SBMPTN Zone : Limit Trigonometri

Berikut ini adalah kumpulan soal mengenai Limit Trigonometri tipe SBMPTN. Jika ingin bertanya soal, silahkan gabung ke grup Facebook atau Telegram.

Tipe:

Standar SBMPTN HOTS

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No. 1

\displaystyle\lim_{x\to0}\dfrac{\tan x}{x^2\csc3x}= ....

1. 1
2. 2
3. 3

4. 4
5. 5

SBMPTN 2017

\(\begin{aligned} \displaystyle\lim_{x\to0}\dfrac{\tan x}{x^2\csc3x}&=\displaystyle\lim_{x\to0}\dfrac{\sin x\sin3x}{x^2\cos x}\\ &=\dfrac{1\cdot3}{1^2\cdot1}\\ &=3 \end{aligned}\)

JAWAB: C

No. 2

\displaystyle\lim_{x\to\infty}\dfrac{x+\sec\dfrac1x}{x^3\sin^2\dfrac1{4x}}= ....

1. 0
2. \dfrac14
3. 4

4. 16
5. 32

SBMPTN 2017

Misal t=\dfrac1x.
Jika x\to\infty maka t\to0

\(\begin{aligned} \displaystyle\lim_{x\to\infty}\dfrac{x+\sec\dfrac1x}{x^3\sin^2\dfrac1{4x}}&=\displaystyle\lim_{t\to0}\dfrac{\dfrac1t+\sec t}{\dfrac1{t^3}\sin^2\dfrac{t}4}\\ &=\displaystyle\lim_{t\to0}\dfrac{t^2+t^3\sec t}{\sin^2\dfrac{t}4}\\ &=\displaystyle\lim_{t\to0}\dfrac{t^2}{\sin^2\dfrac{t}4}\cdot(1+t\sec t)\\ &=\dfrac{1^2}{\left(\dfrac14\right)^2}(1+0)\\ &=16 \end{aligned}\)

JAWAB: D

No. 3

\displaystyle\lim_{x\to0}\dfrac{\sec x+\cos x-2}{x^2\sin x}= ....

1. -\dfrac18
2. -\dfrac14
3. 0

4. \dfrac14
5. \dfrac18

\(\begin{aligned} \displaystyle\lim_{x\to0}\dfrac{\sec x+\cos x-2}{x^2\sin x}&=\displaystyle\lim_{x\to0}\dfrac{\dfrac1{\cos x}+\cos x-2}{x^2\sin x}\\[10pt] &=\displaystyle\lim_{x\to0}\dfrac{1+\cos^2 x-2\cos x}{\cos x\cdot x^2\sin x}\\[10pt] &=\displaystyle\lim_{x\to0}\dfrac{\cos^2 x-2\cos x+1}{\cos x\cdot x^2\sin x}\\[10pt] &=\displaystyle\lim_{x\to0}\dfrac{\left(\cos x-1\right)^2}{\cos x\cdot x^2\sin x}\\[10pt] &=\displaystyle\lim_{x\to0}\dfrac{\left(1-2\sin^2\dfrac12x-1\right)^2}{\cos x\cdot x^2\sin x}\\[10pt] &=\displaystyle\lim_{x\to0}\dfrac{\left(-2\sin^2\dfrac12x\right)^2}{\cos x\cdot x^2\sin x}\\[10pt] &=\displaystyle\lim_{x\to0}\dfrac{4\sin^4\dfrac12x}{\cos x\cdot x^2\sin x}\\[10pt] &=\dfrac{4\left(\dfrac12\right)^4}{\cos0\cdot1^2\cdot1^2}\\[10pt] &=\dfrac{4\left(\dfrac1{16}\right)}{1\cdot1\cdot1}\\[10pt] &=\dfrac14 }\)

Jadi, \displaystyle\lim_{x\to0}\dfrac{\sec x+\cos x-2}{x^2\sin x}=\dfrac14.
JAWAB: B

No. 4

\displaystyle\lim_{x\to\infty}\dfrac{4x^3\sin\dfrac1{x^2}-x\tan^2\dfrac1x-\dfrac4x}{x\sqrt{\cos\dfrac2x}}= ....

