Question

(lim x стремится к 2)x^2-3x+2/x^2+2x-8

Answer 1

Объяснение:

$\displaystyle\\ \lim_{x \to 2} \frac{x^2-3x+2}{x^2+2x-8} = \lim_{x \to 2} \frac{x^2-2x-x+2}{x^2+4x-2x-8} = \lim_{x \to 2} \frac{x*(x-2)-(x-2)}{x*(x+4)-2*(x+4)}=\\\\\\= \lim_{x \to 2} \frac{(x-2)*(x-1)}{(x+4)*(x-2)}= \lim_{n \to \infty} \frac{x-1}{x+4}=\frac{2-1}{2+4} =\frac{1}{6}.$