let, $f(x) =x^\frac{1}{3}$ be a diffrentiable function on $ (0, \infty).$
Given that $$\frac{f(3+h) -f(3)}{h}=f'(3+\theta(h)h)$$ Then find out $\lim_{h\to 0+} \theta(h) =? $
Since, $f$ is diffrentiable at $3$ , I think limit must be $0$ as it tends to $f'(3)$
Please help me
You can obtain explicitly that $$\theta(h)=\frac{1}{h}\left(\left(3\frac{\sqrt[3]{3+h}-\sqrt[3]{3}}{h}\right)^{-3/2}-3\right)$$ Now, by noticing that $h=(\sqrt[3]{3+h}-\sqrt[3]{3})(\sqrt[3]{(3+h)^2}+\sqrt[3]{3(3+h)}+\sqrt[3]{9})$, compute the desired limit as $$\lim_{h\to0^+}\theta(h)=\lim_{h\to0^+}\frac{1}{h}\left(\left(3\frac{\sqrt[3]{3+h}-\sqrt[3]{3}}{h}\right)^{-3/2}-3\right) = \lim_{h\to0^+}\frac{1}{h}\left(\left(\frac{\sqrt[3]{(3+h)^2}+\sqrt[3]{3(3+h)}+\sqrt[3]{9}}{3}\right)^{3/2}-3\right)$$ and now treat this limit as a derivative to finally get $$\lim_{h\to0^+}\theta(h)=\left(\dfrac{\left(\frac{2}{3\sqrt[3]{3+h}}+\frac{1}{ \sqrt[3]{9(3+h)^2}}\right)\sqrt{\sqrt[3]{(3+h)^2}+\sqrt[3]{3(3+h)}+\sqrt[3]{9}}}{2\sqrt{3}} \right)_{h=0} = \frac{\left(\frac{2}{3\sqrt[3]{3}}+\frac{1}{ \sqrt[3]{9\cdot 9}}\right)\sqrt{3\sqrt[3]{9}}}{2\sqrt{3}}=\frac{1}{2}$$
