Надо решить 4 примера по алгебре (лимиты) 1) lim = x/x^2-x, при x -> 0 2) lim =...

Решите задачу:

$1)\; \; \lim\limits _{x \to 0} \frac{x}{x^2-x} = \lim\limits _{x \to 0} \frac{x}{x(x-1)} = \lim\limits _{x \to 0} \frac{1}{x-1} = \frac{1}{0-1} =-1\\\\2)\; \; \lim\limits _{x \to -3} \frac{x^2+7x+12}{x^2-9} = \lim\limits _{x \to -3} \frac{(x+3)(x+4)}{(x+3)(x-3)} = \lim\limits _{x \to -3} \frac{x+4}{x-3} = \frac{-3+4}{-3-3} =- \frac{1}{6}$

$3)\; \; \lim\limits _{x \to 4} \frac{\sqrt{x}-2}{x-4} = \lim\limits _{x \to 4} \frac{\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)} = \lim\limits_{n \to 4}\frac{1}{\sqrt{x}+2} = \frac{1}{2+2} = \frac{1}{4}$

$4)\; \; \lim\limits _{x \to 6} \frac{x-6}{2-\sqrt{x-2}} = \lim\limits _{x \to 6} \frac{(x-6)(2+\sqrt{x-2})}{4-(x-2)} = \lim\limits _{x \to 6} \frac{(x-6)(2+\sqrt{x-2})}{6-x} =\\\\= -\lim\limits _{x \to 6} (2+\sqrt{x-2})=-(2+2)=-4$