Question

if $F^{\prime }\left(3\right)=2$, then $\underset{h\to 0}{lim}\frac{f(3+h^2)-f(3-h^2)}{2h^2}$ is

• A. 1
• B. 2
• C. 3
• D. 1/2

Answer 1

The correct option is B 2

Let, $L=\underset{h\to 0}{lim}\frac{f^{\prime }(3+h^2)-f(3-h^2)}{2h^2}=\frac{0}{0}$ form
Hence using L-hospital rule
$L=\underset{h\to 0}{lim}\frac{f^{\prime }(3+h^2)\times 2h-f^{\prime }(3-h^2)\times -2h}{4h}$
$=\underset{h\to 0}{lim}\frac{f^{\prime }(3+h^2)\times 2h+f^{\prime }(3-h^2)\times 2h}{4h}$
$=\frac{f^{\prime }\left(3\right)\times 2+f^{\prime }\left(3\right)\times 2}{4}$
$=\frac{4+4}{4}=2$