Answer
L'Hospital's rule applies.
limit = $-1$
Work Step by Step
Theorem 11.3 L'Hospital's Rule:
If $f$ and $g$ are differentiable functions such that
substituting $x=a$ in the expression $\displaystyle \frac{f(x)}{g(x)}$ gives
the indeterminate form $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$, then $\displaystyle \lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a} \displaystyle \frac{f^{\prime}(x)}{\mathrm{g}^{\prime}(x)}$.
That is, we can replace $f(x)$ and $g(x)$ with their derivatives and try again to take the limit.
--------------------
When x approaches $-1$,
the numerator approaches 0,
the denominator approaches 0,
$\displaystyle \frac{f(x)}{g(x)}$ gives the indeterminate form $\displaystyle \frac{0}{0}$,
L'Hospital's rule applies.
$\displaystyle \lim_{x\rightarrow-1}\frac{x^{2}+3x+2}{x^{2}+x}=\lim_{x\rightarrow-1} \displaystyle \frac{[x^{2}+3x+2]^{\prime}}{[x^{2}+x]^{\prime}}$
$=\displaystyle \lim_{x\rightarrow 1} \displaystyle \frac{2x+3}{2x+1}=$ .
$=\displaystyle \frac{2(-1)+3}{2(-1)+1}=\frac{1}{-1}=-1$
