Нужно решить 29 задание

Решите задачу:

$1)\; \; \lim\limits _{x \to -2} \frac{3x^2+5x-8}{2x^2+3x-5}= \frac{3\cdot 4-5\cdot 2-8}{2\cdot 4-3\cdot 2-5} = \frac{-6}{-3} =2\\\\ \lim\limits _{x \to 1} \frac{3x^2+5x-8}{2x^2+3x-5}= \lim\limits _{x \to 1}\frac{3(x-1)(x+\frac{8}{3})}{2(x-1)(x+\frac{5}{2})} = \lim\limits _{x \to 1} \frac{3+8x}{2x+5} = \frac{11}{7}\\\\ \lim\limits _{x \to \infty } \frac{3x^2+5x-8}{2x^2+3x-5}= \lim\limits _{x \to \infty} \frac{3+\frac{5}{x}-\frac{8}{x^2}}{2+\frac{3}{x}-\frac{5}{x^2}} = \frac{3}{2}$

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

$3)\; \; \lim\limits _{x \to 0} \frac{sin3x}{tg2x}= \lim\limits _{x \to 0}\frac{3x}{2x}= \frac{3}{2}\\\\4)\; \; \lim\limits _{n \to \infty}\Big (\frac{5n-3}{5n+4}\Big )^{n+4}= \lim\limits _{n \to \infty}\Big (\underbrace {\Big (1+\frac{-7}{5n+4}\Big )^{ \frac{5n+4}{-7}}}_{e}\Big )^{ \frac{-7(n+4)}{5n+4} }=\\\\=e^{ \lim\limits _{n \to \infty} \frac{-7n-28}{5n+4} }=e^{-\frac{7}{5}}$