# Soal dan Solusi Limit Fungsi Aljabar

Berikut ini soal dan solusi limit fungsi

1. Nilai dari

$$\lim_{x\to -3} 3x+5$$

A. -14
B. -4
C. 0
D. 4
E. 14

\begin{align} \lim_{x\to -3} 3x+5 &= 3(-3)+5\\ &=-9+5\\ &=-4 \end{align}

2. Nilai dari

$$\lim_{x\to 1} \frac{2x^2-x-1}{3x^2-x-2}$$

A. $$\frac{5}{3}$$

B. $$\frac{3}{4}$$

C. $$\frac{2}{3}$$

D. $$\frac{3}{5}$$

E. $$\frac{2}{5}$$

\begin{align} \lim_{x\to 1} \frac{2x^2-x-1}{3x^2-x-2}&=\lim_{x\to 1} \frac{(2x+1)(x-1)}{(3x+2)(x-1)}\\ &=\lim_{x\to 1}\frac{2x+1}{3x+2}\\ &=\frac{2(1)+1}{3(1)+2}\\ &=\frac{3}{5}\\ \end{align}

3. Nilai dari

$$\lim_{x\to 4} \frac{x^2-2x-8}{x-4}$$

A. -6
B. -2
C. 0
D. 2
E. 6

\begin{align} \lim_{x\to 4} \frac{x^2-2x-8}{x-4}&=\lim_{x\to 4} \frac{(x-4)(x+2)}{x-4}\\ &=\lim_{x\to 4} {(x+2)}\\ &=(4+2)\\ &=6\\ \end{align}

4. Nilai dari

$$\lim_{x\to 1} \frac{2x^2-2}{x-1}$$

A. 0
B. 1
C. 2
D. 4
E. 6

\begin{align} \lim_{x\to 1} \frac{2x^2-2}{x-1}&=\lim_{x\to 1} \frac{2(x^2-1)}{x-1}\\ &=\lim_{x\to 1} \frac{2(x-1)(x+1)}{x-1}\\ &=\lim_{x\to 1} 2(x+1)\\ &=2(1+1)\\ &=2(2)\\ &=4\\ \end{align}

5. Nilai dari

$$\lim_{x\to 2} \frac{x^2-5x+6}{x^2-3x+3}$$

A. -1
B. $$-\frac{1}{3}$$
C. 0
D. 1
E. -5

\begin{align} \lim_{x\to 2} \frac{x^2-5x+6}{x^2-3x+3}&=\frac{2^2-5(2)+6}{2^2-3(2)+3}\\ &=\frac{4-10+6}{4-6+3}\\ &=\frac{0}{1}\\ &=0 \end{align}

6. Nilai dari

$$\lim_{x\to 3} \frac{2x^2-5x-3}{x-3}$$

A. -7
B. 0
C. 5
D. 7
E.10

\begin{align} \lim_{x\to 3} \frac{2x^2-5x-3}{x-3}&=\lim_{x\to 3} \frac{(2x+1)(x-3)}{x-3}\\ &=\lim_{x\to 3} 2x+1\\ &=2(3)+1\\ &=7\\ \end{align}

7. Nilai dari

$$\lim_{x\to -5} \frac{x^2+3x-10}{x+5}$$

A. -10
B. -7
C. 0
D. 7
E. 10

\begin{align} \lim_{x\to -5} \frac{x^2+3x-10}{x+5}&=\lim_{x\to -5} \frac{(x+5)(x-2)}{x+5}\\ &=\lim_{x\to -5} x-2\\ &=-5-2\\ &=-7\\ \end{align}

8. Nilai dari

$$\lim_{x\to 2} \frac{x^4-16}{x-2}$$

A. -32
B. -16
C. 4
D. 16
E. 32

\begin{align} \lim_{x\to 2} \frac{x^4-16}{x-2}&=\lim_{x\to 2} \frac{(x^2-4)(x^2+4)}{x-2}\\ &=\lim_{x\to 2} \frac{(x-2)(x+2)(x^2+4)}{x-2}\\ &=\lim_{x\to 2}(x+2)(x^2+4)\\ &=(2+2)(2^2+4)\\ &=4(4+4)\\ &=4(8)\\ &=32\\ \end{align}

