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*}
Then $f(x)$ has a horizontal asymptote of $y = 2$.
