Answer
$=0$
Work Step by Step
This limit is also equal to $$\lim_{x\to\infty} e^{-3x}=\lim_{x\to\infty} \frac{1}{e^{3x}}=\frac{1}{\lim_{x\to\infty} e^{3x}}$$ Since the limit at $e^{3x}$ grows infinitely large, $\frac{1}{\infty}$ should approach $0$.