## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 2 - Section 2.2 - The Limit of a Function - 2.2 Exercises: 33

#### Answer

$\lim\limits_{x \to 1}\frac{2-x}{(x-1)^2}=\infty$

#### Work Step by Step

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

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