1) 2)

Решите задачу:

$\displaystyle \lim_{x \to 0} \frac{2-\sqrt{2x+4}}{x}= \lim_{x \to 0} \frac{(2-\sqrt{2x+4})(2+\sqrt{2x+4})}{x(2+\sqrt{2x+4})}=\\\\=\lim_{x \to 0} \frac{4-(2x+4)}{x(2+\sqrt{2x+4})}=\lim_{x \to 0} \frac{-2x}{x(2+\sqrt{2x+4})}=\\\\=\lim_{x \to 0} \frac{-2}{(2+\sqrt{2x+4})}=\frac{-2}{2+\sqrt4}=\frac{-2}{2+2}=-\frac{1}2$

$\displaystyle \lim_{n \to -2} \frac{2x+x^2}{x^2+5x+6}= \left[\begin{array}{ccc}x^2+5x+6=0\\D=25-4*6=25-24=1\\x_1=\frac{-5+1}{2}=-2\\x_2=\frac{-5-1}2=-3\end{array}\right] =\\\\\\ \lim_{n \to -2} \frac{x(x+2)}{(x+2)(x+3)}=\lim_{n \to -2}\frac{x}{x+3}=\frac{-2}{-2+3}=-2$