## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 2 - Limits - Chapter Review Exercises - Page 95: 60

#### Answer

$y = 2$

#### Work Step by Step

Since \begin{align*} \lim _{u \rightarrow \infty} \frac{2 u^{2}-1}{\sqrt{6+u^{4}}}&=\lim _{u \rightarrow \infty} \frac{2-1 / u^{2}}{\sqrt{6 / u^{4}+1}}\\ &=\frac{2}{\sqrt{1}}=2 \end{align*} and \begin{align*} \lim _{u \rightarrow-\infty} \frac{2 u^{2}-1}{\sqrt{6+u^{4}}}&=\lim _{u \rightarrow-\infty} \frac{2-1 / u^{2}}{\sqrt{6 / u^{4}+1}}\\ &=\frac{2}{\sqrt{1}}=2 \end{align*} We see that $f(x)$ has has a horizontal asymptote of $y = 2$.

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