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
Answer & explanation
Correct answer: option 4
$\underset{h→0}{\lim}\frac{f(2h+2+h^2)-f(2)}{f(h-h^2+1)-f(1)}$
$=\underset{h→0}{\lim}\frac{f'(2h+2+h^2)(2+2h)}{f'(h-h^2+1)(1-2h)}=\frac{6×2}{4×1}=3$