Contoh Soal Limit SMA

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Contoh Soal Limit SMA

Category: SMA

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1. Nilai $\lim_{x\rightarrow 2}\left [ \frac{2}{x-2}-\frac{8}{x^{2}-4} \right ]=...$

$\frac{1}{4}$
$\frac{1}{2}$
$2$
$4$
$\infty$

JAWABAN

TUTUP JAWABAN

Jawaban: B

$\lim_{x\rightarrow 2}\left [ \frac{2}{x-2}-\frac{8}{x^{2}-4} \right ]=...$

$=\lim_{x\rightarrow 2}\left [ \frac{2}{x-2}-\frac{8}{(x-2)(x+2)} \right ]$

$\lim_{x\rightarrow 2}\left [ \frac{2(x+2)-8}{(x-2)(x+2)} \right ]=\lim_{x\rightarrow 2}\left [ \frac{2x+4-8}{(x-2)(x+2)} \right ]$

$\lim_{x\rightarrow 2}\left [ \frac{2x-4}{(x-2)(x+2)} \right ]=\lim_{x\rightarrow 2}\left [ \frac{2(x-2)}{(x-2)(x+2)} \right ]$

$\lim_{x\rightarrow 2}\left [ \frac{2}{x+2} \right ]=\frac{2}{2+2}=\frac{1}{2}$

2. Nilai $\lim_{x\rightarrow \infty }\frac{\sqrt{5x+4}-\sqrt{3x+9}}{4x}=...$

$0$
$\frac{1}{2}$
$1$
$2$
$4$

JAWABAN

TUTUP JAWABAN

Jawaban: A

$\lim_{x\rightarrow \infty }\frac{\sqrt{5x+4}-\sqrt{3x+9}}{4x}=...$

$=\lim_{x\rightarrow \infty }\frac{\sqrt{\frac{5}{x}+\frac{4}{x^{2}}}-\sqrt{\frac{3}{x}+\frac{9}{x^{2}}}}{4}$

$=\frac{\sqrt{0+0}-\sqrt{0+0}}{4}=0$

3. $\lim_{x\rightarrow 2}\frac{(2x+1)tan(x-2)}{x^{2}-4}=...$

5
2,5
2
1,5
1,25

JAWABAN

TUTUP JAWABAN

Jawaban: E

$\lim_{x\rightarrow 2}\frac{(2x+1)tan(x-2)}{x^{2}-4}$

$=\lim_{x\rightarrow 2}\frac{(2x+1)tan(x-2)}{(x-2)(x+2)}$

$=\lim_{x\rightarrow 2}\frac{(2x+1)(x-2)}{(x-2)(x+2)}$

$=\lim_{x\rightarrow 2}\frac{(2x+1)}{(x+2)}=\frac{5}{4}=1,25$

4. Nilai $\lim_{x\rightarrow \infty }\left ( \sqrt{x^{2}-6x+9}-(x-2) \right )$ adalah...

-1
-2
-3
-4
-5

JAWABAN

TUTUP JAWABAN

Jawaban: A

$\lim_{x\rightarrow \infty }\left ( \sqrt{x^{2}-6x+9}-(x-2) \right )$

$\lim_{x\rightarrow \infty }\left ( \sqrt{x^{2}-6x+9}-(x-2)^{2} \right )$

$=\lim_{x\rightarrow \infty }\left ( \sqrt{x^{2}-6x+9}-\sqrt{x^{2}-4x+4}\right )$

$=\frac{-6-(-4)}{2\sqrt{1}}=\frac{-2}{2}=1$

5. Nilai $\lim_{x\rightarrow -3}\frac{x^{2}+6x+9}{2-2cos(2x+6)}=...$

$3$
$1$
$\frac{1}{2}$
$\frac{1}{3}$
$\frac{1}{4}$

JAWABAN

TUTUP JAWABAN

Jawaban: E

$\lim_{x\rightarrow -3}\frac{x^{2}+6x+9}{2-2cos(2x+6)}$

$=\lim_{x\rightarrow -3}\frac{(x+3)(x+3)}{2-2cos2(x+3)}$

$=\lim_{x\rightarrow -3}\frac{(x+3)(x+3)}{2(1-cos2(x+3))}$

$=\lim_{x\rightarrow -3}\frac{(x+3)(x+3)}{2(1-(1-2sin^{2})(x+3)))}$

$=\lim_{x\rightarrow -3}\frac{(x+3)(x+3)}{2.2sin^{2}(x+3)}$

$=\lim_{x\rightarrow -3}\frac{1}{4}.\frac{(x+3)^{2}}{sin^{2}(x+3)}=\frac{1}{4}.1=\frac{1}{4}$

6. $\lim_{x\rightarrow \frac{\pi }{4}}\frac{1-tanx}{sinx-cosx}=...$

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

JAWABAN

TUTUP JAWABAN

Jawaban: B
$\lim_{x\rightarrow \frac{\pi }{4}}\frac{1-tanx}{sinx-cosx}$

$=\lim_{x\rightarrow \frac{\pi }{4}}\frac{1-\frac{sinx}{cosx}}{sinx-cosx}$

$=\lim_{x\rightarrow \frac{\pi }{4}}\frac{\frac{cos x-sin x}{cos x}}{sin x-cos x}$

