This is an example of a very **simple real function that is only differentiable a finite number of times**.

Let $f:\mathbb{R} \to \mathbb{R}$ be a function defined by $f(x)=x|x|$. The aim is to show that this function is differentiable but that it is not twice differentiable. Notice that $f(x)$ can also be written as

$$ f(x) = \left\{

\begin{array}{l l}

x^2 & \quad \text{if $x\geq0$}\\

-x^2 & \quad \text{if $x<0$} \end{array} \right.$$ The graph of this function looks as follows

It is clear that $f(x)$ is differentiable for $x\!>\!0$ and $x\!<\!0$ with derivative $f^{\prime}(x)=2x$ and $f^{\prime}(x)=-2x$ respectively. We only need to check that $f(x)$ is differentiable at $x=0$.

By definition a function $g:I \to \mathbb{R}$ where $I \subset \mathbb{R}$ is **differentiable at a point** $x \in I$ if the limit

$$\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}$$

exists. $g$ is called** differentiable** if it is differentiable at every point $x \in I$.

Let’s check that this limit exists for $f$ when $x=0$

$$\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0}\frac{(0+h)|0+h|}{h}$$

$$=\lim_{h\to 0}\frac{h|h|}{h}=\lim_{h\to 0}|h|=0$$

So $f(x)$ is indeed differentiable at $x=0$ and we can write the derivative of $f(x)$ as $f^{\prime}(x)=2|x|$. The graph of the derivative of $f$ looks as follows

We see that $f^{\prime}(x)$ is differentiable when $x \neq 0$ but when we try to find the derivative of $f^{\prime}$ at $x=0$ we have

$$\lim_{h\to 0^{+}}\frac{f^{\prime}(0+h)-f^{\prime}(0)}{h}=\lim_{h\to 0^{+}}\frac{2|h|}{h}=2$$

$$\lim_{h\to 0^{-}}\frac{f^{\prime}(0+h)-f^{\prime}(0)}{h}=\lim_{h\to 0^{-}}\frac{2|h|}{h}=-2$$

The left and right limits are not the same and therefore $f^{\prime}$ is not differentiable at $x=0$. **The conclusion is that $f$ is differentiable but not twice differentiable.**

It is not much more difficult to show (by induction is possibly the easiest way) that the function given by $h(x)=x^{n}|x|$ is $n$-times differentiable but not $n+1$ times differentiable.