## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 2 - Limits - 2.7 Limits at Infinity - Exercises - Page 82: 22

#### Answer

The lines $y=\pm\frac{1}{2}$ are the horizontal asymptotes of the given function.

#### Work Step by Step

To find the horizontal asymptotes, we calculate the following limit \begin{align*} \lim _{t \rightarrow \pm \infty}\frac{t^{1/3}}{(64t^2+9)^{1/6}} &=\lim _{t \rightarrow \pm \infty}\frac{t^{1/3}}{\pm t^{1/3}(64+\frac{9}{t^2})^{1/6}}\\ &=\pm\frac{1}{(64+0)^{1/6}}=\pm\frac{1}{2} \end{align*} Hence, the lines $y=\pm\frac{1}{2}$ are the horizontal asymptotes of the given function.

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