$=\lim_{x\rightarrow \frac{\pi }{4}}\frac{cosx-sinx}{(sinx-cosx)cosx}$

$=\lim_{x\rightarrow \frac{\pi }{4}}\frac{1}{-cosx}=-\frac{1}{\sqrt{\frac{2}{2}}}=-\sqrt{2}$

7. Nilai $\lim_{x\rightarrow -2}\frac{x^{2}4}{x-2}=...$

$-1$
$-\frac{1}{2}$
$0$
$\frac{1}{2}$
$1$

JAWABAN

TUTUP JAWABAN

Jawaban: C

$\lim_{x\rightarrow -2}\frac{x^{2}4}{x-2}$

$=\lim_{x\rightarrow -2}\frac{(-2)^{2}-4}{-2-2}$

$=\frac{4-4}{-4}$

$=\frac{0}{-4}$

$=0$

8. Nilai $\lim_{x\rightarrow \infty }\left ( \frac{x^{2}-5x+4}{2x^{3}+3x-4} \right )=...$

$-\infty$
$-1$
$0$
$1$
$\infty$

JAWABAN

TUTUP JAWABAN

Jawaban: C

Ingat!

$\lim_{x\rightarrow \infty }\frac{ax^{m}+bx^{m-1}+...}{px^{n}+qx^{n-1}\pm ...}\begin{cases} \infty , & \text{ jika } m>n \\ \frac{a}{p}, & \text{jika } m=n \\ 0, & \text{ jika } m<n \end{cases}$

$Jadi, \lim_{x\rightarrow \infty }\left ( \frac{x^{2}-5x+4}{2x^{3}+3x-4} \right )=0$

9. Nilai $\lim_{x\rightarrow 3}\frac{x^{2}+2x-15}{2x-6}=...$

$\frac{1}{2}$
$1$
$\frac{5}{3}$
$4$
$\frac{15}{2}$

JAWABAN

TUTUP JAWABAN

Jawaban: D

$\lim_{x\rightarrow 3}\frac{x^{2}+2x-15}{2x-6}$

$=\lim_{x\rightarrow 3}\frac{(x-3)(x+5)}{2(x-3)}$

$=\lim_{x\rightarrow 3}\frac{x+5}{2}$

$=\frac{3+5}{2}=4$

10. Nilai $\lim_{x\rightarrow \infty }\left ( (3x+2)-\sqrt{9x^{2}-6x+1} \right )=...$

1
2
3
6
9

JAWABAN

TUTUP JAWABAN

Jawaban: C

$\lim_{x\rightarrow \infty }\left ( (3x+2)-\sqrt{9x^{2}-6x+1} \right )$

$=\lim_{x\rightarrow \infty }\left ( \sqrt{(3x+2)^{2}}-\sqrt{9x^{2}-6x+1} \right )$

$=\lim_{x\rightarrow \infty }\left ( \sqrt{9x^{2}+12x+4}-\sqrt{9x^{2}-6x+1} \right )$

$=\frac{12-(-6)}{2\sqrt{9}}$

$=\frac{18}{6}=3$

11. $\lim_{x\rightarrow 5}\frac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}=...$

$\sqrt{3}+\sqrt{2}$
$5-2\sqrt{6}$
$2\sqrt{6}$
$5$
$5+2\sqrt{6}$

JAWABAN

TUTUP JAWABAN

Jawaban: E

$\lim_{x\rightarrow 5}\frac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}$

$=\frac{\sqrt{5+2\sqrt{5+1}}}{\sqrt{5-2\sqrt{5+1}}}$

$=\frac{\sqrt{5+2\sqrt{6}}}{\sqrt{5-2\sqrt{6}}}$

$=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

$=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

$=\frac{3+2\sqrt{6}+2}{3-2}$

$=5+2\sqrt{6}$

12. Nilai $\lim_{x\rightarrow \infty }(3x-\sqrt{4x^{2}+3x-1}-\sqrt{x^{2}-2x+2})$

$-\frac{1}{4}$
$-\frac{1}{2}$
$1$
$\frac{1}{2}$
$\frac{1}{4}$

JAWABAN

TUTUP JAWABAN

Jawaban: E

$\lim_{x\rightarrow \infty }(3x-\sqrt{4x^{2}+3x-1}-\sqrt{x^{2}-2x+2})$

$\lim_{x\rightarrow \infty }(2x+x-\sqrt{4x^{2}+3x-1}-\sqrt{x^{2}-2x+2})$

$\lim_{x\rightarrow \infty }(\sqrt{4x^{2}}+\sqrt{x^{2}}-\sqrt{4x^{2}+3x-1}-\sqrt{x^{2}-2x+2})$

$\lim_{x\rightarrow \infty }(\sqrt{4x^{2}}-\sqrt{4x^{2}+3x-1}+\sqrt{x^{2}}-\sqrt{x^{2}-2x+2})$

$=\frac{0-3}{2\sqrt{4}}+\frac{0-(-2)}{2\sqrt{1}}$

$=\frac{-3}{2.2}+\frac{2}{2}=-\frac{3}{4}+1=\frac{1}{4}$

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