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# Solving limit without L'Hopital's Rule.

Find the limits without using L'Hospital's rule

1.

Solution.

2.

Solution.

Hence, .

Therefore

3.

Solution.

4.

Solution.

5.

Solution.

6.

Solution.

7.

Solution.

Hence, .

Therefore

8.

Solution.

9.

Solution.

10.

Solution.

11. $\lim\limits_{x\rightarrow\infty}\frac{3x^2+2x-1}{4x^3+3x^2+4}.$

Solution.

$$\lim\limits_{x\rightarrow\infty}\frac{3x^2+2x-1}{4x^3+3x^2+4}=\left[\frac{\infty}{\infty}\right]=\lim\limits_{x\rightarrow\infty}\frac{\frac{3x^2}{x^3}+\frac{2x}{x^3}-\frac{1}{x^3}}{\frac{4x^3}{x^3}+\frac{3x^2}{x^3}+\frac{4}{x^3}}=\lim\limits_{x\rightarrow\infty}\frac{\frac{3}{x}+\frac{2}{x^2}-\frac{1}{x^3}}{4+\frac{3}{x}+\frac{4}{x^3}}=\frac{0}{4}=0.$$

12. $\lim\limits_{x\rightarrow\infty}\left(\frac{3x+1}{3x-2}\right)^{2x}.$

Solution.

$$\lim\limits_{x\rightarrow\infty}\left(\frac{3x+1}{3x-2}\right)^{2x}=\lim\limits_{x\rightarrow\infty}\left(\frac{3x-2+3}{3x-2}\right)^{2x}=\lim\limits_{x\rightarrow\infty}\left(1+\frac{3}{3x-2}\right)^{\frac{3x-2}{3}\cdot\frac{3}{3x-2}\cdot 2x}=$$

$$=e^{\lim\limits_{x\rightarrow\infty}\frac{3}{3x-2}\cdot 2x}=e^{\lim\limits_{x\rightarrow\infty}\frac{6x}{3x-2}}=e^{\lim\limits_{x\rightarrow\infty}\frac{6}{3-\frac{2}{x}}}=e^{\frac{6}{3-0}}=e^2.$$

Answer: $e^2$.

13. $\lim\limits_{x\rightarrow\infty}\frac{\sqrt x-6x}{3x+1}.$

Solution.

$$\lim\limits_{x\rightarrow\infty}\frac{\sqrt x-6x}{3x+1}=\left[\frac{\infty}{\infty}\right]=\lim\limits_{x\rightarrow\infty}\frac{\frac{\sqrt x}{x}-\frac{6x}{x}}{\frac{3x}{x}+\frac{1}{x}}=\lim\limits_{x\rightarrow\infty}\frac{\frac{1}{\sqrt x}-6}{3+\frac{1}{x}}=\frac{-6}{3}=-2.$$

Answer: $-2.$

14.

Solution.

15.

Solution.

Therefore,

.

Hence