How to guarantee differentiability? According to my calculus book the following holds. Given that $h\to0$:
$f'(x)=\frac{1}{2}(f'(x)+f'(x))=\frac{1}{2}(\frac{f(x+h)-f(x)}{h}+\frac{f(x)-f(x-h)}{h})=\frac{f(x+h)-f(x-h)}{2h}$
But then the book claims that the existence of the limit in $\frac{f(x+h)-f(x-h)}{2h}$ does not guarantee that $f$ is differentiable at $x$. This can be shown by taking $f(x)=|x|$ at $x=0$. Then the limit gives $0$, while $f'(0)$ should be not defined at $x=0$.
Question: Although I can follow the example with $f(x)=|x|$ above, I do not understand why the existence of the limit in $\frac{f(x+h)-f(x-h)}{2h}$ would not guarantee that $f$ is differentiable at $x$. It seems to say that $f'(x)=\frac{f(x+h)-f(x-h)}{2h}$, then why doesn't it also guarantee that $f$ is differentiable at $x$? Any thoughts?
 A: Let us first see again the defintion of differentiability. $f$ is differentiable at $x$ if  $$  \lim_{h\rightarrow 0^+} \frac{f(x+h)-f(x)}{h} \quad  \text{ and }  \lim_{h\rightarrow 0^-} \frac{f(x+h)-f(x)}{h} $$
exist and equal. Then these limits, both, are  equal to  $f'(x)$. 
Now to see that the existance of  the limit  $$\lim_{h\rightarrow 0} \frac{f(x+h)-f(x-h)}{2h}  $$
doesnot impliy that $f$ is differentiable at $x$,  consider the question : 
Does the existance of this last limit guarantee the existance  and the equality of the limits: $$  \lim_{h\rightarrow 0^+} \frac{f(x+h)-f(x)}{h} \quad  \text{ and }  \lim_{h\rightarrow 0^-} \frac{f(x+h)-f(x)}{h} $$
The answer of this question is no. Check the counter example you give in the question: take  $f(x)=|x|$ and check differentiablity at  $x=0$. Indeed, we have 
$$  \lim_{h\rightarrow 0^+} \frac{|h|-0}{h}=\lim_{h\rightarrow 0^+}  \frac{h}{h}=1 \quad  \text{ and }  \lim_{h\rightarrow 0^-} \frac{|h|-0}{h}=   \lim_{h\rightarrow 0^-} \frac{-h}{h}=-1$$ however 
 $$\lim_{h\rightarrow 0} \frac{|h|-|-h|}{2h}=0  $$
A: Here is the same situation that you can't conclude any thing about $\lim u$ and $\lim v$ even when $\lim u+v$ is exist and finite. For example, take $\lim( \frac{1}{x} - \frac{1}{x})=0$ when $ x\rightarrow 0$ but you can't separate into 2 limit. The differential is the existence and finiteness of the $\lim \frac{f(x+h)-f(x)}{h}$ as $\lim u$ above
A: Just to add an example to illustrate that 'the left limit is not equal to right limit' is not the  intrinsic issue, as had been argued on this page: set
$$ f(x) = \begin{cases} x,\ x\not = 0, \\ 1,\ x =0.\end{cases}$$
Now, $f'(0)$ does not exists, as $f$ is not continuous at $x=0$. On the other hand,
$$ 1 = \lim_{h\to 0} {f(0+h) - f(0-h) \over 2h}. $$
Of course, the left and right limits  implicit in the (non)-definition of $f'(0)$ don't exist either, but...
The point, in the notation of the accepted answer, is the wanted validity of 
$$\lim_{x\to a}\left(\, u(x) + v(x)\, \right) = \lim_{x\to a} u(x) + \lim_{x\to a}v(x).$$
This equation holds if the limits on the right exist. In words, we are saying, "if $u$ is close to something, and $v$ close to something else, then $v+v$ is close to the sum of the two somethings." That is to say, the existence of the limits on the right implies the existence of (and calculates) the limit on the left. 
However, as the counter-example of the accepted answer shows, it is possible for the sum to be "close to something" without the summands being close to anything at all.
Just as for the addition of limits above, the same holds for subtraction, multiplication, and division of limits. (Although, for division, one has to be careful about dividing by $0$, of course...)
Irrelevant Note: For what it's worth: one could say that "limits commutes with addition" (on condition of the existence of limits) is equivalent to the continuity of $+:\mathbb R^2 \to \mathbb R$, if one defines 'continuity' for functions with domain in $\mathbb R^2$.
