Answer
$\displaystyle \frac{2}{5}$
Work Step by Step
when $x\rightarrow+\infty,$
$-3x\rightarrow-\infty\ \ \ ,$ (determinate form $ k(\pm\infty)=\pm\infty$)
$e^{-3x}\rightarrow 0\ \ \ $ (determinate form $k^{-\infty}=0$, k$=e>1$)
$5.3e^{-3x}\rightarrow 0$
$5-5.3e^{-3x}\rightarrow 5-0=5$
$\displaystyle \lim_{x\rightarrow+\infty}f(x)=\frac{2}{5-0}=\frac{2}{5}$
