Finding if a function with cases is differntiable on a point 
Is $g$ differentiable on $x=0$ ?
$$g(x)=\begin{cases}\dfrac{e^x-1}{x}&,x\neq0 \\ 1 &,x=0  \end{cases} $$

The derivative for $x\neq0$: $g'(x)=\dfrac{e^x(x-1)+1}{x^2}$, by taking the limit of $g'$ from both sides of 0 I get $\displaystyle\lim_{x\to0\pm}\frac {e^x}{2}=\frac 1 2$.
Now trying to use the definition of a derivative with $\displaystyle\lim_{h\to 0}\frac {f(x+h)-f(x)}{h}$ to get to the same value with $x=0$: 
$\displaystyle\lim_{h\to 0}\frac {g(h)-g(0)}{h}=
\lim_{h\to 0}\frac{e^{h}-1} {h^2}=\text{(LHR twice)}=
\lim_{h\to 0}\frac{e^{h}} {2h}=
\lim_{h\to 0}\frac{e^{h}} {2}=
(h\to0)=
\frac{e^{0}}{2}=
\frac 1 {2}$
So does that mean that it's differentiable on $x=0$ ?
Is this the general approach for finding out if a function is differentiable on a point ?
 A: I gave a very similar question in an exam. The derivative from the left
$$ g'(0-)= \frac{1}{2} $$
is correct and in order of the function to be differentiable you need to demand that the left derivative equals to the derivative from the right. In your case,
$$ g'(0+) = \frac{1}{2} $$
as well.
It is easy to see that $g$ is continuous 
$$ \lim_{x \to 0} \frac{e^x-1}{x} = 1 = g(0) . $$
Now use Taylor expansion
$$ e^x = 1 + x + \frac{x^2}{2} + o(x^3) $$ to get that
$$ g(x) = \frac{ 1 + x + \frac{x^2}{2} + o(x^3) - 1}{x} = 1 + \frac{x}{2} + o(x^2) $$
so $g$ is differentiable at $0$.
A: As you started to do in the post, we check differentiability at $0$ by using the definition. So we want to find whether 
$$\lim_{h\to 0} \frac{\frac{e^h-1}{h}-1}{h}$$
exists. Minor manipulation transforms this to 
$$\lim_{h\to 0} \frac{e^h-1-h}{h^2}.$$
Now using L'Hospital's Rule twice finishes things. The derivative at $0$ is $\frac{1}{2}$. 
Remarks: $1.$ the original algebraic glitch in the OP has now been corrected. As to the question about whether the method is the general approach, there are many approaches. But writing down an explicit expression for $\frac{g(h)-g(0)}{h}$ and finding the limit of this (if it exists) is certainly the most fundamental approach, but once we have developed some machinery, many approaches become available. 
$2.$ More easily, if one is comfortable with series, write down the power series for $\frac{e^x-1}{x}$. We get a nice expression, and in fact the given function is infinitely differentiable everywhere. 
