Question

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

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

Answer 1

Answer: (A)

$\underset{h\to 0}{\lim }\frac{f(2h+2+h^2)-f(2)}{f(h-h^2+1)-f(1)}$ $\underset{h\to 0}{\lim }\frac{\{f^{\prime }(2h+2+h^2)\}(2+2h)-0}{\{f^{\prime }(h-h^2+1)\}(1-2h)-0}$ (using L Hospital?s rule) $=\frac{f^{\prime }(2).2}{f^{\prime }(1).1}=\frac{6\times 2}{4\times 1}$ $=\frac{12}{4}=3$