4}} \\&= \dfrac{2 (\frac{1}{2} \sqrt{2})+0}{\frac{1}{2} \sqrt{2}} \\&= 2 \end{aligned}$
Contoh soal 2
Nilai dari $\lim\limits_{x \to \frac{\pi}{4}} \dfrac{1- \tan x}{\sin x-\cos x}$ adalah ...
Jawab:
$\begin{aligned} &\lim\limits_{x \to \frac{\pi}{4}} \dfrac{1- \tan x}{\sin x-\cos x}\\&= \lim\limits_{x \to \frac{\pi}{4}} \dfrac{1- \frac{\sin x}{\cos x}}{\sin x-\cos x}\\&= \lim\limits_{x \to \frac{\pi}{4}} \dfrac{\frac{\cos x}{\cos x}- \frac{\sin x}{\cos x}}{\sin x-\cos x}\\&= \lim\limits_{x \to \frac{\pi}{4}} \dfrac{\cos x -\sin x}{(\sin x-\cos x)\cos x}\\&= \lim\limits_{x \to \frac{\pi}{4}} \dfrac{-(\sin x-\cos x) }{(\sin x-\cos x)\cos x}\\&= \lim\limits_{x \to \frac{\pi}{4}} \frac{-1}{\cos x}\\&= \frac{-1}{\cos \frac{\pi}{4}}\\&= \frac{-1}{ \frac{1}{2} \sqrt{2}}\\&= - \sqrt{2} \end{aligned}$
Contoh soal 3
Nilai dari $\lim\limits_{x \to 0} \dfrac{ \tan^{4} 3x}{ (x \sin 2x)^2}$ adalah ...
Jawab:
$\begin{aligned} &\lim\limits_{x \to 0} \dfrac{ \tan^{4} 3x}{ (x \sin 2x)^2}\\&= \lim\limits_{x \to 0} \dfrac{ \tan 3x \times \tan 3x \times \tan 3x \times \tan 3x}{ x \times x \times \sin 2x \times \sin 2x} \\&= \frac{\tan 3x}{x} \times \frac{\tan 3x}{x} \times \frac{\tan 3x}{\sin 2x} \times \frac{\tan 3x}{\sin 2x} \\&= \frac{3}{1} \times \frac{3}{1} \times \frac{3}{2} \times \frac{3}{2}\\&= \frac{81}{4} \end{aligned}$
Contoh soal 4
Nilai dari $\lim\limits_{x \to 0} \dfrac{ \cos 3x-\cos5x}{ 1-\cos 4x}$ adalah ...
Jawab:
$\begin{aligned} &\lim\limits_{x \to 0} \dfrac{ \cos 3x-\cos5x}{ 1-\cos 4x}\\&= \lim\limits_{x \to 0} \dfrac{ -2 \sin \frac{1}{2} (3x+5x) \sin \frac{1}{2} (3x-5x) }{ 1-(1-2 \sin^2 2x)}\\&= \lim\limits_{x \to 0} \dfrac{ -2 \sin 4x \sin (-x) }{ 2 \sin^2 2x}\\&= \lim\limits_{x \to 0} \dfrac{ \sin 4x \sin x }{ \sin^2 2x} \\&= \lim\limits_{x \to 0} \frac{\sin 4x}{\sin 2x} \times \frac{\sin x}{\sin 2x} \\&= \frac{4}{2} \times \frac{1}{2}\\&= 1 \end{aligned}$
Contoh soal 5
Nilai dari $\lim\limits_{x \to 0} \dfrac{ \left ( \frac{5x}{4} \right )^3}{(1- \cos^2 \frac{5}{4}x ) \tan 8x}$ adalah ...
Jawab:
$\begin{aligned} &\lim\limits_{x \to 0} \dfrac{ \left ( \frac{5x}{4} \right )^3}{(1- \cos^2 \frac{5}{4}x ) \tan 8x}\\&= \lim\limits_{x \to 0} \dfrac{ \left ( \frac{5x}{4} \right )^3}{\sin^2 \frac{5}{4}x \tan 8x} \\&=\lim\limits_{x \to 0} \frac{ \frac{5x}{4}}{\sin \frac{5}{4}x} \times \frac{ \frac{5x}{4}}{\sin \frac{5}{4}x} \times \frac{ \frac{5x}{4}}{\tan 8x} \\&= \frac{\frac{5}{4}}{\frac{5}{4}} \times \frac{\frac{5}{4}}{\frac{5}{4}} \times \frac{\frac{5}{4}}{8}\\&= \frac{5}{32} \end{aligned}$
Contoh soal 6
Nilai dari $\lim\limits_{x \to \frac{1}{2}} \dfrac{(4x-2) \sin(x-\frac{1}{2})}{\tan^2 (2x-1)}$ adalah ...
