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: 39

Answer

$\dfrac{1}{2}$

Work Step by Step

We want to find $\lim\limits_{x \to 2}\dfrac{\sqrt{x^2+12}-4}{x-2}$, but we can't use the quotient rule for limits because the limit of the denominator as $x$ approaches 2 is zero, and if we try to substitute $x=2$ directly, we get zero in both the numerator and the denominator. But note that $\dfrac{\sqrt{x^2+12}-4}{x-2}=\left(\dfrac{\sqrt{x^2+12}-4}{x-2}\right)\left(\dfrac{\sqrt{x^2+12}+4}{\sqrt{x^2+12}+4}\right)=\dfrac{x^2+12-16}{(x-2)(\sqrt{x^2+12}+4)}=\dfrac{x^2-4}{(x-2)(\sqrt{x^2+12}+4)}=\dfrac{(x-2)(x+2)}{(x-2)(\sqrt{x^2+12}+4)}=\dfrac{x+2}{\sqrt{x^2+12}+4}$ for all $x \neq 2.$ Thus $\lim\limits_{x \to 2}\dfrac{\sqrt{x^2+12}-4}{x-2}=\lim\limits_{x \to 2}\dfrac{x+2}{\sqrt{x^2+12}+4}=\dfrac{2+2}{\sqrt{4+12}+4}=\dfrac{4}{\sqrt{16}+4}=\dfrac{4}{4+4}=\dfrac{4}{8}=\dfrac{1}{2}.$
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