Answer
determinate,
30
Work Step by Step
See the table: "Some Determinate and lndeterminate Forms" and$, $after plugging in the value for x, recognize
that $e^{-x }$ results in the determinate form $k^{-\infty}=0 $ when $x\rightarrow\infty,$
where k$>1$, $(k=$e$)$,
which makes the limit equal to
$L=\displaystyle \frac{60+0}{2-0}=30$
