Помогите,пожалуйста))

Решите задачу:

$\lim\limits _{x \to 1} \frac{x^2+4}{x-2} = \frac{1+4}{1-2} =-5\\\\\lim\limits _{x\to 1} \frac{x^2-1}{x^2-5x+4} =\lim\limits _{x\to 1} \frac{(x-1)(x+1)}{(x-1)(x-4)} =\lim\limits _{x\to 1} \frac{x+1}{x-4} =-\frac{2}{3}\\\\\lim\limits_{x\to \infty } \frac{5x-2}{x^2-x+4} =\lim\limits _{x\to \infty } \frac{\frac{5}{x}-\frac{2}{x^2}}{1-\frac{1}{x}+\frac{4}{x^2}} = \frac{0}{1}=0$

$\lim\limits _{x\to \infty } \frac{x^3+4}{x^2-3} =\lim\limits _{x\to \infty } \frac{1+\frac{4}{x^3}}{\frac{1}{x}-\frac{3}{x^3}} =\frac{1}{0}=\infty$

$\lim_{x \to \infty} \frac{x^2+4x-5}{x^2-1} = \lim_{x \to \infty}\frac{1+\frac{4}{x}-\frac{5}{x^2}}{1-\frac{1}{x^2}} =\frac{1}{1}=1\\\\\\\lim\limits _{\alpha (x)\to 0}\, \frac{1}{\alpha (x)}=\infty \; \; ;\; \; \; \lim\limits _{\alpha (x)\to \infty }\, \frac{1}{\alpha (x)}=0$