University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises - Page 109: 80



Work Step by Step

$$A= \lim_{x\to\infty}(\sqrt{x+9}-\sqrt{x+4})$$ We cannot examine the behavior of $(\sqrt{x+9}-\sqrt{x+4})$ as $x$ approaches $\infty$ right away, because it will lead to the unsolved situation of $\infty-\infty$, which we would try to avoid. Instead, we would want to turn it into a rational function and apply the usual method, by doing the followings: $$A=\lim_{x\to\infty}\Big[(\sqrt{x+9}-\sqrt{x+4})\times\frac{\sqrt{x+9}+\sqrt{x+4}}{\sqrt{x+9}+\sqrt{x+4}}\Big]$$ $$A=\lim_{x\to\infty}\frac{(x+9)-(x+4)}{\sqrt{x+9}+\sqrt{x+4}}=\lim_{x\to\infty}\frac{5}{\sqrt{x+9}+\sqrt{x+4}}$$ Now we can divide both numerator and denominator by the highest degree of $x$ in the denominator, which is $\sqrt x$: $$A=\lim_{x\to\infty}\frac{\frac{5}{\sqrt x}}{\frac{\sqrt{x+9}}{\sqrt x}+\frac{\sqrt{x+4}}{\sqrt x}}=\lim_{x\to\infty}\frac{\frac{5}{\sqrt x}}{\sqrt{1+\frac{9}{x}}+\sqrt{1+\frac{4}{x}}}$$ $$A=\frac{0}{\sqrt{1+0}+\sqrt{1+0}}=0$$
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