Answer
There will not be any horizontal asymptote.
Work Step by Step
Write $h\left( x \right)$ in the form $\frac{p\left( x \right)}{q\left( x \right)}$.
$\begin{align}
& h\left( x \right)=\frac{15{{x}^{3}}}{3{{x}^{2}}+1} \\
& =\frac{p\left( x \right)}{q\left( x \right)}
\end{align}$
To find the horizontal asymptotes, divide the numerator and denominator by the highest degree of x, which is 3.
$\begin{align}
& h\left( x \right)=\frac{15{{x}^{3}}}{3{{x}^{2}}+1} \\
& =\frac{\frac{15{{x}^{3}}}{{{x}^{2}}}}{\frac{3{{x}^{2}}}{{{x}^{2}}}+\frac{1}{{{x}^{2}}}}
\end{align}$
Now, when x tends to infinity, $\frac{1}{x}$ tends to zero.
$\begin{align}
& g\left( x \right)=\frac{15{{x}^{3}}}{3{{x}^{2}}+1} \\
& =\frac{\frac{15{{x}^{3}}}{{{x}^{2}}}}{\frac{3{{x}^{2}}}{{{x}^{3}}}+\frac{1}{{{x}^{3}}}} \\
& =\frac{15x}{\frac{3}{x}+\frac{1}{{{x}^{3}}}}
\end{align}$
It can be easily observed that, the highest degree of x is in the numerator. Due to this, when x tends to infinity, the numerator also tends to infinity.
Therefore, there is no horizontal asymptote.
