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: 41



Work Step by Step

We want to find $\lim\limits_{x \to -3}\dfrac{2-\sqrt{x^2-5}}{x+3}$, but we can't use the quotient rule for limits because the limit of the denominator as $x$ approaches -3 is zero, and if we try to substitute $x=-3$ directly, we get zero in both the numerator and the denominator. But note that $\dfrac{2-\sqrt{x^2-5}}{x+3}=\left(\dfrac{2-\sqrt{x^2-5}}{x+3}\right)\left(\dfrac{2+\sqrt{x^2-5}}{2+\sqrt{x^2-5}}\right)=\dfrac{4-(x^2-5)}{(x+3)(2+\sqrt{x^2-5})}=\dfrac{4-x^2+5}{(x+3)(2+\sqrt{x^2-5})}=\dfrac{9-x^2}{(x+3)(2+\sqrt{x^2-5})}=\dfrac{-(x^2-9)}{(x+3)(2+\sqrt{x^2-5})}=\dfrac{-(x-3)(x+3)}{(x+3)(2+\sqrt{x^2-5})}=\dfrac{-(x-3)}{2+\sqrt{x^2-5}}$ for all $x \neq -3.$ Thus $\lim\limits_{x \to -3}\dfrac{2-\sqrt{x^2-5}}{x+3}=\lim\limits_{x \to -3}\dfrac{-(x-3)}{2+\sqrt{x^2-5}}=\dfrac{-(-3-3)}{2+\sqrt{9-5}}=\dfrac{-(-6)}{2+\sqrt{4}}=\dfrac{6}{2+2}=\dfrac{6}{4}=\dfrac{3}{2}.$
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