9. Nilai dari

$$\lim_{x\to 2} \frac{x^2-3x+2}{x^-2x}$$

A. $$-1$$

B. $$-\frac{1}{2}$$

C. $$0$$

D. $$\frac{1}{2}$$

E. $$\frac{1}{2}$$

\begin{align} \lim_{x\to 2} \frac{x^2-3x+2}{x^2-2x}&=\lim_{x\to 2} \frac{(x-2)(x-1)}{x(x-2)}\\ &=\lim_{x\to 2} \frac{x-1}{x}\\ &=\frac{2-1}{2}\\ &=\frac{1}{2}\\ \end{align}

10. Nilai dari

$$\lim_{x\to 0} \frac{x^3-2x^2+x}{2x^3+x^2-2x}$$

A. $$-1$$

B. $$-\frac{1}{2}$$

C. $$0$$

D. $$\frac{1}{2}$$

E. $$\frac{1}{2}$$

\begin{align} \lim_{x\to 0} \frac{x^3-2x^2+x}{2x^3+x^2-2x}&=\lim_{x\to 0} \frac{x(x^2-2x+1)}{x(2x^2+x-2)}\\ &=\lim_{x\to 0} \frac{x^2-2x+1}{x(x^2+x-2)}\\ &=\lim_{x\to 0} \frac{(x-1)(x-1)}{(x+2)(x-1)}\\ &=\lim_{x\to 0} \frac{x-1}{x+2}\\ &=\frac {0-1}{0+2}\\ &=\frac{-1}{2}\\ \end{align}

11. Nilai dari

$$\lim_{x\to 1} \frac{x^3-5x+4}{x^3-1}$$

A. $$-\frac{3}{4}$$

B. $$-\frac{2}{3}$$

C. $$\frac{1}{2}$$

D. $$\frac{2}{3}$$

E. Tak terhingga

\begin{align} \lim_{x\to 1} \frac{x^3-5x+4}{x^3-1}&=\lim_{x\to 1} \frac{(x-1)(x^2+x-4)}{(x-1)(x^2+x+1)}\\ &=\lim_{x\to 1} \frac{(x^2+x-4)}{(x^2+x+1)}\\ &=\frac{1^2+1-4}{1^2+1+1}\\ &=\frac{1+1-4}{1+1+1}\\ &=\frac{-2}{3}\\ &=-\frac{2}{3}\\ \end{align}

12. Nilai dari

$$\lim_{x\to -2} \frac{3x+6}{x^3+8}$$

A. $$-\frac{1}{4}$$

B. $$-\frac{1}{6}$$

C. $$0$$

D. $$\frac{1}{4}$$

E. $$1$$

\begin{align} \lim_{x\to -2} \frac{3x+6}{x^3+8}&=\lim_{x\to -2} \frac{3(x+2)}{x^3+2^3}\\ &=\lim_{x\to -2} \frac{3(x+2)}{(x+2)(x^2-2x+4)}\\ &=\lim_{x\to -2} \frac{3}{(x^2-2x+4)}\\ &=\frac{3}{(-2)^2-2(-2)+4)}\\ &=\frac{3}{4+4+4}\\ &=\frac{3}{12}\\ &=\frac{1}{4}\\ \end{align}

13. Nilai dari

$$\lim_{x\to 36} \frac{x-36}{\sqrt{x}-6}$$

A. 0
B. 4
C. 16
D. 36
E. 128

\begin{align} \lim_{x\to 36} \frac{x-36}{\sqrt{x}-6}&=\lim_{x\to 36} \frac{(\sqrt x)^2-6^2}{\sqrt{x}-6}\\ &=\lim_{x\to 36} \frac{(\sqrt x+6)(\sqrt x-6)}{\sqrt{x}-6}\\ &=\lim_{x\to 36} (\sqrt x+6)\\ &=(\sqrt {36}+6)\\ &=(6+6)\\ &=12 \end{align}

