Pembahasan Contoh Soal Materi Limit Fungsi Trigonometri #5

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

Contoh soal 1

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

Jawab:

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

Contoh soal 2

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

Jawab:

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

Contoh soal 3

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

Jawab:

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

Contoh soal 4

Nilai dari $\underset{x\to \frac{\pi }{4}}{lim}\frac{\sec 2x+1}{\tan 2x}$ adalah ...

Jawab:

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

Contoh soal 5

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

Jawab:

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

Contoh soal 6

Nilai dari $\underset{x\to 0}{lim}\frac{24{\tan }^3\frac{2}{3}x}{x{\sin }^{2}6x}$ adalah ...

Jawab:

$\begin{array}{rl}& \underset{x\to 0}{lim}\frac{24{\tan }^3\frac{2}{3}x}{x{\sin }^{2}6x}\\ & =\underset{x\to 0}{lim}24\times \frac{\tan \frac{2}{3}x}{x}\times \frac{\tan \frac{2}{3}x}{\sin 6x}\times \frac{\tan \frac{2}{3}x}{\sin 6x}\\ & =24\times \frac{\frac{2}{3}}{1}\times \frac{\frac{2}{3}}{6}\times \frac{\frac{2}{3}}{6}\\ & =24\times \frac{2}{3}\times \frac{1}{9}\times \frac{1}{9}\\ & =\frac{16}{81}\end{array}$

Contoh soal 7

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

Jawab:

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

Contoh soal 8

Nilai dari $\underset{x\to \frac{\pi }{6}}{lim}\frac{\cot x}{\sin 2x+\cos 2x}$ adalah ...

Jawab:

$\begin{array}{rl}& \underset{x\to \frac{\pi }{6}}{lim}\frac{\cot x}{\sin 2x+\cos 2x}\\ & =\frac{\cot \frac{\pi }{6}}{\sin 2(\frac{\pi }{6})+\cos 2(\frac{\pi }{6})}\\ & =\frac{\sqrt{3}}{\frac{1}{2}\sqrt{3}+\frac{1}{2}}\\ & =3-\sqrt{3}\end{array}$

Contoh soal 9

Nilai dari $\underset{x\to 0}{lim}\frac{36x^2\sin \frac{4x}{3}}{{\tan }^3\frac{2}{3}x}$ adalah ...

Jawab:

$\begin{array}{rl}& \underset{x\to 0}{lim}\frac{36x^2\sin \frac{4x}{3}}{{\tan }^3\frac{2}{3}x}\\ & =\underset{x\to 0}{lim}36\times \frac{x}{\tan \frac{2}{3}x}\times \frac{x}{\tan \frac{2}{3}x}\times \frac{\sin \frac{4x}{3}}{\tan \frac{2}{3}x}\\ & =36\times \frac{1}{\frac{2}{3}}\times \frac{1}{\frac{2}{3}}\times \frac{\frac{4}{3}}{\frac{2}{3}}\\ & =36\times \frac{3}{2}\times \frac{3}{2}\times 2\\ & =162\end{array}$

Contoh soal 10

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

Jawab:

$\begin{array}{rl}& \underset{x\to 0}{lim}\frac{(\frac{4x}{3}{)}^3}{(\cos \frac{2x}{3}\tan \frac{2x}{3}{)}^3}\\ & =\underset{x\to 0}{lim}\frac{(\frac{4x}{3}{)}^3}{(\cos \frac{2x}{3}\frac{\sin \frac{2x}{3}}{\cos \frac{2x}{3}}{)}^3}\\ & =\underset{x\to 0}{lim}\frac{(\frac{4x}{3}{)}^3}{(\sin \frac{2x}{3}{)}^3}\\ & =\frac{\frac{4}{3}}{\frac{2}{3}}\times \frac{\frac{4}{3}}{\frac{2}{3}}\times \frac{\frac{4}{3}}{\frac{2}{3}}\\ & =8\end{array}$

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