Answer
$0$
Work Step by Step
When $x\rightarrow\infty,$ we have:
Numerator: $\quad 2^{-3x}$ has the form $ k^{Big\ negative }= Small$
so the numerator is a small positive number.
Denominator:$\qquad e^{-x}$ has the form $ k^{Big\ negative }= Small$
so we have:
$1+5.3e^{-x}\rightarrow(1+$ small$)\rightarrow 1$
The limit exists and we have:
$L=\displaystyle \frac{0}{1}=0$
