Answer
L'Hospital's rule applies,
limit $=0$
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.
$\displaystyle \lim_{x\rightarrow-\infty}\frac{x^{2}+30x}{2x^{6}+10x}$= $\displaystyle \lim_{x\rightarrow-\infty}\frac{x^{2}}{2x^{6}}$
This limit has form $\displaystyle \frac{\infty}{\infty}$, L'Hospital's rule applies. (*)
$= \displaystyle \lim_{x\rightarrow-\infty}\frac{2x}{2(6x^{5}) }= \displaystyle \lim_{x\rightarrow-\infty}\frac{x}{6x^{5} }$
... again, $\displaystyle \frac{\infty}{\infty}$, L'H
$= \displaystyle \lim_{x\rightarrow-\infty}\frac{1}{6(5x^{4}) }= \displaystyle \lim_{x\rightarrow-\infty}\frac{1}{30x^{4} }$
... determinate form $\displaystyle \frac{k}{\pm\infty}$
$=0$
(*) 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.
