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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• 2

No. 1

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

1. 1
2. 2
3. 3

4. 4
5. 5

SBMPTN 2017

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

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

$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{x+\sec {\displaystyle \frac{1}{x}}}{x^3{\sin }^2{\displaystyle \frac{1}{4x}}}}}& ={\displaystyle \underset{t\to 0}{lim}{\displaystyle \frac{{\displaystyle \frac{1}{t}}+\sec t}{{\displaystyle \frac{1}{t^3}}{\sin }^2{\displaystyle \frac{t}{4}}}}}\\ & ={\displaystyle \underset{t\to 0}{lim}{\displaystyle \frac{t^2+t^3\sec t}{{\sin }^2{\displaystyle \frac{t}{4}}}}}\\ & ={\displaystyle \underset{t\to 0}{lim}{\displaystyle \frac{t^2}{{\sin }^2{\displaystyle \frac{t}{4}}}}\cdot (1+t\sec t)}\\ & ={\displaystyle \frac{1^2}{{\left({\displaystyle \frac{1}{4}}\right)}^2}}(1+0)\\ & =16\end{array}$

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

$\text{Extra close brace or missing open brace}$

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

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

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

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

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

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

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

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

CARA 2 (L'HOPITAL)

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

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}
$\begin{array}{rl}{\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{3x^2-\sin \left(3x^2\right)}{2x^6}}}& ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{3x^2-(3\sin x^2-4{\sin }^3{x}^2)}{2x^6}}}\\ L& ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{3x^2-3\sin x^2+4{\sin }^3{x}^2}{2x^6}}}\\ L& ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{3x^2-3\sin x^2}{2x^6}}+{\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{4{\sin }^3{x}^2}{2x^6}}}}\\ L& =3{\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x^2-\sin x^2}{2x^6}}+{\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{2{\sin }^3{x}^2}{x^6}}}}\end{array}$
Misal x^2=3y^2
$\begin{array}{rl}L& =3{\displaystyle \underset{y\to 0}{lim}{\displaystyle \frac{3y^2-\sin 3y^2}{2{\left(3y^2\right)}^3}}+{\displaystyle \underset{x\to 0}{lim}2\cdot {\displaystyle \frac{\sin x^2}{x^2}}\cdot {\displaystyle \frac{\sin x^2}{x^2}}\cdot {\displaystyle \frac{\sin x^2}{x^2}}}}\\ & =3{\displaystyle \underset{y\to 0}{lim}{\displaystyle \frac{3y^2-\sin 3y^2}{2\cdot 27y^6}}+2\cdot 1\cdot 1\cdot 1}\\ & ={\displaystyle \frac{3}{27}}{\displaystyle \underset{y\to 0}{lim}{\displaystyle \frac{3y^2-\sin 3y^2}{2y^6}}+2}\\ L& ={\displaystyle \frac{1}{9}}L+2\\ {\displaystyle \frac{8}{9}}L& =2\\ L& =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{9}{4}}}}}}\end{array}$

CARA L'HOPITAL

$\begin{array}{rlr}{\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{3x^2-\sin \left(3x^2\right)}{2x^6}}}& ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{6x-6x\cos \left(3x^2\right)}{12x^5}}}& \text{L'hopital}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{1-\cos \left(3x^2\right)}{2x^4}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{6x\sin \left(3x^2\right)}{8x^3}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{3\sin \left(3x^2\right)}{4x^2}}}& \text{L'hopital}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{3}{4}}\cdot 3\cdot {\displaystyle \frac{\sin \left(3x^2\right)}{3x^2}}}\\ & ={\displaystyle \frac{3}{4}}\cdot 3\cdot 1\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{9}{4}}}}}}\end{array}$

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}=

$\begin{array}{rl}k& =\sqrt{{\left({\displaystyle \frac{1}{\sqrt{2}}}\right)}^2+{\left({\displaystyle \frac{1}{\sqrt{2}}}\right)}^2}\\ & =\sqrt{{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}}\\ & =\sqrt{1}\\ & =1\end{array}$

$\begin{array}{rl}\tan \alpha & ={\displaystyle \frac{{\displaystyle \frac{1}{\sqrt{2}}}}{{\displaystyle \frac{1}{\sqrt{2}}}}}\\ & =1\\ \alpha & ={\displaystyle \frac{\pi }{4}}\end{array}$

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

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

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

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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