Answer
$$0$$
Work Step by Step
Since
$$
\lim _{x \rightarrow \infty} \frac{ x^2 }{e^x}=\frac{\infty}{\infty}
$$
is an intermediate form, then we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow \infty} \frac{ x^2 }{e^x}=\lim _{x \rightarrow \infty} \frac{ 2x }{e^x}=\frac{\infty}{\infty}.
$$
Again, by applying L’Hôpital’s Rule, we get
$$
\lim _{x \rightarrow \infty} \frac{ 2x }{e^x}=\lim _{x \rightarrow \infty} \frac{ 2 }{e^x}=\frac{2}{\infty}=0.
$$
