Answer
L'Hospital's rule applies.
limit = $ 2$
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.
$\displaystyle \lim_{x\rightarrow-\infty}\frac{6x^{2}+5x+10}{3x^{2}-9} = \displaystyle \lim_{x\rightarrow-\infty}\frac{6x^{2}}{3x^{2}}$
This limit has form $\displaystyle \frac{\infty}{\infty}$, so L'Hospital's rule applies (*),
$= \displaystyle \lim_{x\rightarrow-\infty}\frac{6x^{2}}{3x^{2}}$ = $= \displaystyle \lim_{x\rightarrow-\infty}\frac{12x}{6x}$
again, the limit has form $\displaystyle \frac{\infty}{\infty}$, so L'Hospital's rule applies
$=\displaystyle \frac{12}{6} =2$
(*) We did not need to apply the LH rule here.
We could have just reduced the $x^{2}$ factors and arrived at the same answer.