Misal \dfrac1x=t
Jika x\to\infty maka t\to0

\(\begin{aligned} \displaystyle\lim_{x\to\infty}\dfrac{4x^3\sin\dfrac1{x^2}-x\tan^2\dfrac1x-\dfrac4x}{x\sqrt{\cos\dfrac2x}}&=\displaystyle\lim_{t\to0}\dfrac{\dfrac4{t^3}\sin t^2-\dfrac1t\tan^2t-4t}{\dfrac1t\sqrt{\cos2t}}\cdot\dfrac{t}t\\[10pt] &=\displaystyle\lim_{t\to0}\dfrac{4\dfrac{\sin t^2}{t^2}-\tan^2t-4t^2}{\sqrt{\cos2t}}\\[10pt] &=\dfrac{4(1)-\tan^2(0)-4(0)^2}{\sqrt{\cos2(0)}}\\[10pt] &=\dfrac{4-0-0}{\sqrt{1}}\\[10pt] &=4 \end{aligned}\)

Jadi, \displaystyle\lim_{x\to\infty}\dfrac{4x^3\sin\dfrac1{x^2}-x\tan^2\dfrac1x-\dfrac4x}{x\sqrt{\cos\dfrac2x}}=4.

No. 5

\displaystyle\lim_{h\to0}\dfrac{\cos(x+2h)-\cos(x-2h)}{h\sqrt{4-h^2}}=

Grup Telegram, SBMPTN 2016

\(\eqalign{\displaystyle\lim_{h\to0}\dfrac{\cos(x+2h)-\cos(x-2h)}{h\sqrt{4-h^2}}&=\displaystyle\lim_{h\to0}\dfrac{\cos x\cos2h-\sin x\sin2h-(\cos x\cos2h+\sin x\sin2h)}{h\sqrt{4-h^2}}\\&=\displaystyle\lim_{h\to0}\dfrac{\cancel{\cos x\cos2h}-\sin x\sin2h-\cancel{\cos x\cos2h}-\sin x\sin2h}{h\sqrt{4-h^2}}\\&=\displaystyle\lim_{h\to0}\dfrac{-2\sin x\sin2h}{h\sqrt{4-h^2}}\\&=\displaystyle\lim_{h\to0}\dfrac{-2\sin x}{\sqrt{4-h^2}}\cdot\displaystyle\lim_{h\to0}\dfrac{\sin2h}h\\&=\dfrac{-2\sin x}{\sqrt{4-0^2}}\cdot2\\&=\dfrac{-4\sin x}2\\&=-2\sin x}\)

Jadi, \displaystyle\lim_{h\to0}\dfrac{\cos(x+2h)-\cos(x-2h)}{h\sqrt{4-h^2}}=-2\sin x.

No. 6

\displaystyle\lim_{a\to0}\dfrac1a\left(\dfrac{\sin^32a}{\cos2a}+\sin2a\cos2a\right) sama dengan

1. 0
2. \dfrac12
3. 1

4. 2
5. \infty

UM UGM 2004

\(\begin{aligned} \displaystyle\lim_{a\to0}\dfrac1a\left(\dfrac{\sin^32a}{\cos2a}+\sin2a\cos2a\right)&=\displaystyle\lim_{a\to0}\dfrac{\sin2a}a\left(\dfrac{\sin^22a}{\cos2a}+\cos2a\right)\\ &=2\left(\dfrac{\sin^22(0)}{\cos2(0)}+\cos2(0)\right)\\ &=2\left(\dfrac01+1\right)\\ &=\boxed{\boxed{2}} \end{aligned}\)

Jadi, \displaystyle\lim_{a\to0}\dfrac1a\left(\dfrac{\sin^32a}{\cos2a}+\sin2a\cos2a\right)=2.
JAWAB: D

No. 7

\displaystyle\lim_{x\to\frac14\pi}\dfrac{1-2\sin x\cos x}{\sin x-\cos x} =

1. \dfrac12
2. \dfrac12\sqrt2
3. 1

4. 0
5. -1

SNMPTN 08 Kode 201

CARA 1

\(\begin{aligned} \displaystyle\lim_{x\to\frac14\pi}\dfrac{1-2\sin x\cos x}{\sin x-\cos x}&=\displaystyle\lim_{x\to\frac14\pi}\dfrac{\sin^2x+\cos^2x-2\sin x\cos x}{\sin x-\cos x}\\[8pt] &=\displaystyle\lim_{x\to\frac14\pi}\dfrac{\sin^2x-2\sin x\cos x+\cos^2x}{\sin x-\cos x}\\[8pt] &=\displaystyle\lim_{x\to\frac14\pi}\dfrac{(\sin x-\cos x)^2}{\sin x-\cos x}\\[8pt] &=\displaystyle\lim_{x\to\frac14\pi}(\sin x-\cos x)\\ &=\sin\dfrac14\pi-\cos\dfrac14\pi\\ &=\dfrac12\sqrt2-\dfrac12\sqrt2\\ &=\boxed{\boxed{0}} \end{aligned}\)

