Does the function $f(x) = |x|e^x$ have a derivative at $x = 0$? Does the function $f(x) \ = \ |x|e^x$ have a derivative at $x = 0$? 
If we split it into two sides 
$$f(x) =
\begin{cases}
xe^x,  & \text{if $\geq 0$} \\
-xe^x, & \text{if $x<0$}
\end{cases}
$$
& 
$$f'(x) =
\begin{cases}
(x+1)e^x,  & \text{if $x \geq 0$} \\
-(x+1)e^x, & \text{if $x<0$}
\end{cases}
$$
so 
$\lim\limits_{x \to 0^+} \ (x+1)e^x = 1$,  
and
$\lim\limits_{x \to 0^-} \ -(x+1)e^x = -1$  
so we can say that the function doesn't have derivative a derivative at zero? Is this correct?
 A: $f$ is not differentiable at $x=0$ because
$$
 \frac{f(x)-f(0)}{x-0} = \frac{|x|e^x}{x} =
\begin{cases}
e^x & \text{if $x > 0$}\\
-e^x & \text{if $x < 0$}
\end{cases}
$$
does not have a limit for $x \to 0$.
It is not necessary to compute $f'(x)$ for $x\ne 0$ in order to draw this conclusion, or to investigate
its limit behavior at $x=0$. And actually the implication
$$
 \lim_{x \to 0}f'(x) \text{ does not exist } \implies
 \text{ $f$  is not differentiable at $x=0$ }
$$
is wrong, as the following example shows:
$$
 f(x) = \begin{cases}
x^2 \cos(\frac 1x) & \text{if $x \ne 0$}\\
0 & \text{if $x = 0$}
\end{cases}
$$
It is however true that if $f'(x)$ has both left-sided limit and
right-sided limit at $x=0$, and if those one-sided limits are different,
then $f'(0)$ does not exist. That follows e.g. from the
“intermediate value property for derivatives.”
But again: it is not needed here.
A: You are correct that the function does not have a derivative at $x = 0$.  Since the derivative of $xe^x, x \geq 0$ is defined in the open interval $(0, \infty)$ and the derivative of $-xe^x, x < 0$ is defined in the open interval $(-\infty, 0)$, we cannot write 
$$f'(x) = 
\begin{cases}
(x + 1)e^x & \text{if $x \color{red}{\geq} 0$}\\
-(x + 1)e^x & \text{if $x < 0$}
\end{cases}
$$
What you should have written is 
$$f'(x) = 
\begin{cases}
(x + 1)e^x & \text{if $x > 0$}\\
-(x + 1)e^x & \text{if $x < 0$}
\end{cases}
$$
then tested the left- and right-sided limits, as you later did, to determine whether $f'(0)$ exists. 
A: If $f$ were differentiable at $0$, then so would $f(x)e^{-x} = |x|$ be. That's a contradiction, though.
A: The function is not differentiable at $x=0$ because the right derivative is $1$ while the left derivative is $-1$.
However, the function is continuous at  $x=0$ thus it is a good example of a continuous function which is not differentiable at a point.
