Answer
determinate,
$-60$
Work Step by Step
See the table: "Some Determinate and lndeterminate Forms" and$, $after plugging in the value for x, recognize
that the term $e^{x} $ in the denominator is of the form $k^{-\infty}=0$ (determinate) where $k=e>1.$
So,
$\displaystyle \lim_{x\rightarrow-\infty}\frac{60}{e^{x}-1}=\frac{60}{0-1}=-60$
