Найти указанные пределы. Задания во вложениях

Решите задачу:

$1)\; \; \lim\limits _{x \to \infty} \frac{3x-x^6}{x^2-2x+5}=\Big [\frac{\infty }{\infty }\Big ]= \Big [\frac{:n^6}{:n^6}\Big ]=\lim\limits _{x \to \infty} \frac{\frac{3}{x^5}-1}{\frac{1}{x^4}-\frac{2}{x^5}+\frac{5}{x^6}}=\\\\=\Big [\frac{0-1}{0-0+0}=\frac{-1}{0}\Big ]= \infty \\\\2)\; \; \lim\limits _{x \to -\infty} \frac{2x^2-x+7}{3x^4-5x^2+10} = \lim\limits _{x \to -\infty} \frac{\frac{2}{x^2}-\frac{1}{x^3}+\frac{7}{x^4}}{3-\frac{5}{x^2}+\frac{10}{x^4}} =\frac{0}{3}=0$

$3)\; \; \lim\limits _{x \to -2} \frac{\sqrt{2-x}-\sqrt{x+6}}{x^2-x+6} =\Big [\frac{0}{0}\Big ]= \lim\limits _{x \to -2} \frac{(\sqrt{2-x}-\sqrt{x+6})(\sqrt{2-x}+\sqrt{x+6})}{(x+2)(x-3)(\sqrt{2-x}+\sqrt{x+6})} =\\\\= \lim\limits _{x \to -2} \frac{(2-x)-(x+6)}{(x+2)(x-3)(\sqrt{2-x}+\sqrt{x+6})} = \lim\limits _{x \to -2} \frac{-2(x+2)}{(x+2)(x-3)(\sqrt{2-x}+\sqrt{x+6})} =\\\\= \lim\limits _{x \to -2} \frac{-2}{(x-3)(\sqrt{2-x}+\sqrt{x+6})} =\frac{-2}{-5(\sqrt4+\sqrt4)}=\frac{1}{10}=0,1$