Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $1$

Limit is in the form $\frac{0}{0}$ when we substitute h=0

Apply L-Hopitals rule

$\Rightarrow li{m}_{h\to 0}\frac{f^{{}^{\prime }}(3+h^2).2h-f^{{}^{\prime }}(3-h^2).-2h}{4h}$

$\Rightarrow li{m}_{h\to 0}\frac{f^{{}^{\prime }}(3+h^2).2+f^{{}^{\prime }}(3-h^2).2}{4}$

Now substitute h=0;

$\Rightarrow \frac{f^{{}^{\prime }}\left(3\right)+f^{{}^{\prime }}\left(3\right)}{4}$

$\Rightarrow \frac{2+2}{4}$

$=1$