Question

lim=(х^2-х-2)/(1-х^2)
х (стремится) -1

Answer 1

$\lim_{x \to -1} \frac{x^2-x-2}{1-x^2} = \left[\begin{array}{ccc} \frac{0}{0}\end{array}\right]= \lim_{x \to -1}\frac{(x+1)(x-2)}{(1-x)(1+x)}= \lim_{x \to -1}\frac{x-2}{1-x}=\\=\frac{-1-2}{1+1}=\frac{-3}2=-1.5$