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}

\(\begin{aligned} \displaystyle\lim_{x\to0}\dfrac{\cos3x-\cos x}{\tan^22x}&=\displaystyle\lim_{x\to0}\dfrac{-2\sin\left(\dfrac{3x+x}2\right)\sin\left(\dfrac{3x-x}2\right)}{\tan^22x}\\[8pt] &=\displaystyle\lim_{x\to0}\dfrac{-2\sin2x\sin x}{\tan^22x}\\[8pt] &=-2\displaystyle\lim_{x\to0}\dfrac{\sin2x}{\tan2x}\cdot\dfrac{\sin x}{\tan2x}\\[8pt] &=-2\cdot\dfrac22\cdot\dfrac12\\ &=\boxed{\boxed{-1}} \end{aligned}\)

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

\(\eqalign{ \displaystyle\lim_{x\to0}\dfrac{1-\cos x}{\sin x}&=\displaystyle\lim_{x\to0}\dfrac{1-\cos x}{\sin x}\cdot\dfrac{1+\cos x}{1+\cos x}\\[4pt] &=\displaystyle\lim_{x\to0}\dfrac{1-\cos^2x}{\sin x(1+\cos x)}\\[4pt] &=\displaystyle\lim_{x\to0}\dfrac{\sin^2x}{\sin x(1+\cos x)}\\[4pt] &=\displaystyle\lim_{x\to0}\dfrac{\sin x}{1+\cos x}\\[4pt] &=\dfrac{\sin 0}{1+\cos 0}\\[4pt] &=\dfrac0{1+1}\\ &=0 }\)

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

\(\eqalign{ \displaystyle\lim_{x\to0}\dfrac{2\sin^2\dfrac12x}{x\tan x}&=\dfrac{2\left(\dfrac12\right)^2}{(1)(1)}\\[4pt] &=\dfrac{2\left(\dfrac14\right)}1\\[4pt] &=\dfrac12 }\)

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

\(\eqalign{ \displaystyle\lim_{x\to-2}\dfrac{\left(x^2-4\right)\tan(x+2)}{\sin^2(x+2)}&=\displaystyle\lim_{x\to-2}\dfrac{(x-2)\cancel{(x+2)}\cancel{\tan(x+2)}}{\cancel{\sin^2(x+2)}}\\ &=-2-2\\ &=-4 }\)

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}= ....

\(\eqalign{\displaystyle\lim_{x\to0}\dfrac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}&=\displaystyle\lim_{x\to0}\dfrac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}\cdot\dfrac{\sqrt{1+\tan x}+\sqrt{1+\sin x}}{\sqrt{1+\tan x}+\sqrt{1+\sin x}}\\&=\displaystyle\lim_{x\to0}\dfrac{(1+\tan x)-(1+\sin x)}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x})}\\&=\displaystyle\lim_{x\to0}\dfrac{1+\tan x-1-\sin x}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x})}\\&=\displaystyle\lim_{x\to0}\dfrac{\dfrac{\sin x}{\cos x}-\sin x}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x})}\\&=\displaystyle\lim_{x\to0}\dfrac{\dfrac{\sin x}{\cos x}(1-\cos x)}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x})}\cdot\dfrac{1+\cos x}{1+\cos x}\\&=\displaystyle\lim_{x\to0}\dfrac{\sin x\left(1-\cos^2 x\right)}{x^3\cos x(\sqrt{1+\tan x}+\sqrt{1+\sin x})(1+\cos x)}\\&=\displaystyle\lim_{x\to0}\dfrac{\sin x\left(\sin^2 x\right)}{x^3\cos x(\sqrt{1+\tan x}+\sqrt{1+\sin x})(1+\cos x)}\\&=\displaystyle\lim_{x\to0}\dfrac{\sin^3 x}{x^3\cos x(\sqrt{1+\tan x}+\sqrt{1+\sin x})(1+\cos x)}\\&=\displaystyle\lim_{x\to0}\dfrac{\sin^3 x}{x^3}\cdot\displaystyle\lim_{x\to0}\dfrac1{\cos x(\sqrt{1+\tan x}+\sqrt{1+\sin x})(1+\cos x)}\\&=1^3\cdot\dfrac1{\cos 0(\sqrt{1+\tan 0}+\sqrt{1+\sin 0})(1+\cos 0)}\\&=\dfrac1{1(\sqrt{1+0}+\sqrt{1+0})(1+1)}\\&=\dfrac1{1(1+1)(2)}\\&=\dfrac14}\)

