Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - Chapter Review Exercises - Page 95: 63

Answer

$y = 1$

Work Step by Step

Since \begin{align*} \lim _{t \rightarrow \infty} \frac{t^{1 / 3}-t^{-1 / 3}}{\left(t-t^{-1}\right)^{1 / 3}}&=\lim _{t \rightarrow \infty} \frac{1-t^{-2 / 3}}{\left(1-t^{-2}\right)^{1 / 3}}\\ &= \frac{1}{1^{1 / 3}}=1 \end{align*} and \begin{align*} \lim _{t \rightarrow-\infty} \frac{t^{1 / 3}-t^{-1 / 3}}{\left(t-t^{-1}\right)^{1 / 3}}&=\lim _{t \rightarrow-\infty} \frac{1-t^{-2 / 3}}{\left(1-t^{-2}\right)^{1 / 3}}\\ &=\frac{1}{1^{1 / 3}}=1 \end{align*} Then $f(x)$ has a horizontal asymptote of $y = 1$.
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