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

Solution :
$${\text{Let }}L = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {2h + 2 + {h^2}} \right) - f\left( 2 \right)}}{{f\left( {h - {h^2} + 1} \right) - f\left( 1 \right)}}\,\,\,\,\,\,\,\left[ {\frac{0}{0}{\text{form}}} \right]$$
$$\therefore $$ Applying L'Hospital rule, we get
$$\eqalign{ & L = \mathop {\lim }\limits_{h \to 0} \frac{{f'\left( {2h + 2 + {h^2}} \right).\left( {2 + 2h} \right)}}{{f'\left( {h - {h^2} + 1} \right).\left( {1 - 2h} \right)}} \cr & = \frac{{f'\left( 2 \right).2}}{{f'\left( 1 \right).1}} \cr & = \frac{{6 \times 2}}{{4 \times 1}} \cr & = 3 \cr} $$