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 $\underset{x\to \frac{\pi }{4}}{lim}{\displaystyle \frac{2\sin x+\cos 2x}{\cos x}}$ adalah ...

Jawab:

$\begin{array}{rl}& \underset{x\to \frac{\pi }{4}}{lim}{\displaystyle \frac{2\sin x+\cos 2x}{\cos x}}\\ & ={\displaystyle \frac{2\sin \frac{\pi }{4}+\cos (\frac{\pi }{2})}{\cos \frac{\pi }{4}}}\\ & ={\displaystyle \frac{2(\frac{1}{2}\sqrt{2})+0}{\frac{1}{2}\sqrt{2}}}\\ & =2\end{array}$

Contoh soal 2

Nilai dari $\underset{x\to \frac{\pi }{4}}{lim}{\displaystyle \frac{1-\tan x}{\sin x-\cos x}}$ adalah ...

Jawab:

$\begin{array}{rl}& \underset{x\to \frac{\pi }{4}}{lim}{\displaystyle \frac{1-\tan x}{\sin x-\cos x}}\\ & =\underset{x\to \frac{\pi }{4}}{lim}{\displaystyle \frac{1-\frac{\sin x}{\cos x}}{\sin x-\cos x}}\\ & =\underset{x\to \frac{\pi }{4}}{lim}{\displaystyle \frac{\frac{\cos x}{\cos x}-\frac{\sin x}{\cos x}}{\sin x-\cos x}}\\ & =\underset{x\to \frac{\pi }{4}}{lim}{\displaystyle \frac{\cos x-\sin x}{(\sin x-\cos x)\cos x}}\\ & =\underset{x\to \frac{\pi }{4}}{lim}{\displaystyle \frac{-(\sin x-\cos x)}{(\sin x-\cos x)\cos x}}\\ & =\underset{x\to \frac{\pi }{4}}{lim}\frac{-1}{\cos x}\\ & =\frac{-1}{\cos \frac{\pi }{4}}\\ & =\frac{-1}{\frac{1}{2}\sqrt{2}}\\ & =-\sqrt{2}\end{array}$

Contoh soal 3

Nilai dari $\underset{x\to 0}{lim}{\displaystyle \frac{{\tan }^{4}3x}{(x\sin 2x{)}^2}}$ adalah ...

Jawab:

$\begin{array}{rl}& \underset{x\to 0}{lim}{\displaystyle \frac{{\tan }^{4}3x}{(x\sin 2x{)}^2}}\\ & =\underset{x\to 0}{lim}{\displaystyle \frac{\tan 3x\times \tan 3x\times \tan 3x\times \tan 3x}{x\times x\times \sin 2x\times \sin 2x}}\\ & =\frac{\tan 3x}{x}\times \frac{\tan 3x}{x}\times \frac{\tan 3x}{\sin 2x}\times \frac{\tan 3x}{\sin 2x}\\ & =\frac{3}{1}\times \frac{3}{1}\times \frac{3}{2}\times \frac{3}{2}\\ & =\frac{81}{4}\end{array}$

Contoh soal 4

Nilai dari $\underset{x\to 0}{lim}{\displaystyle \frac{\cos 3x-\cos 5x}{1-\cos 4x}}$ adalah ...

Jawab:

$\begin{array}{rl}& \underset{x\to 0}{lim}{\displaystyle \frac{\cos 3x-\cos 5x}{1-\cos 4x}}\\ & =\underset{x\to 0}{lim}{\displaystyle \frac{-2\sin \frac{1}{2}(3x+5x)\sin \frac{1}{2}(3x-5x)}{1-(1-2{\sin }^{2}2x)}}\\ & =\underset{x\to 0}{lim}{\displaystyle \frac{-2\sin 4x\sin (-x)}{2{\sin }^{2}2x}}\\ & =\underset{x\to 0}{lim}{\displaystyle \frac{\sin 4x\sin x}{{\sin }^{2}2x}}\\ & =\underset{x\to 0}{lim}\frac{\sin 4x}{\sin 2x}\times \frac{\sin x}{\sin 2x}\\ & =\frac{4}{2}\times \frac{1}{2}\\ & =1\end{array}$

Contoh soal 5

Nilai dari $\underset{x\to 0}{lim}{\displaystyle \frac{{\left(\frac{5x}{4}\right)}^3}{(1-{\cos }^2\frac{5}{4}x)\tan 8x}}$ adalah ...

