SOAL SBMPTN LIMIT FUNGSI TRIGONOMETRI DAN PEMBAHASAN

Nilai $\displaystyle \lim_{x\to\frac{\pi}{4}}\sin\left ( \frac{\pi}{4}-x \right )\tan\left( x+\frac{\pi}{4} \right )$
adalah...

[Mat-IPA UTUL UGM 2010]
1. 2
2. 1
3. 0
4. -1
5. -2

Jawaban B

$\begin{align*} &= \lim_{x\to\frac{\pi}{4}}\sin\left ( \frac{\pi}{4}-x \right )\tan\left( x+\frac{\pi}{4} \right )\\ &= \lim_{x-\frac{\pi}{4}\to 0}\left ( -\sin\left ( x-\frac{\pi}{4} \right ) \right )\tan\left ( \frac{\pi}{2}+ \left ( x-\frac{\pi}{4} \right ) \right )\\ &= \lim_{y\to0}\left ( -sin\left ( y \right ) \right ) \left ( -cot\left ( y \right ) \right )\\ &= \lim_{y\to0}\sin y -\cot y\\ &= \lim_{y\to0}\cos y\\ &= \cos 0\\ &= 1 \end{align}$

$$\lim_{x\to0}\frac{1-cos^2x}{x^2 \tan \left( x+\frac{\pi}{3} \right )}=\cdots$$

[Mat-IPA SNMPTN 2012]
1. $-\sqrt{3}$
2. 0
3. $\frac{\sqrt{3}}{3}$
4. $\frac{\sqrt{3}}{2}$
5. $\sqrt{3}$

Jawaban C

$\lim_{x\to0}\frac{1-cos^2x}{x^2 \tan \left( x+\dfrac{\pi}{3} \right )}\\ =\lim_{x\to0}\frac{\sin^2x}{x^2\tan \left( x+\dfrac{\pi}{3} \right )}\\ =\lim_{x\to0}\frac{\sin^2x}{x^2}\cdot \lim_{x\to0}\frac{1}{\tan\left( x+ \dfrac{\pi}{3} \right )}\\ =1\cdot\frac{1}{\tan\left( 0+\dfrac{\pi}{3} \right )}\\ =\dfrac{1}{\sqrt{3}}\\ =\dfrac{\sqrt{3}}{3}$

$$\lim_{x\to0}\dfrac{x^2\sqrt{4-x}}{\cos x-\cos 3x}=\cdots$$

[Mat-IPA SBMPTN 2013]
1. -2
2. $-\frac{1}{2}$
3. $\frac{1}{2}$
4. 1
5. 2

Jawaban C

$\lim_{x\to0}\dfrac{x^2\sqrt{4-x}}{\cos x-\cos 3x}\\ =\lim_{x\to0}\dfrac{x^2\sqrt{4-x}}{2\sin 2x \sin x}\\ =\dfrac{1}{2}\cdot\dfrac{1}{2}\lim_{x\to0}\dfrac{2x}{\sin 2x}\cdot\lim_{x\to0}\dfrac{x}{\sin x}\cdot\lim_{x\to0}\sqrt{4-x}\\ =\dfrac{1}{4}\sqrt{4-0}\\ =\dfrac{1}{2} \end{aligned}$

$$\lim_{x\to0}\left \{ \dfrac{1}{x\tan x} -\dfrac{\cos^2 x}{x\sin x} \right \}=\cdots$$

[Mat-IPA SNMPTN 2007]
1. $\frac{1}{4}$
2. $-\frac{1}{4}$
3. 0
4. $-\frac{1}{2}$
5. $\frac{1}{2}$

Jawaban E

$\lim_{x\to0}\left \{ \dfrac{1}{x\tan x} -\dfrac{\cos^2 x}{x\sin x} \right \}\\ =\lim_{x\to0}\left \{ \dfrac{\cos x-\cos^2 x}{x\sin x} \right \}\\ =\lim_{x\to0}\left \{ \dfrac{\cos x-\left ( 1-\sin^2 x \right )}{x\sin x} \right \}\\ =\lim_{x\to0}\left \{ \dfrac{\cos x-1+sin^2x}{x\sin x} \right \}\\ =\lim_{x\to0}\left \{ \dfrac{-2\sin^2 \dfrac{1}{2}x+sin^2x}{x\sin x} \right \}\\ =\frac{-2\cdot\left ( \dfrac{1}{2}x \right )^2+x^2}{x\cdot x}\\ =\frac{-\dfrac{1}{2}x^2+x^2}{x^2}\\ =\dfrac{1}{2}$

