Решите пределы пожалуйста!!!!

Решите задачу:

$1. \displaystyle \lim_{n \to \infty} \frac{2n^3 + 7n -2}{3n^3 - n -2} = \lim_{n \to \infty} \frac{n^3(2+ \frac{7}{n^2} - \frac{2}{n^3}) }{n^3(3- \frac{1}{n^2} - \frac{2}{n^3}) } = \\ \\ \\ \frac{ \lim_{n \to \infty} (2+ \frac{7}{n^2} - \frac{2}{n^3 }) }{ \lim_{n \to \infty} (3- \frac{1}{n^2} - \frac{2}{n^3}) } = \frac{2}{3}$

$2. \displaystyle \lim_{x \to -1} \frac{ \sqrt{5+x}-2 }{ \sqrt{8-x}-3 } = \lim_{x \to -1} \frac{(\sqrt{5+x}-2)*(\sqrt{8-x}+3)}{(\sqrt{8-x}-3)(\sqrt{8-x} + 3)} = \\ \\ \\ \lim_{x \to -1} \frac{(\sqrt{5+x}-2)*(\sqrt{8-x}+3)}{-x-1} = \lim_{x \to -1} \frac{(2-\sqrt{5+x})*(\sqrt{8-x}+3)}{x+1} \\ \\ \\ \lim_{x \to -1} \frac{(2-\sqrt{5+x})*(\sqrt{8-x}+3)}{x+1}* \frac{2+ \sqrt{5+x} }{2+ \sqrt{5+x}} = \\ \\ \\ \lim_{x \to -1} \frac{(4 - 5 -x)(\sqrt{8-x}+3)}{(x+1)(2+ \sqrt{5+x})} =$

$\displaystyle \lim_{x \to -1} -\frac{(x+1)(\sqrt{8-x}+3)}{(x+1)(2+ \sqrt{5+x})} = - \lim_{x \to -1} \frac{\sqrt{8-x}+3}{2+\sqrt{5+x}} = \frac{ \sqrt{9}+3 }{2+ \sqrt{4} } = -\frac{3}{2}$