Question

Помогите найти пределы.

Answer 1

$\displaystyle \lim_{x\to-1}\frac{7x^2+8x+1}{2x+2}=\frac{0}{0}=\lim_{x\to-1}\frac{(x+1)(7x+1)}{2(x+1)}=\lim_{x\to-1}\frac{7x+1}{2}=-3$

$\displaystyle \lim_{x\to\infty}(x-\sqrt{x^2+3x})=\infty-\infty=\displaystyle \lim_{x\to\infty}\frac{-3x}{(x+\sqrt{x^2+3x})}=\frac{\infty}{\infty}=\\=\lim_{x\to\infty}\frac{-3x}{x(1+\sqrt{1+\frac{3}{x}_{\to0}})}=-\frac{3}{2}$

$\displaystyle \lim_{x\to0}\frac{tg6x}{tg3x}=\frac{0}{0}=\lim_{x\to0}\frac{6x}{3x}=2$

$\displaystyle \lim_{x\to\infty}(\frac{5x+2}{5x-3})^{2x+1}=1^\infty=\lim_{x\to\infty}(\frac{5x-3+5}{5x-3})^{2x+1}=\\=[\lim_{x\to\infty}(1+\frac{5}{5x-3})^\frac{5x-3}{5}]^{\frac{5(2x+1)}{5x-3}}=e^{\displaystyle5\lim_{x\to\infty}\frac{2x+1}{5x-3}}=e^2$