Q. $\lim_{h \rightarrow 0}\frac{f\left( 2h + 2 + h^{2} \right) - f(2)}{f\left( h - h^{2} + 1 \right) - f(1)}$, given that $f^{'}(2) =$ and $f^{'}(1) = 4$

A) does not exist
B) is equal to -3/2
C) is equal to 3/2
D) is equal to 3
Explanation

(d)
$\lim_{h \rightarrow 0}\frac{f\left( 2h + 2 + h^{2} \right) - f(2)}{f\left( h - h^{2} + 1 \right) - f(1)}$
$= \lim_{h \rightarrow 0}\frac{\left\{ f'\left( 2h + 2 + h^{2} \right) \right\}.(2 + 2h) - 0}{\left\{ f^{'}\left( h - h^{2} + 1 \right) \right\}.(1 - 2h) - 0}$
[using L’ Hospital’s rule]
$= \frac{f^{'}(2).2}{f^{'}(1).1} = \frac{6.2}{4.1} = 3$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.