Помогите с предалами,даю 50 баллов

[latex]1)\quad \lim\limits _{x \to 0} \frac{x+2}{sinx} =\frac{0+2}{sin0}=\left [\; \frac{2}{0}\; \right ]=\infty \\\\2)\quad \lim\limits _{x \to 2} \frac{x^2-3x+2}{\sqrt{5-x}-\sqrt{x+1}} = \lim\limits_{x \to 2} \frac{(x-1)(x-2)(\sqrt{5-x}+\sqrt{x+1})}{(\sqrt{5-x}-\sqrt{x+1})(\sqrt{5-x}+\sqrt{x+1})} =\\\\= \lim\limits _{x \to 2} \frac{(x-1)(x-2)(\sqrt{5-x})(\sqrt{x+1})}{5-x-(x+1)} =\lim\limits _{x \to 2} \frac{(x-1)(x-2)(\sqrt{5-x}+\sqrt{x+1})}{-2(x-2)}=[/latex]    [latex]=\lim\limits _{x\to 2}\frac{(x-1)(\sqrt{5-x}+\sqrt{x+1})}{-2}=\frac{\sqrt3+\sqrt3}{-2}=-\frac{2\sqrt3}{2}=-\sqrt3[/latex] [latex]3)\quad \lim\limits _{x \to \infty} (\sqrt{x^3+3}-2\sqrt{x^2+1})=\\\\=\lim\limits _{x\to \infty } \frac{(\sqrt{x^3+3}-2\sqrt{x^2+1})(\sqrt{x^3+3}+2\sqrt{x^2+1})}{\sqrt{x^3+3}+2\sqrt{x^2+1}} =\lim\limits _{x\to \infty }\frac{x^3+3-4(x^2+1)}{\sqrt{x^2+3}+2\sqrt{x^2+1}}=\\\\=\lim\limits _{x\to \infty }\frac{x^3}{x^{3/2}}=\lim\limits _{x\to \infty }x^{3/2}=\infty [/latex] [latex]4)\quad \lim\limits _{x \to \infty} \left (\frac{x+3-1+1}{x+2}\right )^{x}= \lim\limits _{x \to \infty} \left (\frac{(x+2)+1}{x+2}\right )^{x}=\\\\=\lim\limits _{x\to \infty }\left (\left (1+\frac{1}{x+2}\right )^{x+2}\right )^{\frac{x}{x+2}}=e^{\lim\limits _{x\to \infty }\frac{x}{x+2}}=e^1=e[/latex] [latex]5)\quad \lim\limits _{x \to 4}(5-x)^{\frac{3x}{x-4}}=\lim\limits _{x\to 4}(1+(4-x))^{\frac{3x}{x-4}} =\\\\=\lim\limits _{x\to 4}\left (\left (1+\underbrace {(4-x)}_{\to 0 }\right )^{\frac{1}{4-x}}\right )^{\frac{(4-x)3x}{-(4-x)}}=e^{\lim\limits _{x\to 4}(-3x)}=e^{-12}[/latex]