Answer
L'Hospital's rule applies.
limit = $ \displaystyle \frac{3}{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.
= $\displaystyle \lim_{x\rightarrow-\infty}\frac{3x^{2}+10x-1}{2x^{2}-5x} = \displaystyle \lim_{x\rightarrow-\infty}\frac{3x^{2}}{2x^{2}}$
This limit has form $\displaystyle \frac{\infty}{\infty}$, so L'Hospital's rule applies.
$= \displaystyle \lim_{x\rightarrow-\infty}\frac{[3x^{2}]^{\prime}}{[2x^{2}]^{\prime}}= \displaystyle \lim_{x\rightarrow-\infty}\frac{3\cdot 2x}{2\cdot 2x}= \displaystyle \lim_{x\rightarrow-\infty}\frac{3\cdot 2x}{2\cdot 2x}= \displaystyle \lim_{x\rightarrow-\infty}\frac{3x}{2x}$=.
This limit has form $\displaystyle \frac{\infty}{\infty}$, so L'Hospital's rule applies.
$= \displaystyle \lim_{x\rightarrow-\infty}\frac{[3x]^{\prime}}{[2x]^{\prime} }= \displaystyle \lim_{x\rightarrow-\infty}\frac{3}{2 }=\frac{3}{2}$
