Answer
0
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),
$\displaystyle \frac{h^{2}}{h+h^{2}}=\frac{h(h)}{h(1+h)}=\frac{h}{1+h}=g(h)$
$g(0)=0$
(0 is in the domain of g, g is closed-form), so
$\displaystyle \lim_{h\rightarrow 0} f(h)=\displaystyle \lim_{h\rightarrow 0} g(h)=g(h)=0$