CARA 2 (L'HOPITAL)

\(\begin{aligned} \displaystyle\lim_{x\to\frac14\pi}\dfrac{1-2\sin x\cos x}{\sin x-\cos x}&=\displaystyle\lim_{x\to\frac14\pi}\dfrac{1-\sin2x}{\sin x-\cos x}\\[8pt] &=\displaystyle\lim_{x\to\frac14\pi}\dfrac{-2\cos2x}{\cos x+\sin x}\\[8pt] &=\dfrac{-2\cos2\left(\dfrac14\pi\right)}{\cos\dfrac14\pi+\sin\dfrac14\pi}\\[8pt] &=\dfrac{-2\cos\dfrac12\pi}{\dfrac12\sqrt2+\dfrac12\sqrt2}\\[8pt] &=\dfrac{-2(0)}{\sqrt2}\\ &=\boxed{\boxed{0}} \end{aligned}\)

Jadi, \displaystyle\lim_{x\to\frac14\pi}\dfrac{1-2\sin x\cos x}{\sin x-\cos x}=0.
JAWAB: D

No. 8

Nilai dari \displaystyle\lim_{x\to0}\dfrac{3x^2-\sin\left(3x^2\right)}{2x^6}

1. 0
2. \dfrac94
3. \dfrac92

4. \dfrac34
5. 9

CARA BIASA

Misal L=\displaystyle\lim_{x\to0}\dfrac{3x^2-\sin\left(3x^2\right)}{2x^6}
\(\begin{aligned} \displaystyle\lim_{x\to0}\dfrac{3x^2-\sin\left(3x^2\right)}{2x^6}&=\displaystyle\lim_{x\to0}\dfrac{3x^2-\left(3\sin x^2-4\sin^3x^2\right)}{2x^6}\\[8pt] L&=\displaystyle\lim_{x\to0}\dfrac{3x^2-3\sin x^2+4\sin^3x^2}{2x^6}\\[8pt] L&=\displaystyle\lim_{x\to0}\dfrac{3x^2-3\sin x^2}{2x^6}+\displaystyle\lim_{x\to0}\dfrac{4\sin^3x^2}{2x^6}\\[8pt] L&=3\displaystyle\lim_{x\to0}\dfrac{x^2-\sin x^2}{2x^6}+\displaystyle\lim_{x\to0}\dfrac{2\sin^3x^2}{x^6} \end{aligned}\)
Misal x^2=3y^2
\(\begin{aligned} L&=3\displaystyle\lim_{y\to0}\dfrac{3y^2-\sin3y^2}{2\left(3y^2\right)^3}+\displaystyle\lim_{x\to0}2\cdot\dfrac{\sin x^2}{x^2}\cdot\dfrac{\sin x^2}{x^2}\cdot\dfrac{\sin x^2}{x^2}\\[8pt] &=3\displaystyle\lim_{y\to0}\dfrac{3y^2-\sin3y^2}{2\cdot27y^6}+2\cdot1\cdot1\cdot1\\[8pt] &=\dfrac3{27}\displaystyle\lim_{y\to0}\dfrac{3y^2-\sin3y^2}{2y^6}+2\\[8pt] L&=\dfrac19L+2\\[8pt] \dfrac89L&=2\\ L&=\boxed{\boxed{\dfrac94}} \end{aligned}\)

CARA L'HOPITAL

\(\begin{aligned} \displaystyle\lim_{x\to0}\dfrac{3x^2-\sin\left(3x^2\right)}{2x^6}&=\displaystyle\lim_{x\to0}\dfrac{6x-6x\cos\left(3x^2\right)}{12x^5}\qquad&\color{red}{\text{L'hopital}}\\[8pt] &=\displaystyle\lim_{x\to0}\dfrac{1-\cos\left(3x^2\right)}{2x^4}\\[8pt] &=\displaystyle\lim_{x\to0}\dfrac{6x\sin\left(3x^2\right)}{8x^3}\\[8pt] &=\displaystyle\lim_{x\to0}\dfrac{3\sin\left(3x^2\right)}{4x^2}\qquad&\color{red}{\text{L'hopital}}\\[8pt] &=\displaystyle\lim_{x\to0}\dfrac34\cdot3\cdot\dfrac{\sin\left(3x^2\right)}{3x^2}\\ &=\dfrac34\cdot3\cdot1\\ &=\boxed{\boxed{\dfrac94}} \end{aligned}\)

