Question

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

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

Answer 1

The correct option is A 3

${lim}_{h\to 0}\frac{f(2h+2+h^2)-f\left(2\right)}{f(h-h^2+1)-f\left(1\right)}={lim}_{h\to 0}\frac{f(2h+2+h^2)-f\left(2\right)}{2h+2+h^2-2}\times \frac{h(2+h)}{h(1-h)}\times \frac{(h-h^2+1)-1}{f(h-h^2+1)-f\left(1\right)}=f^{\prime }\left(2\right)\times {lim}_{h\to 0}\frac{2+h}{1-h}\times \frac{1}{f^{\prime }\left(1\right)}=6\times 2\times \frac{1}{4}=3$
Hence, option 'A' is correct.