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