$\lim_{x\to0}\dfrac{\sqrt{\sin 2x+1}-\sqrt{\tan 2x+1}}{x^3}=\cdots$

[Mat-IPA SIMAK UI 2010]
1. -8
2. -4
3. -2
4. 2
5. 4

Jawaban C

$\lim_{x\to0}\dfrac{\sqrt{\sin 2x+1}-\sqrt{\tan 2x+1}}{x^3}\\ =\lim_{x\to0}\dfrac{\sqrt{\sin 2x+1}-\sqrt{\tan 2x+1}}{x^3}\times\dfrac{\sqrt{\sin 2x+1}+\sqrt{\tan 2x+1}}{\sqrt{\sin 2x+1}+\sqrt{\tan 2x+1}}\\ =\lim_{x\to0}\dfrac{\sin 2x+1-\left ( \tan 2x+1 \right )}{x^3 \left (\sqrt{\sin 2x+1}+\sqrt{\tan 2x+1} \right )}\\ =\lim_{x\to0}\dfrac{\sin 2x-\tan 2x}{x^3 \left (\sqrt{\sin 2x+1}+\sqrt{\tan 2x+1} \right )}\\ =\lim_{x\to0}\dfrac{\tan 2x\left ( \cos 2x-1 \right )}{x^3 \left (\sqrt{\sin 2x+1}+\sqrt{\tan 2x+1} \right )}\\ =\lim_{x\to0}\dfrac{\tan 2x\left ( -2\sin^2x \right )}{x^3 \left (\sqrt{\sin 2x+1}+\sqrt{\tan 2x+1} \right )}\\ =-2\lim_{x\to0}2\cdot\frac{\tan 2x}{2x}\cdot\frac{\sin^2x}{x^2}\times \frac{1}{ \left (\sqrt{\sin 2x+1}+\sqrt{\tan 2x+1} \right )}\\ =-2\cdot 2\cdot 1\cdot 1\cdot \frac{1}{\left (\sqrt{\sin 2\left (0 \right )+1}+\sqrt{\tan 2\left (0 \right )+1} \right )}\\ =-4\cdot \frac{1}{2}\\ =-2$

$\lim_{x\to0}\frac{3\sin^2 x-x^2 \cos^2 x}{x\tan x}=\cdots$

[Mat-IPA SBMPTN 2013]
1. 2
2. $\frac{1}{2}$
3. -1
4. $-\frac{1}{2}$
5. -2

Jawaban A

$\lim_{x\to0}\dfrac{3\sin^2 x-x^2 \cos^2 x}{x\tan x}\\ =\lim_{x\to0}\dfrac{3\sin^2 x-x^2\left ( 1-2\sin^2 x \right )}{x\tan x}\\ =\lim_{x\to0}\dfrac{3\sin^2 x-x^2+x^2\sin^2 x}{x\tan x}\\ =\lim_{x\to0}\dfrac{3\cdot \sin x\cdot\sin x}{x\tan x}-\lim_{x\to0}\dfrac{x\cdot x}{x\tan x}+\lim_{x\to0}\dfrac{x\cdot x\cdot \sin x \cdot \sin x}{x\tan x}\\ =3-1+0\\ =2$

$\lim_{a\to b}\dfrac{\tan a-\tan b}{1+\left ( 1-\dfrac{a}{b} \right )\tan a \tan b-\dfrac{a}{b}}=\cdots$

[Mat-IPA SIMAK UI 2012]
1. $\frac{1}{b}$
2. b
3. -b
4. $-\frac{1}{b}$
5. 1

Jawaban C

PENJELASAN

$\begin{align*} \lim_{x\to0}\dfrac{x\tan 3x}{\cos 6x-1}=\cdots \end{align}$

Mat-IPA SBMPTN 2013
1. 2
2. $\frac{1}{2}$
3. $-\frac{1}{6}$
4. $-\frac{1}{2}$
5. -6

Jawaban C

$$\begin{align*} &=\lim_{x\to0}\dfrac{x\tan 3x}{\cos 6x-1}\\ &=\lim_{x\to0}\dfrac{x\tan 3x}{-2\sin^2 3x}\\ &=\lim_{x\to0}\dfrac{x\cdot3x}{-2\cdot\left ( 3x \right )^2}\\ &=\lim_{x\to0}\dfrac{3x^2}{-18x^2}\\ &=-\dfrac{1}{6} \end{align*}$$

$$\lim_{x\to 0}\dfrac{\cos x \sin x - \tan x}{x^2 \sin x}=\cdots$$
[Mat-IPA SIMAK UI 2013]
1. -1
2. $-\frac{1}{2}$
3. 0
4. $\frac{1}{2}$
5. 1

Jawaban A

$\begin{align*} &= \lim_{x\to0}\dfrac{\cos x \sin x - \tan x}{x^2 \sin x}\\ &= \lim_{x\to 0}\dfrac{\tan x \left( \cos^2 x-1 \right)}{x^2 \sin x}\\ &=\lim_{x\to 0}\dfrac{\tan x\cdot-\sin x}{x^2\sin x}\\ &=-\lim_{x\to 0}\dfrac{\tan x}{x}\cdot\lim_{x\to 0}\dfrac{\sin^2 x}{x^2}\cdot\lim_{x\to0}\dfrac{x}{\sin x}\\ &=-1\cdot1\cdot1\\ &=-1 \end{align}$