14. Nilai dari

$$\lim_{x\to 1} \left(\frac{1}{x^2-1}-\frac{2}{x^4-1}\right)$$

A. $$-\frac{3}{4}$$

B. $$-\frac{2}{3}$$

C. $$0$$

D. $$\frac{1}{2}$$

E. $$\frac{2}{3}$$

\begin{align} \lim_{x\to 1} \left(\frac{1}{x^2-1}-\frac{2}{x^4-1}\right)&=\lim_{x\to 1} \left(\frac{1(x^4-1)-2(x^2-1)}{(x^4-1)(x^2-1)}\right)\\ &=\lim_{x\to 1} \left(\frac{x^4-1-2x^2+2}{(x^4-1)(x^2-1)}\right)\\ &=\lim_{x\to 1} \left(\frac{x^4-2x^2+1}{(x^2-1)(x^2+1)(x^2-1)}\right)\\ &=\lim_{x\to 1} \left(\frac{(x^2-1)(x^2-1)}{(x^2-1)(x^2+1)(x^2-1)}\right)\\ &=\lim_{x\to 1} \frac{1}{x^2+1}\\ &=\frac{1}{(1^2+1)}\\ &=\frac{1}{2}\\ \end{align}

15. Nilai dari

$$\lim_{x\to 2} \left(\frac{6-x}{x^2-4}-\frac{1}{x-2}\right)$$

A. $$-\frac{1}{2}$$

B. $$-\frac{1}{4}$$

C. $$0$$

D. $$\frac{1}{4}$$

E. $$\frac{1}{2}$$

\begin{align} \lim_{x\to 2} \left(\frac{6-x}{x^2-4}-\frac{1}{x-2}\right)&=\lim_{x\to 2} \left(\frac{(6-x)-1(x-2)}{x^2-4}\right)\\ &=\lim_{x\to 2} \left(\frac{6-x-x+2}{x^2-4}\right)\\ &=\lim_{x\to 2} \left(\frac{4-2x}{(x-2)(x+2)}\right)\\ &=\lim_{x\to 2} \left(\frac{2(2-x)}{(x-2)(x+2)}\right)\\ &=\lim_{x\to 2} \left(\frac{-2}{(x+2)}\right)\\ &=\frac{-2}{2+2}\\ &=\frac{-2}{4}\\ &=-\frac{1}{2}\\ \end{align}

16. Nilai dari

$$\lim_{x\to 1} \left(\frac{1}{1-x}-\frac{2}{x-x^3}\right)$$

A. $$-\frac{3}{2}$$

B. $$-\frac{2}{3}$$

C. $$\frac{2}{3}$$

D. $$1$$

E. $$\frac{3}{2}$$

\begin{align} \lim_{x\to 1} \left(\frac{1}{1-x}-\frac{2}{x-x^3}\right)&=\lim_{x\to 1} \left(\frac{1}{1-x}-\frac{2}{x(1-x^2)}\right)\\ &=\lim_{x\to 1} \left(\frac{1}{1-x}-\frac{2}{x(1-x)(1+x))}\right)\\ &=\lim_{x\to 1} \left(\frac{(x+x^2)-2}{x(1-x)(1+x)}\right)\\ &=\lim_{x\to 1} \left(\frac{x^2+x-2}{x(1-x)(1+x)}\right)\\ &=\lim_{x\to 1} \left(\frac{(x+2)(x-1)}{x(1-x)(1+x)}\right)\\ &=\lim_{x\to 1} \left(\frac{-(x+2)}{x(1+x)}\right)\\ &=\frac{-(1+2)}{1(1+1)}\\ &=-\frac{3}{2}\\ \end{align}

17. Nilai dari

$$\lim_{x\to 1} \frac{x-1}{\sqrt{x+3}-2}$$ adalah …

A. $$\frac{1}{4}$$

B. $$\frac{1}{2}$$

C. $$1$$

D. $$2$$

E. $$4$$

\begin{align} \lim_{x\to 1} \frac{x-1}{\sqrt{x+3}-2}&=\lim_{x\to 1} \frac{x-1}{\sqrt{x+3}-2}\times\frac{\sqrt{x+3}+2}{\sqrt{x+3}+2}\\ &=\lim_{x\to 1} \frac{(x-1)(\sqrt{x+3}+2)}{x+3-4}\\ &=\lim_{x\to 1} \frac{(x-1)(\sqrt{x+3}+2)}{x-1}\\ &=\lim_{x\to 1} (\sqrt{x+3}+2)\\ &=(\sqrt{1+3}+2)\\ &=(\sqrt{4}+2)\\ &=2+2\\ &=4\\ \end{align}

