Question

If $f\left(x\right)$ is differentiable function and $f^{\prime }\left(1\right)=4,f^{\prime }\left(2\right)=6,f^{\prime }\left(c\right)$ means the derivative of $f$ at $c$, then $\underset{h\to 0}{lim}\frac{f(2+2h+h^2)-f\left(2\right)}{f(1+h-h^2)-f\left(1\right)}$

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

Answer 1

Answer: (C)

$3$

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

C-hospital rule

$\underset{h\to 0}{lim}\frac{f^{\prime }(2+2h+h^2)\times (2+2h)-0}{f^{\prime }(1+h-h^2)\times (1-2h)-0}$

$=\frac{f^{\prime }\left(2\right)\times 2}{f^{\prime }\left(1\right)\times 1}=\frac{6\times 2}{4}=3$