Question

Определите пределы: 2lim x->1 (2-x) -1 0 ---------- = -- x-1 0

Answer 1

$\lim_{x \to 1} \frac{(2-x)^2-1}{x-1}= \lim_{x \to 1} \frac{4-4x+x^2-1}{x-1} = \lim_{x \to 1} \frac{x^2-4x+3}{x-1} =\\\\ \lim_{x \to 1} \frac{(x^2-2x+1)-2x+2}{x-1}= \lim_{x \to 1} \frac{(x-1)^2-2(x-1)}{x-1}= \lim_{x \to 1} ( \frac{(x-1)^2}{x-1}- \frac{2(x-1)}{x-1})=\\\\= \lim_{x \to 1} (x-1-2)= \lim_{x \to 1} (x-3)=1-3=-2$