Jawab:

$\begin{array}{rl}& \underset{x\to 0}{lim}{\displaystyle \frac{{\left(\frac{5x}{4}\right)}^3}{(1-{\cos }^2\frac{5}{4}x)\tan 8x}}\\ & =\underset{x\to 0}{lim}{\displaystyle \frac{{\left(\frac{5x}{4}\right)}^3}{{\sin }^2\frac{5}{4}x\tan 8x}}\\ & =\underset{x\to 0}{lim}\frac{\frac{5x}{4}}{\sin \frac{5}{4}x}\times \frac{\frac{5x}{4}}{\sin \frac{5}{4}x}\times \frac{\frac{5x}{4}}{\tan 8x}\\ & =\frac{\frac{5}{4}}{\frac{5}{4}}\times \frac{\frac{5}{4}}{\frac{5}{4}}\times \frac{\frac{5}{4}}{8}\\ & =\frac{5}{32}\end{array}$

Contoh soal 6

Nilai dari $\underset{x\to \frac{1}{2}}{lim}{\displaystyle \frac{(4x-2)\sin (x-\frac{1}{2})}{{\tan }^2(2x-1)}}$ adalah ...

Jawab:

$\begin{array}{rl}& \underset{x\to \frac{1}{2}}{lim}{\displaystyle \frac{(4x-2)\sin (x-\frac{1}{2})}{{\tan }^2(2x-1)}}\\ & =\underset{x\to \frac{1}{2}}{lim}{\displaystyle \frac{4(x-\frac{1}{2})\sin (x-\frac{1}{2})}{\tan 2(x-\frac{1}{2})\tan 2(x-\frac{1}{2})}}\times \frac{2(x-\frac{1}{2})}{2(x-\frac{1}{2})}\\ & =\underset{x\to \frac{1}{2}}{lim}(\frac{2}{2}\times \frac{2(x-\frac{1}{2})}{\tan 2(x-\frac{1}{2})}\times \frac{\sin (x-\frac{1}{2})}{x-\frac{1}{2}}\times \frac{2(x-\frac{1}{2})}{\tan 2(x-\frac{1}{2})})\\ & =\frac{2}{2}\underset{(x-\frac{1}{2})\to 0}{lim}(\frac{2(x-\frac{1}{2})}{\tan 2(x-\frac{1}{2})}\times \frac{\sin (x-\frac{1}{2})}{x-\frac{1}{2}}\times \frac{2(x-\frac{1}{2})}{\tan 2(x-\frac{1}{2})})\\ & =1\end{array}$

Contoh soal 7

Nilai dari $\underset{x\to \frac{\pi }{4}}{lim}{\displaystyle \frac{1-\sqrt{2}\sin x}{\sin x\cos 2x}}$ adalah ...

Jawab:

