Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Review - Exercises: 14

Answer

$$\lim\limits_{x\to-\infty}\frac{\sqrt{x^2-9}}{2x-6}=-\frac{1}{2}$$

Work Step by Step

$$A=\lim\limits_{x\to-\infty}\frac{\sqrt{x^2-9}}{2x-6}$$ Divide both numerator and denominator by $x$ (the highest power of $x$ in the denominator), we have: - Numerator: Here since $x\to-\infty$, we only consider the values of $x\lt0$. So, $\sqrt{x^2}=|x|=-x$. Which means $x=-\sqrt{x^2}$ Therefore, $$\frac{\sqrt{x^2-9}}{x}=-\sqrt{\frac{x^2-9}{x^2}}=-\sqrt{1-\frac{9}{x^2}}$$ - Denominator: $$\frac{2x-6}{x}=2-\frac{6}{x}$$ Therefore, $$A=\lim\limits_{x\to\infty}\frac{-\sqrt{1-\frac{9}{x^2}}}{2-\frac{6}{x}}$$$$A=-\frac{\sqrt{1-9\times0}}{2-6\times0}$$$$A=-\frac{1}{2}$$
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