## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x \to 1}\frac{2-x}{(x-1)^2}=\infty$
$\lim\limits_{x \to 1^+}\frac{2-x}{(x-1)^2}=\frac{1^-}{(0^+)^2}=\frac{1^-}{0^+}=\infty$ $\lim\limits_{x \to 1^-}\frac{2-x}{(x-1)^2}=\frac{1^+}{(0^-)^2}=\frac{1^-}{0^+}=\infty$ Limits from both sides are equal, so $\therefore \lim\limits_{x \to 1}\frac{2-x}{(x-1)^2}=\infty$