Answer
$$
\lim _{x \rightarrow \infty} x^{3} e^{-x^{2}}=0
$$
Work Step by Step
$$
\lim _{x \rightarrow \infty} x^{3} e^{-x^{2}}
$$
We can deal with it by writing the product $ x^{3} e^{-x^{2}} $ as a quotient $\frac{x^{3} }{e^{x^{2}} }$.
Since
$$
\lim _{x \rightarrow \infty} ( x^{3})=\infty
$$
and
$$
\lim _{x \rightarrow \infty } (e^{x^{2} })=\infty
$$
So, we find that, this convert the given limit into an indeterminate form of type $\frac{\infty}{\infty }$ and we can apply l’Hospital’s Rule:
$$
\begin{aligned}
\lim _{x \rightarrow \infty} x^{3} e^{-x^{2}} & =\lim _{x \rightarrow \infty} \frac{x^{3}}{e^{x^{2}}} \rightarrow \frac{\infty}{\infty} \\
& \stackrel{\mathrm{H}}{=} \lim _{x \rightarrow \infty} \frac{3 x^{2}}{2 x e^{x^{2}}} \\
&=\lim _{x \rightarrow \infty} \frac{3 x}{2 e^{x^{2}}} \rightarrow \frac{\infty}{\infty} \\
& \stackrel{\mathrm{H}}{=} \lim _{x \rightarrow \infty} \frac{3}{4 x e^{x^{2}}} \\
&=0
\end{aligned}
$$
