Answer
$$0$$
Work Step by Step
Given $$\lim _{x \rightarrow \infty} e^{-3 x}$$
Then
\begin{align*}
\lim _{x \rightarrow \infty} e^{-3 x}&=\lim _{x \rightarrow \infty} \frac{1}{e^{3 x}}\\
&=\lim _{x \rightarrow \infty}\frac{1}{\infty}\\
&=0
\end{align*}