$\begin{align*} \lim_{x\to0}\dfrac{3\sin^2 x-x^2\cos^2 x}{x\tan x}=\cdots \end{align}$

[Mat-IPA SBMPTN 2013]
1. 2
2. $\frac{1}{2}$
3. -1
4. $-\frac{1}{2}$
5. -2

Jawaban A

$\begin{align*} &=\lim_{x\to0}\dfrac{3\sin^2 x-x^2\cos^2 x}{x\tan x}\\ &=\lim_{x\to0}\dfrac{3\sin^2 x-x^2\left(1-\sin^2 x\right)}{x\tan x}\\ &=\lim_{x\to0}\dfrac{3\sin^2 x-x^2+x^2\sin^2 x}{x\tan x}\\ &=\lim_{xto0}\dfrac{3\cdot x^2-x^2 + x^2\cdot x^2}{x \cdot x}\\ &=\lim_{x\to0}\frac{2x^2+x^4}{x^2}\\ &=\lim_{x\to0}2+x^2\\ &= 2+0^2\\ &=2 \end{align}$

$\begin{align*} \lim_{a\to b}\dfrac{\tan a-\tan b}{1+\left ( 1-\dfrac{a}{b} \right )\tan a\tan b-\dfrac{a}{b}}=\cdots \end{align}$

[Mat-IPA SIMAK UI 2012]
1. $\frac{1}{b}$
2. b
3. -b
4. $-\frac{1}{b}$
5. 1

Jawaban C

$\begin{align*} &=\lim_{a\to b}\dfrac{\tan a-\tan b}{1+\left ( 1-\dfrac{a}{b} \right )\tan a\tan b-\dfrac{a}{b}}\\ &=\lim_{a-b\to 0}\dfrac{\tan a-\tan b}{1+\left ( 1-\dfrac{a}{b} \right )\tan a\tan b-\dfrac{a}{b}}\\ &=\lim_{a-b\to0}\dfrac{\tan a-\tan b}{\left ( 1-\dfrac{a}{b} \right )+\left ( 1-\dfrac{a}{b} \right )\tan a\tan b}\\ &=\lim_{a-b\to0}\dfrac{\tan a-\tan b}{\left ( 1-\dfrac{a}{b} \right )\left ( 1+\tan a\tan b \right )}\\ &=\lim_{a-b\to0}\dfrac{\tan\left ( a-b \right )}{-\dfrac{1}{b}\left ( a-b \right )}\\ &=\lim_{a-b\to0}\dfrac{-b\tan \left ( a-b \right )}{\left ( a-b \right )}\\ &=-b \end{align}$

$\begin{align*} \lim_{x\to0}\dfrac{\sin^2 x-\cos x+1}{x\tan x}=\cdots \end{align}$

Mat-IPA SBMPTN 2013

$\frac{3}{2}$
$\frac{1}{2}$
$-\frac{1}{2}$
-1
-2

Jawaban A

$\begin{align*} &= \lim_{x\to0}\dfrac{\sin^2 x-\cos x+1}{x\tan x}\\ &= \lim_{x\to0}\dfrac{1-\cos^2 x-\cos x+1}{x\tan x}\\ &= \lim_{x\to0}\dfrac{\cos^2 x+\cos x-2}{x\tan x}\\ &= \lim_{x\to0}\dfrac{\left ( \cos x \right )\left ( \cos x-1 \right )}{x\tan x}\\ &=-\dfrac{\left ( 1+2 \right )\left ( -\dfrac{1}{2}x^2 \right )}{x\cdot x}\\ &=\dfrac{3}{2} \end{align}$

$\begin{align*} \lim_{x\to0}\sqrt{\dfrac{x\tan x}{\sin^2 x-\cos 2x+1}}=\cdots \end{}$

Mat-IPA SBMPTN 2013

3
$\sqrt{3}$
$-\frac{\sqrt{3}}{3}$
$\frac{1}{3}$
$\frac{\sqrt{3}}{2}$

Jawaban C

$\begin{align*} &=\lim_{x\to0}\sqrt{\dfrac{x\tan x}{\sin^2 x-\cos 2x+1}}\\ &=\lim_{x\to0}\sqrt{\dfrac{x\tan x}{\sin^2 x+2\sin^2 x}}\\ &=\lim_{x\to0}\sqrt{\dfrac{x\tan x}{3\sin^3 x}}\\ &=\lim_{x\to0}\sqrt{\dfrac{x\cdot x}{3\cdot x^2}}\\ &= \sqrt{\dfrac{1}{3}}\\ &=\dfrac{1}{3}\sqrt{3} \end{}$

Cari Blog Ini

Laporkan Penyalahgunaan

Kikiholmes

PENGATURAN

SOAL SBMPTN LIMIT FUNGSI TRIGONOMETRI DAN PEMBAHASAN

Related Posts

Posting Komentar

Kategori

Terpopuler