## 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 - Page 57: 35

#### Answer

$\dfrac{1}{6}$

#### Work Step by Step

We want to find $\lim\limits_{x \to 9}\dfrac{\sqrt{x}-3}{x-9}$, but we can't use the quotient rule for limits because the limit of the denominator as $x$ approaches 9 is zero, and if we try to substitute $x=9$ directly, we get zero in both the numerator and the denominator. But note that $\dfrac{\sqrt{x}-3}{x-9}=\left(\dfrac{\sqrt{x}-3}{x-9}\right)\left(\dfrac{\sqrt{x}+3}{\sqrt{x}+3}\right)=\dfrac{x-9}{(x-9)(\sqrt{x+3})}=\dfrac{1}{\sqrt{x}+3}$ for all $x \neq 9.$ Thus $\lim\limits_{x \to 9}\dfrac{\sqrt{x}-3}{x-9}=\lim\limits_{x \to 9}\dfrac{1}{\sqrt{x}+3}=\dfrac{1}{\sqrt{9}+3}=\dfrac{1}{3+3}=\dfrac{1}{6}.$

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