Question

$\underset{h\to 0}{lim}\frac{f(2h+2+h^2)-f\left(2\right)}{f(h-h^2+1)-f\left(1\right)},\left(giventhat{f}^{{}^{\prime }}\right(2)=6and{f}^{{}^{\prime }}(1)=4)$ is equal to

• A. Does not exist
• B. Is equal to
• C. Is equal to
• D. Is equal to 3

Answer 1

The correct option is D Is equal to 3

$\underset{h\to 0}{lim}\{\frac{f(2h+2+h^2)-f\left(2\right)}{(2h+2+h^2)-2}.\frac{\left(\right(h-h^2+1)-1)}{f(h-h^2+1)-f\left(1\right)}\times \frac{2h+h^2}{(h-h^2)}\}=\frac{f^{{}^{\prime }}\left(2\right)}{f^{{}^{\prime }}\left(1\right)}.2=\frac{6.2}{4}=3$