Answer
indeterminate,
0
Work Step by Step
See the table: "Some Determinate and lndeterminate Forms" and$, $after plugging in the value for x, recognize
that the limit is initially in the form
$\displaystyle \frac{\infty}{\infty}$ (indeterminate)
Applying the "Strategy for Evaluating Limits Algebraically", case 2 of step 2,
after simplifying (reducing) with $x^{3},$ the limit becomes
$\displaystyle \lim_{x\rightarrow-\infty}\frac{-1}{3x^{3}}.$
(when x$\rightarrow-\infty$,
$ x^{3}\rightarrow-\infty$
$3x^{3}\rightarrow-\infty $
$\displaystyle \frac{-1}{3x^{3}}\rightarrow 0$, because it is of the determinate form $ \displaystyle \pm\frac{k}{\infty}=0)$
so, $\displaystyle \lim_{x\rightarrow-\infty}\frac{-1}{3x^{3}}$=0
