Given that f'(2) = 6 and f'(1) = 4, then $\underset{h→0}{\lim}\frac{f(2h+2+h^2)-f(2)}{f(h-h^2+1)-f(1)}$ is equal to

does not exist $-\frac{3}{2}$ $\frac{3}{2}$ 3

3

$\underset{h→0}{\lim}\frac{f(2h+2+h^2)-f(2)}{f(h-h^2+1)-f(1)}$ $=\underset{h→0}{\lim}\frac{f'(2h+2+h^2)(2+2h)}{f'(h-h^2+1)(1-2h)}=\frac{6×2}{4×1}=3$