## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 2: Limits and Continuity - Section 2.2 - Limit of a Function and Limit Laws - Exercises 2.2: 31

#### Answer

$-1$

#### Work Step by Step

We want to find $\lim\limits_{x \to 1}\dfrac{\frac{1}{x}-1}{x-1}$, but we can't use the quotient rule for limits because the limit of the denominator as $x$ approaches 1 is zero, and if we try to substitute $x=1$ directly, we get zero in both the numerator and the denominator. Note that $\dfrac{\frac{1}{x}-1}{x-1}=\left(\dfrac{\frac{1}{x}-1}{x-1}\right)\left(\dfrac{x}{x}\right)=\dfrac{1-x}{x(x-1)}=\dfrac{-(x-1)}{x(x-1)}=\dfrac{-1}{x}$ for all $x \neq 1.$ Thus $\lim\limits_{x \to 1}\dfrac{\frac{1}{x}-1}{x-1}=\lim\limits_{x \to 1}\dfrac{-1}{x}=\dfrac{-1}{1}=-1.$

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