Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.2 - Limit of a Function and Limit Laws - Exercises 2.2 - Page 57: 42



Work Step by Step

We want to find $\lim\limits_{x \to 4}\dfrac{4-x}{5-\sqrt{x^2+9}}$, but we can't use the quotient rule for limits because the limit of the denominator as $x$ approaches 4 is zero, and if we try to substitute $x=4$ directly, we get zero in both the numerator and the denominator. But note that $\dfrac{4-x}{5-\sqrt{x^2+9}}=\left(\dfrac{4-x}{5-\sqrt{x^2+9}}\right)\left(\dfrac{5+\sqrt{x^2+9}}{5+\sqrt{x^2+9}}\right)=\dfrac{(4-x)(5+\sqrt{x^2+9})}{25-(x^2+9)}=\dfrac{(4-x)(5+\sqrt{x^2+9})}{25-x^2-9}=\dfrac{(4-x)(5+\sqrt{x^2+9})}{16-x^2}=\dfrac{(4-x)(5+\sqrt{x^2+9})}{(4-x)(4+x)}=\dfrac{5+\sqrt{x^2+9}}{4+x}$ for all $x \neq 4.$ Thus $\lim\limits_{x \to 4}\dfrac{4-x}{5-\sqrt{x^2+9}}=\lim\limits_{x \to 4}\dfrac{5+\sqrt{x^2+9}}{4+x}=\dfrac{5+\sqrt{16+9}}{4+4}=\dfrac{5+\sqrt{25}}{8}=\dfrac{5+5}{8}=\dfrac{10}{8}=\dfrac{5}{4}.$
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