Answer
$\displaystyle \frac{1}{2}$
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: $f(0)$, (plugging h=0) results in an indeterminate form 0/0.
So, we see that h=0 is NOT in the domain of f.
We apply case 2 (of step 2). Simplifying f(h),
$\displaystyle \frac{h^{2}+h}{h^{2}+2h}=\frac{h(h+1)}{h(h+2)}=\frac{h+1}{h+2}=g(h)$
$g(0)=\displaystyle \frac{1}{2}$
(0 is in the domain of g, g is closed-form), so
$\displaystyle \lim_{h\rightarrow 0}f(h)=\lim_{h\rightarrow 0}g(h)=g(h)=\frac{1}{2}$