18. Nilai dari

$$\lim_{x\to 3} \frac{x-\sqrt{2x+3}}{9-x^2}$$

A. $$-\frac{1}{9}$$

B. $$-\frac{1}{8}$$

C. $$\frac{1}{3}$$

D. $$\frac{1}{2}$$

E. $$\frac{2}{3}$$

\begin{align} \lim_{x\to 3} \frac{x-\sqrt{2x+3}}{9-x^2}&=\lim_{x\to 3} \frac{x-\sqrt{2x+3}}{9-x^2}\times \frac{x+\sqrt{2x+3}}{x+\sqrt{2x+3}}\\ &=\lim_{x\to 3} \frac{x^2-{(2x+3)}}{(9-x^2)(x+\sqrt{2x+3})}\\ &=\lim_{x\to 3} \frac{x^2-2x-3}{(9-x^2)(x+\sqrt{2x+3})}\\ &=\lim_{x\to 3} \frac{(x-3)(x+1)}{(3-x)(3+x)(x+\sqrt{2x+3})}\\ &=\lim_{x\to 3} \frac{-(x+1)}{(3+x)(x+\sqrt{2x+3})}\\ &=\frac{-(3+1)}{(3+3)(3+\sqrt{6+3})}\\ &=\frac{-4}{(6)(3+\sqrt{9})}\\ &=\frac{-4}{(6)(3+3)}\\ &=\frac{-4}{(6)(6)}\\ &=\frac{-4}{36}\\ &=-\frac{1}{9}\\ \end{align}

19. Nilai dari

$$\lim_{x\to 0} \frac{x^2}{1-\sqrt{1+x^2}}$$

A. 2
B. 0
C. -1
D. -2
E. -3

\begin{align} \lim_{x\to 0} \frac{x^2}{1-\sqrt{1+x^2}}&=\lim_{x\to 0} \frac{x^2}{1-\sqrt{1+x^2}}\times \frac{1+\sqrt{1+x^2}}{1+\sqrt{1+x^2}}\\ &=\lim_{x\to 0} \frac{x^2(1+\sqrt{1+x^2})}{(1-\sqrt{1+x^2})(1+\sqrt{1+x^2})}\\ &=\lim_{x\to 0} \frac{x^2(1+\sqrt{1+x^2})}{1-({1+x^2})}\\ &=\lim_{x\to 0} \frac{x^2(1+\sqrt{1+x^2})}{-x^2}\\ &=\lim_{x\to 0}-(1+\sqrt{1+x^2})\\ &=-(1+\sqrt{1+0^2})\\ &=-(1+1)\\ &=-2\\ \end{align}

20. Nilai dari

$$\lim_{x\to 2} \left(\frac{2}{x^2-4}-\frac{3}{x^2+2x-8}\right)$$

A. $$-\frac{7}{12}$$

B. $$-\frac{1}{4}$$

C. $$-\frac{1}{12}$$

D. $$-\frac{1}{24}$$

E. $$0$$

\begin{align} \lim_{x\to 2} \left(\frac{2}{x^2-4}-\frac{3}{x^2+2x-8}\right)&=\lim_{x\to 1} \left(\frac{2}{(x-2)(x+2)}-\frac{3}{(x+4)(x-2)}\right)\\ &=\lim_{x\to 2} \left(\frac{2(x+4)-3(x+2)}{(x-2)(x+2)(x+4)}\right)\\ &=\lim_{x\to 2} \left(\frac{(2x+8-3x-6)}{(x-2)(x+2)(x+4)}\right)\\ &=\lim_{x\to 2} \left(\frac{(-x+2)}{(x-2)(x+2)(x+4)}\right)\\ &=\lim_{x\to 2} \left(\frac{-(x-2)}{(x-2)(x+2)(x+4)}\right)\\ &=\lim_{x\to 2} \left(\frac{-1}{(x+2)(x+4)}\right)\\ &=\frac{-1}{(2+2)(2+4)}\\ &=-\frac{1}{(4)(6)}\\ &=-\frac{1}{24}\\ \end{align}