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. $37$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $2$

$\underset{h\to 0}{lim}\frac{f(3+h^2)-f(3-h^2)}{{2h}^2}$

$=\frac{1}{2}\left[\underset{h\to 0}{lim}\frac{f(3+h^2)-f\left(3\right)+f\left(3\right)-f(3-h^2)}{h^2}\right]$

$=\frac{1}{2}[f^{\prime }\left(3\right)+f^{\prime }\left(3\right)]$$=2$