$\lim\limits_{h \rightarrow 0} \frac{f\left(2 h+2+h^2\right)-f(2)}{f\left(h-h^2+1\right)-f(1)}$ given that f'(2) = 6 and f'(1) = 4

is equal to 3

We have,

$\lim\limits_{h \rightarrow 0} \frac{f\left(2 h+2+h^2\right)-f(2)}{f\left(h-h^2+1\right)-f(1)}$

$=\lim\limits_{h \rightarrow 0} \frac{f\left(2 h+2+h^2\right)-f(2)}{\left(2 h+2+h^2\right)-2} \times \frac{\left(2 h+2+h^2\right)-2}{\left(h-h^2+1\right)-1} \times \frac{\frac{1}{f\left(h-h^2+1\right)-f(1)}}{\left(h-h^2+1\right)-1}$

$=f'(2) \times \lim\limits_{h \rightarrow 0} \frac{2+h}{1-h} \times \frac{1}{f'(1)}$

$=\frac{2 f'(2)}{f'(1)}=\frac{12}{4}=3$

ALITER

Using De 'L' Hospital's rule, we have

$\lim\limits_{h \rightarrow 0} \frac{f\left(2 h+2+h^2\right)-f(2)}{f\left(h-h^2+1\right)-f(1)}$

$=\lim\limits_{h \rightarrow 0} \frac{f'\left(2 h+2+h^2\right) \times(2+2 h)-0}{f'\left(h-h^2+1\right)(1-2 h)-0}$

$=\frac{f'(2) \times 2}{f'(1)}=\frac{12}{4}=3$