Answer
L'Hospital's rule applies,
the limit diverges to $-\infty$
Work Step by Step
By Theorem 10.2 (see section 10-3 ),
we can calculate this limit by ignoring all powers of $x$ except the highest in both the numerator and denominator.
$\displaystyle \lim_{x\rightarrow-\infty}\frac{x^{3}-100}{2x^{2}+500}$= $\displaystyle \lim_{x\rightarrow-\infty}\frac{x^{3}}{2x^{2}}$
This limit has form $\displaystyle \frac{\infty}{\infty}$, L'Hospital's rule applies. (*)
$= \displaystyle \lim_{x\rightarrow-\infty}\frac{[x^{3}]^{\prime}}{[2x^{2}]^{\prime} }= \displaystyle \lim_{x\rightarrow-\infty}\frac{3x^{2}}{2\cdot 2x }=$
... again, $\displaystyle \frac{\infty}{\infty}$, L'H
$= \displaystyle \lim_{x\rightarrow-\infty}\frac{3(2x)}{2(2) }= \displaystyle \lim_{x\rightarrow-\infty}\frac{3x}{ 2}$
diverges to $-\infty $
(*) We did not need to apply the LH rule here.
We could have just reduced the $x^{3}$ factors and arrived at the same answer.