Question

If $f\text{'}\left(3\right)=2,$then $li{m}_{h\to 0}\frac{\left[f\right(3+h^2)-f(3-h^2\left)\right]}{2h^2}$ is

Answer 1

Finding the value of $\mathit{l}\mathit{i}{\mathit{m}}_{\mathbf{h}\mathbf{\to }\mathbf{0}}\mathbf{}\frac{\mathbf{\left[}\mathbf{f}\mathbf{\right(}\mathbf{3}\mathbf{}\mathbf{+}{\mathbf{h}}^{\mathbf{2}}\mathbf{)}\mathbf{-}\mathbf{f}\mathbf{(}\mathbf{3}\mathbf{-}{\mathbf{h}}^{\mathbf{2}}\mathbf{\left)}\mathbf{\right]}}{\mathbf{2}{\mathbf{h}}^{\mathbf{2}}}$:

Given that $f\text{'}\left(3\right)=2,$

$li{m}_{h\to 0}\frac{\left[f\right(3+h^2)-f(3-h^2\left)\right]}{2h^2}$

When we apply the limit , the value will be $\frac{f\left(3\right)-f\left(3\right)}{0}=\frac{0}{0}$form

Therefore applying L' Hospital Rule

$li{m}_{h\to 0}\frac{\left[f\right(3+h^2)-f(3-h^2\left)\right]}{2h^2}$

Differentiate the numerator and denominator with respect to $h$

$$\begin{gathered} \Rightarrow li{m}_{h\to 0}\frac{[f\text{'}(3+h^2)\times 2h+2h\times f\text{'}(3-h^2\left)\right]}{4h} \\ \Rightarrow li{m}_{h\to 0}\frac{[f\text{'}(3+h^2)+f\text{'}(3-h^2\left)\right]}{2} \\ \Rightarrow \frac{2f\text{'}\left(3\right)}{2}(\because f\text{'}(3)=2) \\ \Rightarrow 2 \end{gathered}$$

Hence, the value of $\mathit{l}\mathit{i}{\mathit{m}}_{\mathbf{h}\mathbf{\to }\mathbf{0}}\mathbf{}\frac{\mathbf{\left[}\mathbf{f}\mathbf{\right(}\mathbf{3}\mathbf{}\mathbf{+}{\mathbf{h}}^{\mathbf{2}}\mathbf{)}\mathbf{-}\mathbf{f}\mathbf{(}\mathbf{3}\mathbf{-}{\mathbf{h}}^{\mathbf{2}}\mathbf{\left)}\mathbf{\right]}}{\mathbf{2}{\mathbf{h}}^{\mathbf{2}}}$ is $\mathbf{2}\mathbf{.}$