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.
$\displaystyle \lim_{x\rightarrow-\infty}\frac{2x^{4}+20x^{3}}{1000x^{3}+6}$ = $\displaystyle \lim_{x\rightarrow-\infty}\frac{2x^{4}}{1000x^{3}}$ = $\displaystyle \lim_{x\rightarrow-\infty}\frac{x^{4}}{500x^{3}}$
This limit has form $\displaystyle \frac{\infty}{\infty}$, L'Hospital's rule applies. (*)
$= \displaystyle \lim_{x\rightarrow-\infty}\frac{[x^{4}]^{\prime}}{[500x^{3}]^{\prime} }= \displaystyle \lim_{x\rightarrow-\infty}\frac{4x^{3}}{1500x^{2} }= \displaystyle \qquad(\frac{\infty}{\infty},\ LH)$
$= \displaystyle \lim_{x\rightarrow-\infty}\frac{4(3x^{2})}{1500(2x) }= \displaystyle \lim_{x\rightarrow-\infty}\frac{x^{2}}{250x }= \displaystyle \qquad(\frac{\infty}{\infty},\ LH)$
$= \displaystyle \lim_{x\rightarrow-\infty}\frac{2x}{250 }$
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.
