Найти предел функции

Решите задачу:

$\displaystyle \lim_{x \to 3} \frac{x^2+x-12}{ \sqrt{3+x} - \sqrt{2x} } =\lim_{x \to 3} \frac{(x+4)(x-3)( \sqrt{3+x}+ \sqrt{2x} ) }{3-x}=\\ \\ \\ =\lim_{x \to 3}-(x+4)( \sqrt{3+x} + \sqrt{2x} )=-14 \sqrt{6}$

$\displaystyle \lim_{x \to 0}\frac{1-\cos 4x}{\sin^27x} = \lim_{x \to 0} \frac{2\sin^22x}{\sin^27x}= \lim_{x \to 0}\frac{2\cdot (2x)^2}{(7x)^2} = \frac{8}{49}$

$\displaystyle \lim_{x \to \infty}\bigg( \frac{7x+3}{7x+5}\bigg)^{7x+4}=\lim_{x \to \infty}\bigg(1- \frac{2}{7x+5}\bigg)^\big{ \frac{7x+5}{2} - \frac{1}{2} } =e^\big{- \frac{1}{2} }$