If $ \lim_{h\to 0}\frac{f(x+h)-f(x-h)}{h}$ exists then $f$ is differentiable TRUE or FALSE :

If $f:\mathbb R\to \mathbb R$ be such that $\displaystyle \lim_{h\to 0}\frac{f(x+h)-f(x-h)}{h}$ exists for all $x\in \mathbb R$ then $f$ is differentiable.

Let , $l=\displaystyle \lim_{h\to 0}\frac{f(x+h)-f(x-h)}{h}=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}+\lim_{h\to 0}\frac{f(x)-f(x-h)}{h}=f'(x)+\lim_{h\to 0}\frac{f(x)-f(x-h)}{h}$.
Then ..??
 A: Consider $f(x)=|x|$. At $x=0$ the limit is just $0$, although the function isn't differentiable there. Away from $x=0$ is not hard, just write it out.
A: The answer is false. Consider $f(x)=
  \begin{cases} 
      \hfill 0    \hfill & x\neq 0 \\
      \hfill  1 \hfill & x=0 \\
  \end{cases}
$.
Clearly this function is not differentiable at $x=0$ because it is not continuous at that point but for any $h\neq0$, $f(0+h)=f(0-h)=0$ so the limit does exist.

The problem is basically that the limit expression makes no reference to $f(a)$ where $a$ is some point at which we would like to know if the function is differentiable, so the existence of the limit doesn't tell us anything about how the function behaves at that point.
A: Hint: Does this limit existing guarantee that your function is continuous at $x$?
A: Let's consider two problems here.

Problem (for calculus students):  If $f:\mathbb R\to\mathbb{R}$ is a
  continuous function that has a symmetric derivative at a point $x_0$,
  i.e. $$SD\, f(x_0) = \lim_{h\to 0} \frac{f(x_0+h)-f(x_0-h)}{h}$$
  exists, does it follow that $f'(x_0)$ exists?
Problem (for analysis students):  If $f:\mathbb R \to\mathbb{R}$ is a
  measurable function that has a symmetric derivative at every point  of
  a set $E$, i.e. $$SD\, f(x_0) = \lim_{h\to 0}
  \frac{f(x_0+h)-f(x_0-h)}{h}$$ exists for each $x_0\in E$, does it
  follow that $f' $ exists almost everywhere in $E$?

We see the first problem all too often.  Many people jump in with obvious counterexamples for the first problem.  One can easily get the impression that this "symmetric derivative"  is something that arises in first year calculus problems and is hardly noteworthy otherwise.  The more advanced version of the problem is something to keep in mind.  While symmetric differentiability doesn't imply differentiability for calculus students it does (in this sense) for analysis students!
A: No. Take $f(x)=|x|$ it's not differentiable at $0$ but $\lim \limits_{h\to 0}\dfrac{f(h)-f(-h)}{h}$ exists.
