Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 3 - Applications of Differentiation - 3.4 Limits at Infinity; Horizontal Asymptotes - 3.4 Exercises - Page 242: 21


$$\frac{1}{6} $$

Work Step by Step

Given $$ \lim _{x \rightarrow \infty}(\sqrt{9 x^{2}+x}-3 x)$$ Then \begin{aligned} \lim _{x \rightarrow \infty}(\sqrt{9 x^{2}+x}-3 x) &=\lim _{x \rightarrow \infty} \frac{(\sqrt{9 x^{2}+x}-3 x)(\sqrt{9 x^{2}+x}+3 x)}{\sqrt{9 x^{2}+x}+3 x}\\ &=\lim _{x \rightarrow \infty} \frac{(\sqrt{9 x^{2}+x})^{2}-(3 x)^{2}}{\sqrt{9 x^{2}+x}+3 x} \\ &=\lim _{x \rightarrow \infty} \frac{\left(9 x^{2}+x\right)-9 x^{2}}{\sqrt{9 x^{2}+x}+3 x}\\ &=\lim _{x \rightarrow \infty} \frac{x}{\sqrt{9 x^{2}+x}+3 x} \cdot \frac{1 / x}{1 / x} \\ &=\lim _{x \rightarrow \infty} \frac{x / x}{\sqrt{9 x^{2} / x^{2}+x / x^{2}}}\\ &=\lim _{x \rightarrow \infty} \frac{1}{\sqrt{9+1 / x}+3}\\ &=\frac{1}{\sqrt{9}+3}=\frac{1}{3+3}=\frac{1}{6} \end{aligned}
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