$\begin{array}{rl}& \underset{x\to \frac{\pi }{4}}{lim}{\displaystyle \frac{1-\sqrt{2}\sin x}{\sin x\cos 2x}}\\ & =\underset{x\to \frac{\pi }{4}}{lim}{\displaystyle \frac{1-\sqrt{2}\sin x}{\sin x\cos 2x}}\times \frac{1+\sqrt{2}\sin x}{1+\sqrt{2}\sin x}\\ & =\underset{x\to \frac{\pi }{4}}{lim}{\displaystyle \frac{1-2{\sin }^{2}x}{\sin x\cos 2x(1+\sqrt{2}\sin x)}}\\ & =\underset{x\to \frac{\pi }{4}}{lim}{\displaystyle \frac{\cos 2x}{\sin x\cos 2x(1+\sqrt{2}\sin x)}}\\ & =\underset{x\to \frac{\pi }{4}}{lim}{\displaystyle \frac{1}{\sin x(1+\sqrt{2}\sin x)}}\\ & =\frac{1}{\sin \frac{\pi }{4}(1+\sqrt{2}\sin \frac{\pi }{4})}\\ & =\frac{1}{\frac{1}{2}\sqrt{2}(1+\sqrt{2}\times \frac{1}{2}\sqrt{2})}\\ & =\frac{1}{\frac{1}{2}\sqrt{2}\times 2}\\ & =\frac{1}{2}\sqrt{2}\end{array}$

Contoh soal 8

Nilai dari $\underset{x\to \frac{3\pi }{4}}{lim}{\displaystyle \frac{\sin x+\cos x}{1+\tan x}}$ adalah ...

Jawab:

$\begin{array}{rl}& \underset{x\to \frac{3\pi }{4}}{lim}{\displaystyle \frac{\sin x+\cos x}{1+\tan x}}\\ & =\underset{x\to \frac{3\pi }{4}}{lim}{\displaystyle \frac{\sin x+\cos x}{1+\frac{\sin x}{\cos x}}}\\ & =\underset{x\to \frac{3\pi }{4}}{lim}{\displaystyle \frac{\sin x+\cos x}{\frac{\cos x+\sin x}{\cos x}}}\\ & =\underset{x\to \frac{3\pi }{4}}{lim}(\sin x+\cos x)\times \frac{\cos x}{\sin x+\cos x}\\ & =\underset{x\to \frac{3\pi }{4}}{lim}\cos x\\ & =\cos \frac{3\pi }{4}\\ & =-\frac{1}{2}\sqrt{2}\end{array}$

Contoh soal 9

Nilai dari $\underset{x\to 0}{lim}\left(\frac{1-{\cos }^{3}x}{{\sin }^{2}x}\right)$ adalah ...

Jawab:

$\begin{array}{rl}& \underset{x\to 0}{lim}\left(\frac{1-{\cos }^{3}x}{{\sin }^{2}x}\right)\\ & =\underset{x\to 0}{lim}\left(\frac{1-{\cos }^{3}x}{1-{\cos }^{2}x}\right)\\ & =\underset{x\to 0}{lim}\left(\frac{(1-\cos x)(1+\cos x+{\cos }^{2}x)}{(1-\cos x)(1+\cos x)}\right)\\ & =\underset{x\to 0}{lim}\left(\frac{1+\cos x+{\cos }^{2}x}{1+\cos x}\right)\\ & =\frac{1+\cos 0+{\cos }^{2}0}{1+\cos 0}\\ & =\frac{1+1+1}{1+1}\\ & =\frac{3}{2}\end{array}$

Contoh soal 10

Nilai dari $\underset{x\to 0}{lim}\frac{\sin 6x-\sin 2x}{\sin 8x}$ adalah ...

Jawab:

$\begin{array}{rl}& \underset{x\to 0}{lim}\frac{\sin 6x-\sin 2x}{\sin 8x}\\ & =\underset{x\to 0}{lim}\frac{2\cos \frac{1}{2}(6x+2x)\sin \frac{1}{2}(6x-2x)}{\sin 2\times 4x}\\ & =\underset{x\to 0}{lim}\frac{2\cos 4x\sin 2x}{2\sin 4x\cos 4x}\\ & =\underset{x\to 0}{lim}\frac{\sin 2x}{\sin 4x}\\ & =\frac{2}{4}\\ & =\frac{1}{2}\end{array}$

Demikianlah beberapa contoh soal dan pembahasan pada materi limit fungsi trigonomteri. Semoga bermanfaat.