Ответ:

a) [tex]\displaystyle +\infty[/tex]

в) [tex]\displaystyle -\frac{1}{2}[/tex]

Объяснение:

a)

[tex]\displaystyle \lim_{x\to +\infty } \frac{x^2-3x-5}{x+1}= \left[\frac{\infty}{\infty} \right] =\lim_{x\to +\infty}\frac{(x^2-3x-5)'}{(x+1)'}=\lim_{x\to +\infty}\frac{2x-3}{1}=+\infty[/tex]

в)

[tex]\displaystyle \lim_{x\to 0} \frac{3x^2+x}{ln(1-2x)}= \left[\frac{0}{0} \right] =\lim_{x\to 0} \frac{(3x^2+x)'}{(ln(1-2x))'}=\\\\\\ \lim_{x\to 0} \frac{6x+1}{\frac{1}{1-2x}\cdot (1-2x)'}=\lim_{x\to 0} \frac{6x+1}{\frac{1}{1-2x}\cdot (-2)}=\\\\\\ \lim_{x\to 0} \frac{6x+1}{-\frac{2}{1-2x}}=\lim_{x\to 0} \frac{-(6x+1)(1-2x)}{2}=\\\\\\ \frac{-(6\cdot 0+1)\cdot (1-2\cdot 0)}{2}=\frac{-(0+1)\cdot (1-0)}{2}=\frac{-1\cdot 1}{2}=\frac{-1}{2}=-\frac{1}{2}[/tex]