Exercise Zone : Limit Trigonometri

Berikut ini adalah kumpulan soal mengenai Limit Trigonometri tingkat dasar. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

Tingkat Kesulitan:

Dasar SBMPTN Olimpiade

• 1
• 2

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No. 1

{\displaystyle\lim_{x\to0}\dfrac{\cos3x-\cos x}{\tan^22x}=} ....

1. {1}
2. {-\dfrac12}
3. {-1}

4. {\dfrac12}
5. {3}

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

Jadi, {\displaystyle\lim_{x\to0}\dfrac{\cos3x-\cos x}{\tan^22x}=-1}.
JAWAB: C

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No. 2

\displaystyle\lim_{x\to0}\dfrac{1-\cos x}{\sin x}= ....

1. 0
2. \dfrac14
3. \dfrac12

4. 1
5. 2

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

Jadi, \displaystyle\lim_{x\to0}\dfrac{1-\cos x}{\sin x}=0.
JAWAB: A

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No. 3

Nilai \displaystyle\lim_{x\to0}\dfrac{2\sin^2\dfrac12x}{x\tan x}= ....

1. -2
2. -1
3. -\dfrac12

4. >\dfrac12
5. 1

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

Jadi, \displaystyle\lim_{x\to0}\dfrac{2\sin^2\dfrac12x}{x\tan x}=\dfrac12.
JAWAB: D

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No. 4

>Nilai dari \displaystyle\lim_{x\to-2}\dfrac{\left(x^2-4\right)\tan(x+2)}{\sin^2(x+2)}= ....

1. -4
2. -3
3. 0

4. 4
5. \infty

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

Jadi, \displaystyle\lim_{x\to-2}\dfrac{\left(x^2-4\right)\tan(x+2)}{\sin^2(x+2)}=-4.
JAWAB: A

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No. 5

\displaystyle\lim_{x\to0}\dfrac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}= ....

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

Jadi, \displaystyle\lim_{x\to0}\dfrac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}=\dfrac14.

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No. 6

\displaystyle\lim_{x\to2}\dfrac{\sin(2x-4)}{2-\sqrt{6-x}}= ....

1. -8
2. -2
3. 0

4. 2
5. 8

CARA 1

$\begin{array}{rl}{\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{\sin (2x-4)}{2-\sqrt{6-x}}}}& ={\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{\sin (2x-4)}{2-\sqrt{6-x}}}\cdot {\displaystyle \frac{2+\sqrt{6-x}}{2+\sqrt{6-x}}}}\\ & ={\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{\sin (2x-4)(2+\sqrt{6-x})}{4-(6-x)}}}\\ & ={\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{\sin (2x-4)(2+\sqrt{6-x})}{4-6+x}}}\\ & ={\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{\sin 2(x-2)(2+\sqrt{6-x})}{-2+x}}}\\ & ={\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{\sin 2(x-2)}{x-2}}\cdot (2+\sqrt{6-x})}\\ & =2\cdot (2+\sqrt{6-2})\\ & =2\cdot (2+\sqrt{4})\\ & =2\cdot (2+2)\\ & =2(4)\\ & =8\end{array}$

CARA 2 : L'hopital

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

Jadi, \displaystyle\lim_{x\to2}\dfrac{\sin(2x-4)}{2-\sqrt{6-x}}=8.

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No. 7

\displaystyle\lim_{x\to0}\dfrac{\tan 5x-\tan 3x-\tan 2x}{x^3}= ....

Grup WhatsApp

$\begin{array}{rl}\tan (5x-3x)& ={\displaystyle \frac{\tan 5x-\tan 3x}{1+\tan 5x\tan 3x}}\\ \tan 2x& ={\displaystyle \frac{\tan 5x-\tan 3x}{1+\tan 5x\tan 3x}}\\ \tan 5x-\tan 3x& =\tan 2x+\tan 2x\tan 5x\tan 3x\end{array}$

$\begin{array}{rl}{\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{\tan 5x-\tan 3x-\tan 2x}{x^3}}}& ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{\tan 2x+\tan 2x\tan 5x\tan 3x-\tan 2x}{x^3}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{\tan 2x\tan 5x\tan 3x}{x^3}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{\tan 2x}{x}}\cdot {\displaystyle \frac{\tan 5x}{x}}\cdot {\displaystyle \frac{\tan 3x}{x}}}\\ & =2\cdot 5\cdot 3\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 30}}}}\end{array}$

Jadi, \displaystyle\lim_{x\to0}\dfrac{\tan 5x-\tan 3x-\tan 2x}{x^3}=30.

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No. 8

\displaystyle\lim_{x\to-1}\dfrac{\sin(2x+2)}{\sqrt{3x+4}-\sqrt{x+2}}=

1. \dfrac12
2. 1
3. 2

4. 4
5. 6

CARA 1 : KALI SEKAWAN

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

CARA 2 : L'HOPITAL

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

Jadi, \displaystyle\lim_{x\to-1}\dfrac{\sin(2x+2)}{\sqrt{3x+4}-\sqrt{x+2}}=2.
JAWAB: C

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No. 9

\displaystyle\lim_{x\to3}\dfrac{\sin\left(\sqrt{x+1}\right)-2}{x-3}

Grup Telegram

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

Jadi, \displaystyle\lim_{x\to3}\dfrac{\sin\left(\sqrt{x+1}\right)-2}{x-3}=\dfrac14.

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No. 10

Nilai dari \displaystyle\lim_{x\to\pi}\left(\sin x+\cos x\right) adalah

1. -1
2. 1
3. -2

4. 1
5. 3

$\begin{array}{rl}{\displaystyle \underset{x\to \pi }{lim}(\sin x+\cos x)}& =\sin \pi +\cos \pi \\ & =0+(-1)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -1}}}}\end{array}$

Jadi, \displaystyle\lim_{x\to\pi}\left(\sin x+\cos x\right)=-1.
JAWAB: A