Jawab:
$\begin{aligned} &\lim\limits_{x \to \frac{1}{2}} \dfrac{(4x-2) \sin(x-\frac{1}{2})}{\tan^2 (2x-1)} \\&= \lim\limits_{x \to \frac{1}{2}} \dfrac{4(x-\frac{1}{2}) \sin(x-\frac{1}{2})}{\tan 2(x-\frac{1}{2}) \tan 2(x-\frac{1}{2})} \times \frac{2(x-\frac{1}{2})}{2(x-\frac{1}{2})}\\&= \lim\limits_{x \to \frac{1}{2}} \left ( \frac{2}{2} \times \frac{2(x-\frac{1}{2})}{ \tan 2(x-\frac{1}{2})} \times \frac{\sin (x-\frac{1}{2})}{ x-\frac{1}{2}} \times \frac{2(x-\frac{1}{2})}{ \tan 2(x-\frac{1}{2})} \right ) \\&= \frac{2}{2} \lim\limits_{(x-\frac{1}{2}) \to 0} \left ( \frac{2(x-\frac{1}{2})}{ \tan 2(x-\frac{1}{2})} \times \frac{\sin (x-\frac{1}{2})}{ x-\frac{1}{2}} \times \frac{2(x-\frac{1}{2})}{ \tan 2(x-\frac{1}{2})} \right ) \\&= 1 \end{aligned}$
Contoh soal 7
Nilai dari $\lim\limits_{x \to \frac{\pi}{4}} \dfrac{1-\sqrt{2} \sin x}{\sin x \cos 2x}$ adalah ...
Jawab:
$\begin{aligned} &\lim\limits_{x \to \frac{\pi}{4}} \dfrac{1-\sqrt{2} \sin x}{\sin x \cos 2x}\\&= \lim\limits_{x \to \frac{\pi}{4}} \dfrac{1-\sqrt{2} \sin x}{\sin x \cos 2x} \times \frac{1+\sqrt{2} \sin x}{1+\sqrt{2} \sin x} \\&= \lim\limits_{x \to \frac{\pi}{4}} \dfrac{1-2 \sin^2x}{\sin x \cos 2x (1+\sqrt{2} \sin x)} \\&= \lim\limits_{x \to \frac{\pi}{4}} \dfrac{\cos2x}{\sin x \cos 2x (1+\sqrt{2} \sin x)} \\&= \lim\limits_{x \to \frac{\pi}{4}} \dfrac{1}{\sin x (1+\sqrt{2} \sin x)}\\&= \frac{1}{\sin \frac{\pi}{4} (1+\sqrt{2} \sin \frac{\pi}{4})}\\&= \frac{1}{\frac{1}{2} \sqrt{2} (1+\sqrt{2} \times \frac{1}{2} \sqrt{2})} \\&= \frac{1}{\frac{1}{2} \sqrt{2} \times 2}\\&=\frac{1}{2} \sqrt{2} \end{aligned}$
Contoh soal 8
Nilai dari $\lim\limits_{x \to \frac{3\pi}{4}} \dfrac{ \sin x + \cos x}{1+\tan x}$ adalah ...
Jawab:
$\begin{aligned} &\lim\limits_{x \to \frac{3\pi}{4}} \dfrac{ \sin x + \cos x}{1+\tan x}\\&= \lim\limits_{x \to \frac{3\pi}{4}} \dfrac{ \sin x + \cos x}{1+\frac{\sin x}{\cos x}}\\&= \lim\limits_{x \to \frac{3\pi}{4}} \dfrac{ \sin x + \cos x}{\frac{\cos x+\sin x}{\cos x}} \\&= \lim\limits_{x \to \frac{3\pi}{4}} (\sin x + \cos x) \times \frac{\cos x}{\sin x+\cos x}\\&= \lim\limits_{x \to \frac{3\pi}{4}} \cos x\\&= \cos \frac{3\pi}{4}\\&= -\frac{1}{2} \sqrt{2} \end{aligned}$
Contoh soal 9
Nilai dari $\lim\limits_{x \to 0} \left (\frac{1-\cos^3 x}{\sin^2 x} \right )$ adalah ...
Jawab:
$\begin{aligned} &\lim\limits_{x \to 0} \left (\frac{1-\cos^3 x}{\sin^2 x} \right )\\&= \lim\limits_{x \to 0} \left (\frac{1-\cos^3 x}{1-\cos^2 x} \right ) \\&= \lim\limits_{x \to 0} \left (\frac{(1-\cos x)(1+\cos x+ \cos^2 x)}{(1-\cos x)(1+\cos x)} \right )\\&= \lim\limits_{x \to 0} \left (\frac{1+\cos x+ \cos^2 x}{1+\cos x} \right )\\&= \frac{1+\cos 0+ \cos^2 0}{1+\cos 0}\\&= \frac{1+1+1}{1+1}\\&= \frac{3}{2} \end{aligned}$
Contoh soal 10
Nilai dari $\lim\limits_{x \to 0} \frac{\sin 6x- \sin 2x}{\sin 8x}$ adalah ...
Jawab:
$\begin{aligned} &\lim\limits_{x \to 0} \frac{\sin 6x- \sin 2x}{\sin 8x}\\&= \lim\limits_{x \to 0} \frac{2 \cos \frac{1}{2}(6x+2x) \sin \frac{1}{2}(6x-2x)}{\sin 2 \times 4x}\\&= \lim\limits_{x \to 0} \frac{2 \cos 4x \sin 2x}{2 \sin 4x \cos 4x}\\&= \lim\limits_{x \to 0} \frac{ \sin 2x}{ \sin 4x }\\&= \frac{2}{4}\\&= \frac{1}{2} \end{aligned}$
Demikianlah beberapa contoh soal dan pembahasan pada materi limit fungsi trigonomteri. Semoga bermanfaat.
