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: 37

Answer

4

Work Step by Step

We want to find $\lim\limits_{x \to 1}\dfrac{x-1}{\sqrt{x+3}-2}$, 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. But note that $\dfrac{x-1}{\sqrt{x+3}-2}=\left(\dfrac{x-1}{\sqrt{x+3}-2}\right)\left(\dfrac{\sqrt{x+3}+2}{\sqrt{x+3}+2}\right)=\dfrac{(x-1)(\sqrt{x+3}+2)}{x+3-4}=\dfrac{(x-1)(\sqrt{x+3}+2)}{x-1}=\sqrt{x+3}+2$ for all $x \neq 1.$ Thus $\lim\limits_{x \to 1}\dfrac{x-1}{\sqrt{x+3}-2}=\lim\limits_{x \to 1}\sqrt{x+3}+2=\sqrt{1+3}+2=\sqrt{4}+2=2+2=4.$
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