Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Review - Exercises: 13

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: $$\frac{\sqrt{x^2-9}}{x}=\sqrt{\frac{x^2-9}{x^2}}=\sqrt{1-\frac{9}{x^2}}$$ - 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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