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

\(\eqalign{\displaystyle\lim_{x\to2}\dfrac{\sin(2x-4)}{2-\sqrt{6-x}}&=\displaystyle\lim_{x\to2}\dfrac{\sin(2x-4)}{2-\sqrt{6-x}}\cdot\dfrac{2+\sqrt{6-x}}{2+\sqrt{6-x}}\\&=\displaystyle\lim_{x\to2}\dfrac{\sin(2x-4)\left(2+\sqrt{6-x}\right)}{4-(6-x)}\\&=\displaystyle\lim_{x\to2}\dfrac{\sin(2x-4)\left(2+\sqrt{6-x}\right)}{4-6+x}\\&=\displaystyle\lim_{x\to2}\dfrac{\sin2(x-2)\left(2+\sqrt{6-x}\right)}{-2+x}\\&=\displaystyle\lim_{x\to2}\dfrac{\sin2(x-2)}{x-2}\cdot\left(2+\sqrt{6-x}\right)\\&=2\cdot\left(2+\sqrt{6-2}\right)\\&=2\cdot\left(2+\sqrt4\right)\\&=2\cdot\left(2+2\right)\\&=2(4)\\&=8}\)

CARA 2 : L'hopital

\(\eqalign{ \displaystyle\lim_{x\to2}\dfrac{\sin(2x-4)}{2-\sqrt{6-x}}&=\displaystyle\lim_{x\to2}\dfrac{2\cos(2x-4)}{-\dfrac{-1}{2\sqrt{6-x}}}\\[4pt] &=\displaystyle\lim_{x\to2}4\cos(2x-4)\sqrt{6-x}\\[4pt] &=4\cos(2(2)-4)\sqrt{6-2}\\ &=4\cos(4-4)\sqrt4\\ &=4\cos0(2)\\ &=4(1)(2)\\ &=8 }\)

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}= ....

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\(\begin{aligned} \tan(5x-3x)&=\dfrac{\tan5x-\tan3x}{1+\tan5x\tan3x}\\ \tan2x&=\dfrac{\tan5x-\tan3x}{1+\tan5x\tan3x}\\ \tan5x-\tan3x&=\tan2x+\tan2x\tan5x\tan3x \end{aligned}\)

\(\begin{aligned} \displaystyle\lim_{x\to0}\dfrac{\tan 5x-\tan 3x-\tan 2x}{x^3}&=\displaystyle\lim_{x\to0}\dfrac{\tan2x+\tan2x\tan5x\tan3x-\tan 2x}{x^3}\\[4pt] &=\displaystyle\lim_{x\to0}\dfrac{\tan2x\tan5x\tan3x}{x^3}\\[4pt] &=\displaystyle\lim_{x\to0}\dfrac{\tan2x}x\cdot\dfrac{\tan5x}x\cdot\dfrac{\tan3x}x\\[4pt] &=2\cdot5\cdot3\\ &=\boxed{\boxed{30}} \end{aligned}\)

