Example: The function $f \colon$ $\mathbb{R}$$\rightarrow \mathbb{R}$ given by $f(x) = \sqrt[3]{x}$ is not differentiable at $x=0$.
Solution: We are asked to show that $f'(0)$ does not exist. By definition,
$$\begin{array}{ll}
f'(0) &= \displaystyle\lim_{h \rightarrow 0} \dfrac{\sqrt[3]{0+h} - \sqrt[3]{0}}{h} \\
&= \displaystyle\lim_{h \rightarrow 0} \dfrac{\sqrt[3]{h}}{h} \\
&= \displaystyle\lim_{h \rightarrow 0} \dfrac{1}{h^{\frac{2}{3}}} \\
&= \infty,
\end{array}$$
which shows that the limit (and hence the derivative) does not exist. $\blacksquare$
You should take note that this situation is different than that of the function $g \colon [0,\infty) \rightarrow \mathbb{R}$ given by $g(x) = \sqrt{x}$. This function is not defined on all of $\mathbb{R}$ because we don't allow "imaginary numbers". Here, we say that $g'(0)$ does not exist, but that is because $\displaystyle\lim_{x \rightarrow 0} g(x)$ does not exist and hence $g$ is not continuous at $0$. So we deduce that $g'(0)$ doesn't exist here because of the theorem "if $f$ is differentiable at $a$, then $f$ is continuous at $a$".
