Pembahasan Contoh Soal Materi Limit Fungsi Trigonometri #3

Berikut ini mimin sajikan beberapa contoh soal dan pembahasan pada materi limit fungsi trigonometri.  Selamat membaca, sobat. Semoga bermanfaat. 

Contoh soal 1

Nilai dari $\underset{x\to 0}{lim}{\displaystyle \frac{3x\tan 4x}{1-{\cos }^{2}x}}$ adalah ...

Jawab:

$\begin{array}{rl}& \underset{x\to 0}{lim}{\displaystyle \frac{3x\tan 4x}{1-{\cos }^{2}x}}\\ & =\underset{x\to 0}{lim}{\displaystyle \frac{3x\tan 4x}{{\sin }^{2}x}}\\ & =\underset{x\to 0}{lim}{\displaystyle \frac{3x}{\sin x}}\times \frac{\tan 4x}{\sin x}\\ & =\frac{3}{1}\times \frac{4}{1}\\ & =12\end{array}$

Contoh soal 2

Nilai dari $\underset{x\to 0}{lim}{\displaystyle \frac{1-{\cos }^{2}2x}{-x\sin 2x}}$ adalah ...

Jawab:

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

Contoh soal 3

Nilai dari $\underset{x\to 0}{lim}{\displaystyle \frac{1-\cos x}{4x\sin 2x}}$ adalah ...

Jawab:

$\begin{array}{rl}& \underset{x\to 0}{lim}{\displaystyle \frac{1-\cos x}{4x\sin 2x}}\\ & =\underset{x\to 0}{lim}{\displaystyle \frac{1-(1-2{\sin }^2\frac{1}{2}x)}{4x\sin 2x}}\\ & =\underset{x\to 0}{lim}{\displaystyle \frac{2{\sin }^2\frac{1}{2}x}{4x\sin 2x}}\\ & =\underset{x\to 0}{lim}2{\displaystyle \frac{\sin \frac{1}{2}x}{4x}}\times \frac{\sin \frac{1}{2}x}{\sin 2x}\\ & =2\times \frac{1}{8}\times \frac{1}{4}\\ & =\frac{1}{16}\end{array}$

Contoh soal 4

Nilai dari $\underset{x\to 0}{lim}{\displaystyle \frac{\sin x+\sin 3x}{x\cos x}}$ adalah ...

Jawab:

Ingat rumus:

$\sin A+\sin B=2\sin \frac{1}{2}(A+B)\cos \frac{1}{2}(A-B)$ dan

$\cos (-x)=cosx$

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

Contoh soal 5

Nilai dari $\underset{x\to 0}{lim}{\displaystyle \frac{2x\tan 4x}{1-{\cos }^{2}2x}}$ adalah ...

Jawab:

Ingat rumus:

$1-{\cos }^{2}A={\sin }^{2}A$

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

Contoh soal 6

Nilai dari $\underset{x\to 0}{lim}{\displaystyle \frac{1-{\cos }^{2}x}{x^2\cot (x+\frac{\pi }{4})}}$ adalah ...

Jawab:

Ingat rumus:

$1-{\cos }^{2}A={\sin }^{2}A$

$\begin{array}{rl}& \underset{x\to 0}{lim}{\displaystyle \frac{1-{\cos }^{2}x}{x^2\cot (x+\frac{\pi }{4})}}\\ & =\underset{x\to 0}{lim}{\displaystyle \frac{{\sin }^{2}x}{x^2\cot (x+\frac{\pi }{4})}}\\ & =\underset{x\to 0}{lim}{\displaystyle \frac{{\sin }^{2}x.\tan (x+\frac{\pi }{4})}{x^2}}\\ & =\underset{x\to 0}{lim}{\displaystyle \frac{{\sin }^{2}x}{x^2}}\times \tan (x+\frac{\pi }{4})\\ & ={\left(\frac{1}{1}\right)}^2\times \tan \left(\frac{\pi }{4}\right)\\ & =1\times 1\\ & =1\end{array}$

Contoh soal 7

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

Jawab:

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

Contoh soal 8

Nilai dari $\underset{x\to 0}{lim}{\displaystyle \frac{\cot 2x}{\cot 4x}}$ adalah ...

Jawab:

$\begin{array}{rl}& \underset{x\to 0}{lim}{\displaystyle \frac{\cot 2x}{\cot 4x}}\\ & =\underset{x\to 0}{lim}{\displaystyle \frac{\frac{1}{\tan 2x}}{\frac{1}{\tan 4x}}}\\ & =\underset{x\to 0}{lim}{\displaystyle \frac{\tan 4x}{\tan 2x}}\\ & =\frac{4}{2}\\ & =2\end{array}$ 

Contoh soal 9

Nilai dari $\underset{x\to 0}{lim}{\displaystyle \frac{2x+3x\cos 2x}{\sin x\cos x}}$ adalah ...

Jawab:

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

Contoh soal 10

Nilai dari $\underset{x\to -3}{lim}{\displaystyle \frac{1-\cos (x+3)}{(x+3{)}^2}}$ adalah ...

Jawab:

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

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