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

\(\begin{aligned} \displaystyle\lim_{x\to-1}\dfrac{\sin(2x+2)}{\sqrt{3x+4}-\sqrt{x+2}}&=\displaystyle\lim_{x\to-1}\dfrac{\sin(2x+2)}{\sqrt{3x+4}-\sqrt{x+2}}\cdot\dfrac{\sqrt{3x+4}+\sqrt{x+2}}{\sqrt{3x+4}+\sqrt{x+2}}\\[8pt] &=\displaystyle\lim_{x\to-1}\dfrac{\sin(2x+2)\left(\sqrt{3x+4}+\sqrt{x+2}\right)}{(3x+4)-(x+2)}\\[8pt] &=\displaystyle\lim_{x\to-1}\dfrac{\sin(2x+2)\left(\sqrt{3x+4}+\sqrt{x+2}\right)}{3x+4-x-2}\\[8pt] &=\displaystyle\lim_{x\to-1}\dfrac{\sin(2x+2)\left(\sqrt{3x+4}+\sqrt{x+2}\right)}{2x+2}\\[8pt] &=\displaystyle\lim_{x\to-1}\dfrac{\sin(2x+2)}{2x+2}\left(\sqrt{3x+4}+\sqrt{x+2}\right)\\[8pt] &=1\left(\sqrt{3(-1)+4}+\sqrt{-1+2}\right)\\ &=\sqrt{1}+\sqrt1\\ &=\boxed{\boxed{2}} \end{aligned}\)

CARA 2 : L'HOPITAL

\(\begin{aligned} \displaystyle\lim_{x\to-1}\dfrac{\sin(2x+2)}{\sqrt{3x+4}-\sqrt{x+2}}&=\displaystyle\lim_{x\to-1}\dfrac{2\cos(2x+2)}{\dfrac3{2\sqrt{3x+4}}-\dfrac1{2\sqrt{x+2}}}\\[23pt] &=\dfrac{2\cos(2(-1)+2)}{\dfrac3{2\sqrt{3(-1)+4}}-\dfrac1{2\sqrt{-1+2}}}\\[23pt] &=\dfrac{2\cos(-2+2)}{\dfrac3{2\sqrt{-3+4}}-\dfrac1{2\sqrt1}}\\[23pt] &=\dfrac{2\cos0}{\dfrac3{2\sqrt1}-\dfrac1{2(1)}}\\[23pt] &=\dfrac{2(1)}{\dfrac3{2(1)}-\dfrac12}\\[23pt] &=\dfrac2{\dfrac32-\dfrac12}\\[23pt] &=\dfrac2{\dfrac22}\\[23pt] &=\dfrac21\\ &=\boxed{\boxed{2}} \end{aligned}\)

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}

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\(\begin{aligned} \displaystyle\lim_{x\to3}\dfrac{\sin\left(\sqrt{x+1}-2\right)}{x-3}&=\displaystyle\lim_{x\to3}\dfrac{\sin\left(\sqrt{x+1}-2\right)}{x-3}\cdot\dfrac{\sqrt{x+1}-2}{\sqrt{x+1}-2}\cdot\dfrac{\sqrt{x+1}+2}{\sqrt{x+1}+2}\\[8pt] &=\displaystyle\lim_{x\to3}\dfrac{\sin\left(\sqrt{x+1}-2\right)}{\sqrt{x+1}-2}\cdot\dfrac{x+1-4}{(x-3)\left(\sqrt{x+1}+2\right)}\\[8pt] &=\displaystyle\lim_{x\to3}\dfrac{\sin\left(\sqrt{x+1}-2\right)}{\sqrt{x+1}-2}\cdot\dfrac{x-3}{(x-3)\left(\sqrt{x+1}+2\right)}\\[8pt] &=\displaystyle\lim_{x\to3}\dfrac{\sin\left(\sqrt{x+1}-2\right)}{\sqrt{x+1}-2}\cdot\dfrac1{\sqrt{x+1}+2}\\[8pt]
&=1\cdot\dfrac1{\sqrt{3+1}+2}\\[8pt] &=\dfrac1{\sqrt{4}+2}\\[8pt] &=\dfrac1{2+2}\\ &=\boxed{\boxed{\dfrac14}} \end{aligned}\)

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

\(\begin{aligned} \displaystyle\lim_{x\to\pi}\left(\sin x+\cos x\right)&=\sin\pi+\cos\pi\\ &=0+(-1)\\ &=\boxed{\boxed{-1}} \end{aligned}\)

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