Find out limit of the following question 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
$$\lim_{h\to0^+}\theta(h)=\left(\dfrac{\left(\frac{2}{3\sqrt[3]{3+h}}+\frac{1}{
\sqrt[3]{9(3+h)^2}}\right)\sqrt{\sqrt[3]{(3+h)^2}+\sqrt[3]{3(3+h)}+\sqrt[3]{9}}}{2\sqrt{3}}
\right)_{h=0} = \frac{\left(\frac{2}{3\sqrt[3]{3}}+\frac{1}{
\sqrt[3]{9\cdot 9}}\right)\sqrt{3\sqrt[3]{9}}}{2\sqrt{3}}=\frac{1}{2}$$
