Solving limit without L'Hopital's Rule.
Find the limits without using L'Hospital's rule
Solution.
Answer: 0.
2. 
Solution.
Hence, 
.
Therefore
Answer:0.
3. 
Solution.
Answer: 
4. 
Solution.
Answer: 
5. 
Solution.
Answer: 
6. 
Solution.
Answer: -5.
7. 
Solution.
Hence, 
.
Therefore
Answer: 1.
8. 
Solution.
Answer: 36.
9. 
Solution.
Answer: 0.
10. 
Solution.
Answer: 
11. $\lim\limits_{x\rightarrow\infty}\frac{3x^2+2x-1}{4x^3+3x^2+4}.$
Solution.
$$\lim\limits_{x\rightarrow\infty}\frac{3x^2+2x-1}{4x^3+3x^2+4}=\left[\frac{\infty}{\infty}\right]=\lim\limits_{x\rightarrow\infty}\frac{\frac{3x^2}{x^3}+\frac{2x}{x^3}-\frac{1}{x^3}}{\frac{4x^3}{x^3}+\frac{3x^2}{x^3}+\frac{4}{x^3}}=\lim\limits_{x\rightarrow\infty}\frac{\frac{3}{x}+\frac{2}{x^2}-\frac{1}{x^3}}{4+\frac{3}{x}+\frac{4}{x^3}}=\frac{0}{4}=0.$$
Answer: 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.
Answer: 3
15. 
Solution.
Therefore,
.
Hence
Answer: -2.