Jadi, \displaystyle\lim_{x\to0}\dfrac{3x^2-\sin\left(3x^2\right)}{2x^6}=\dfrac94.
JAWAB: B

No. 9

\displaystyle\lim_{x\to\dfrac{\pi}4}\dfrac{\dfrac1{\sqrt2}\sin\left(\dfrac{\pi}4-2x\right)+\dfrac1{\sqrt2}\cos\left(\dfrac{\pi}4-2x\right)}{4x-\pi}=

\(\begin{aligned} k&=\sqrt{\left(\dfrac1{\sqrt2}\right)^2+\left(\dfrac1{\sqrt2}\right)^2}\\ &=\sqrt{\dfrac12+\dfrac12}\\ &=\sqrt1\\ &=1 \end{aligned}\)

\(\begin{aligned} \tan\alpha&=\dfrac{\dfrac1{\sqrt2}}{\dfrac1{\sqrt2}}\\ &=1\\ \alpha&=\dfrac{\pi}4 \end{aligned}\)

\(\begin{aligned} \displaystyle\lim_{x\to\dfrac{\pi}4}\dfrac{\dfrac1{\sqrt2}\sin\left(\dfrac{\pi}4-2x\right)+\dfrac1{\sqrt2}\cos\left(\dfrac{\pi}4-2x\right)}{4x-\pi}&=\displaystyle\lim_{x\to\dfrac{\pi}4}\dfrac{\cos\left(\dfrac{\pi}4-2x-\dfrac{\pi}4\right)}{4x-\pi}\\ &=\displaystyle\lim_{x\to\dfrac{\pi}4}\dfrac{\cos\left(-2x\right)}{4x-\pi}\\ &=\displaystyle\lim_{x\to\dfrac{\pi}4}\dfrac{\cos\left(2x\right)}{-2\left(-2x+\dfrac{\pi}2\right)}\\ &=\displaystyle\lim_{x\to\dfrac{\pi}4}\dfrac{\sin\left(\dfrac{\pi}2-2x\right)}{-2\left(\dfrac{\pi}2-2x\right)}\\ &=\dfrac1{-2}\\ &=\boxed{\boxed{-\dfrac12}} \end{aligned}\)

Jadi, \displaystyle\lim_{x\to\dfrac{\pi}4}\dfrac{\dfrac1{\sqrt2}\sin\left(\dfrac{\pi}4-2x\right)+\dfrac1{\sqrt2}\cos\left(\dfrac{\pi}4-2x\right)}{4x-\pi}=-\dfrac12.

No. 10

\displaystyle\lim_{x\to\infty}\dfrac{\sin\dfrac3x}{x^2\tan\dfrac2x\left(1-\cos\dfrac2x\right)}=

1. \dfrac34
2. \dfrac32
3. 3

4. 1
5. \dfrac14

Misal y=\dfrac1x
Jika x\to\infty maka
y=\dfrac1x\to\dfrac1{\infty}=0

\(\begin{aligned} \displaystyle\lim_{x\to\infty}\dfrac{\sin\dfrac3x}{x^2\tan\dfrac2x\left(1-\cos\dfrac2x\right)}&=\displaystyle\lim_{y\to0}\dfrac{\sin3y}{\dfrac1{y^2}\tan2y\left(1-\cos2y\right)}\\[8pt] &=\displaystyle\lim_{y\to0}\dfrac{y^2\sin3y}{\tan2y\left(1-\left(1-2\sin^2y\right)\right)}\\[8pt] &=\displaystyle\lim_{y\to0}\dfrac{y^2\sin3y}{\tan2y\left(1-1+2\sin^2y\right)}\\[8pt] &=\displaystyle\lim_{y\to0}\dfrac{y^2\sin3y}{\tan2y\left(2\sin^2y\right)}\\[8pt] &=\displaystyle\lim_{y\to0}\dfrac{y^2\sin3y}{2\tan2y\sin^2y}\\[8pt] &=\displaystyle\lim_{y\to0}\dfrac12\cdot\dfrac{y}{\tan2y}\cdot\dfrac{y}{\sin x}\cdot\dfrac{\sin3y}{\sin y}\\[8pt] &=\dfrac12\cdot\dfrac12\cdot1\cdot3\\ &=\boxed{\boxed{\dfrac34}} \end{aligned}\)

Jadi, \displaystyle\lim_{x\to\infty}\dfrac{\sin\dfrac3x}{x^2\tan\dfrac2x\left(1-\cos\dfrac2x\right)}=\dfrac34.
JAWAB: A

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