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 108: 28


$$\lim_{x\to\infty}\frac{2+\sqrt x}{2-\sqrt x}=-1$$

Work Step by Step

We would treat these exercises just like we treat the limits of rational functions: we would divide both numerator and denominator by the highest power of $x$ in the denominator. Also remember that $$\lim_{x\to\infty}\frac{a}{x^n}=\lim_{x\to-\infty}\frac{a}{x^n}=a\times0=0\hspace{1cm}(a\in R)$$ $$A=\lim_{x\to\infty}\frac{2+\sqrt x}{2-\sqrt x}$$ The highest power of $x$ in the denominator here is $\sqrt x$, so we divide both numerator and denominator by $\sqrt x$: $$A=\lim_{x\to\infty}\frac{\frac{2}{\sqrt x}+1}{\frac{2}{\sqrt x}-1}=\lim_{x\to\infty}\frac{\frac{2}{x^{1/2}}+1}{\frac{2}{x^{1/2}}-1}=\frac{\lim_{x\to\infty}\frac{2}{x^{1/2}}+\lim_{x\to\infty}1}{\lim_{x\to\infty}\frac{2}{x^{1/2}}-\lim_{x\to\infty}1}$$ $$A=\frac{0+1}{0-1}=\frac{1}{-1}=-1$$
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