Answer
$$
\lim _{x \rightarrow-\infty}\left(x^{2}-x^{3}\right) e^{2 x} =0
$$
Work Step by Step
$$
\lim _{x \rightarrow-\infty}\left(x^{2}-x^{3}\right) e^{2 x}
$$
Rewrite this lime as the following:
$$
\lim _{x \rightarrow-\infty}\left(x^{2}-x^{3}\right) e^{2 x} =\lim _{x \rightarrow-\infty} \frac{x^{2}-x^{3}}{e^{-2 x}}
$$
Notice that the limit on the right side is now indeterminate of type $\frac{\infty}{\infty} $ we use l’Hospital’s Rule we have :
l
$$\begin{aligned}
\lim _{x \rightarrow-\infty}\left(x^{2}-x^{3}\right) e^{2 x} &=\lim _{x \rightarrow-\infty} \frac{x^{2}-x^{3}}{e^{-2 x}} \quad\left[\frac{\infty}{\infty} \text { form }\right] \\
&\stackrel{\mathrm{H}}{=} \lim _{x \rightarrow-\infty} \frac{2 x-3 x^{2}}{-2 e^{-2 x}} \quad \left[\frac{\infty}{\infty} \text { form }\right] \\
& \stackrel{\mathrm{H}}{=} \lim _{x \rightarrow-\infty} \frac{2-6 x}{4 e^{-2 x}}\left[\frac{\infty}{\infty} \text { form }\right] \\
& \stackrel{\mathrm{H}}{=} \lim _{x \rightarrow-\infty} \frac{-6}{-8 e^{-2 x}} \\
&=0
\end{aligned}$$
