Lim (x^3-x^2+2x)/(x^2+x), x->0 lim (x^3-3x^2+2)/(x^2-7x+6), x->1

Решите задачу:

$\lim_{x\to0} \frac{x^3-x^2+2x}{x^2+x}=\left\{\frac{0}{0}\right\}=\lim_{x\to0} \frac{x(x^2-x+2)}{x(x+1)}=\lim_{x\to0} \frac{x^2-x+2}{x+1}=\\ =\frac{2}{1}=2.\\ \lim_{x\to1}\frac{x^3-3x^2+2}{x^2-7x+6}=\left\{\frac{0}{0}\right\}=\lim_{x\to1}\frac{(x-1)(x^2-2x-2)}{(x-1)(x-6)}=\\ =\lim_{x\to1}\frac{x^2-2x-2}{x-6}=\frac{-3}{-5}=\frac{3}{5}=0,6.$