basic application of limits

Divide and multiply the limit by (2h+h^2) and (h-h^2). By putting(2h+h^2) tends to 0 in numerator and (h-h^2) in denominator, we get (f'(2)/f'(1))(2h+h^2)/(h-h^2)=6x2/4=3

We have the following limit $\lim_{h \to 0} \frac {f(2h+2+h^2)-f(2)}{f(h-h^2+1)-f(1)}$

and directly substituting $h=0$ we get $\frac {f(2*0+2+0^2)-f(2)}{f(0-0^2+1)-f(1)}$ which evaluates to $\frac {f(2)-f(2)}{f(1)-f(1)}$ $\Rightarrow \frac{0}{0}$

That is not something we want.

The first thing that comes to my mind now is L'Hôpital's rule We differentiate numerator and denominator with respect to $h$ $\lim_{h \to 0} \frac {\frac{d}{dh}f(2h+2+h^2)-f(2)}{\frac{d}{dh} f(h-h^2+1)-f(1)}$

Using the chain rule, $\frac{d}{dx}f(g(x))=\frac{d}{d g(x)}f(g(x)) * \frac{d}{dx}g(x)$

$\lim_{h \to 0} \frac {f'(2h+2+h^2) * (2+2h) - 0}{f'(h-h^2+1) * (1-2h) - 0}$ Substituting h=0 $\frac {f'(2*0+2+0^2) * (2+2*0)}{f'(0-0^2+1) * (1-2*0)}$ $=\frac {f'(2) * 2}{f'(1) * 1}$ Putting values for f'(2) and f'(1)

$\frac{6*2}{4}$ $=\boxed{3}$