Question

Find $\underset{h\to 0}{lim}\frac{f(2h+2+h^2)-f\left(2\right)}{f(h-h^2+1)-f\left(1\right)}$ if given that $f^{\prime }\left(2\right)=6$ and $f^{\prime }\left(1\right)=4$

• A. does not exist
• B. is equal to $\frac{3}{2}$
• C. is equal to $-\frac{3}{2}$
• D. is equal to $3$

Answer 1

Answer: (D)

is equal to $3$

Here $\underset{h\to 0}{lim}\frac{f(2h+2+h^2)-f\left(2\right)}{f(h-h^2+1)-f\left(1\right)}$

$(\because f^{\prime }\left(2\right)=6$ and $f^{\prime }\left(1\right)=4$ given$)$

Applying L'Hospital rule ,
$=\underset{h\to 0}{lim}\frac{\left\{f^{\prime }(2h+2+h^2)\right\}.(2+2h)-0}{\left\{f^{\prime }(h-h^2+1)\right\}.(1-2h)-0}$

$=\frac{f^{\prime }\left(2\right).2}{f^{\prime }\left(1\right).1}=\frac{6.2}{4.1}=3$