Answer
$g(x)$ approaches $\infty$ and zero.
Work Step by Step
The end behavior for $e^{-2x}$ on $(-\infty,+\infty)$ is given by $$\lim\limits_{x \to -\infty} e^{-2x}=e^{\infty}=\infty$$, i.e. $g(x)$ increases without bound and$$\lim\limits_{x \to \infty} e^{-2x}=\frac{1}{e^{\infty}}=0$$, i.e. $g(x)$ approaches horizontal asymptote $y=0$.
