Answer
$0$
Work Step by Step
We have
$$
\lim _{x \rightarrow \infty}e^{-x}(x^3-x^2+9)=\lim _{x \rightarrow \infty}\frac{x^3-x^2+9}{e^x}=\frac{\infty}{\infty}
$$
which is an intermediate form, so we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow \infty}\frac{x^3-x^2+9}{e^x}\\=\lim _{x \rightarrow \infty}\frac{3x^2-2x}{e^x}=\frac{\infty}{\infty}$$
which is an intermediate form, so we can apply L’Hôpital’s Rule as follows
$$\lim _{x \rightarrow \infty}\frac{3x^2-2x}{e^x}=\lim _{x \rightarrow \infty}\frac{6x-2}{e^x}=\frac{0}{1+0}=0.
$$
which is an intermediate form, so we can apply L’Hôpital’s Rule as follows
$$\lim _{x \rightarrow \infty}\frac{6x-2}{e^x}=\lim _{x \rightarrow \infty}\frac{6}{e^x}=\frac{6}{\infty}=0.
$$
