Answer
$0$
Work Step by Step
When $x\rightarrow-\infty,$ we have:
Numerator: $\quad 3^{2x}$ has the form $ k^{Big\ negative }= Small$
so the numerator is a small negative number.
Denominator:$\qquad e^{x}$ has the form $ k^{Big\ negative }= Small$,
so, $2+e^{x}\rightarrow(2+$ small$)\rightarrow 2$
The limit exists and
$L=\displaystyle \frac{-0}{2}=0$
