Ну,пожалуйста,решите

Решите задачу:

$\displaystyle 1)\lim_{n\to\infty} \frac{2n^2-n-10}{n^2+3n+2} = \lim_{n\to\infty} \frac{2-\frac{1}{n} - \frac{10}{n^2}}{1+\frac{3}{n}+\frac{2}{n^2}} = \frac{2}{1} = 2 \\ \\ 2)\lim_{x\to-1}\frac{\sqrt{3-x} - \sqrt{5+x}}{x+1} = \lim_{x\to-1}\frac{(3-x)-(5+x)}{(x+1)(\sqrt{3-x} + \sqrt{5+x})}= \\\\ =\lim_{x\to-1}\frac{-2(x+1)}{(x+1)(\sqrt{3-x} + \sqrt{5+x})} = \lim_{x\to-1}\frac{-2}{\sqrt{3-x} + \sqrt{5+x}} = \\ \\ = \frac{-2}{2+2} = -0.5\\\\$
$\displaystyle 3) \lim_{x\to2}\frac{x^2-x-2}{x-2}=\lim_{x\to2}\frac{(x-2)(x+1)}{x-2} = \lim_{x\to2}x+1 = 2+1 = 3\\\\ 4) \lim_{x\to0}\frac{2x^2}{\sin^24x}= \frac{1}{8}\lim_{x\to0}\frac{16x^2}{\sin^24x}=\frac{1}{8}\lim_{x\to0}\left(\frac{4x}{\sin4x}\right)^2 = \frac{1}{8}$