Answer
-1
Work Step by Step
1. f is a closed function (we know by Th.10.1 that it is continuous), that is, L= $\displaystyle \lim_{x\rightarrow a}f(x)$ = $f(a)$,
for all a from the domain of f.
2. evaluating: $f(-1)$, (plugging $x=-1$) results in an indeterminate form 0/0.
So, we see that $x=-1$ is NOT in the domain of f.
We apply case 2 (of step 2). Simplifying $f(x)$,
$\displaystyle \frac{x^{2}+3x+1}{x^{2}+x}=$... factor...
$=\displaystyle \frac{(x+1)(x+2)}{x(x+1)}=$...reduce...$=\displaystyle \frac{x+2}{x}=g(x)$
$g(-1)=\displaystyle \frac{-1+2}{-1}=-1$
($-1$ is in the domain of g, g is closed-form), so
$\displaystyle \lim_{x\rightarrow-1}f(x)=\lim_{x\rightarrow-1}g(x)=g(-1)=-1$
