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.
