Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Section 2.6 - Limits at Infinity; Horizontal Asymptotes - 2.6 Exercises: 22

Answer

$$\lim\limits_{x\to\infty}\frac{x^2}{\sqrt{x^4+1}}=1$$

Work Step by Step

$$A=\lim\limits_{x\to\infty}\frac{x^2}{\sqrt{x^4+1}}$$$$A=\lim\limits_{x\to\infty}\frac{X}{Y}$$ Divide both numerator and denominator by $x^2$, we have: - In the numerator: $X=\frac{x^2}{x^2}=1$ - In the denominator: $Y=\frac{\sqrt{x^4+1}}{x^2}=\frac{\sqrt{x^4+1}}{\sqrt{x^4}}=\sqrt{\frac{x^4+1}{x^4}}=\sqrt{1+\frac{1}{x^4}}$ Therefore, $$A=\lim\limits_{x\to\infty}\frac{1}{\sqrt{1+\frac{1}{x^4}}}$$$$A=\frac{1}{\lim\limits_{x\to\infty}\sqrt{1+\frac{1}{x^4}}}$$$$A=\frac{1}{\sqrt{1+\lim\limits_{x\to\infty}(\frac{1}{x^4})}}$$$$A=\frac{1}{\sqrt{1+0}}$$$$A=1$$
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