About the differentiability of $|x|$? The derivative of a function is another function which gives the slope of the tangent line at a certain point of the function. And then, taking $f(x)=|x|$, I'd have:
$$[f(x)]'=\frac{\sqrt{(x+h)^2}-\sqrt{x^2}}{h}=\frac{(x+h)-x}{h}=\frac{h}{h}=1$$
Then that should be the function that would indicate the slope of $f(x)=|x|$, it works when $x>0$ because the slope is $1$ but fails when $x<0$ because the slope is $-1$ and then, I guess that the need to have two different derivatives for one function indicates that some point might be broken.
The problem here is that it's actually possible to express the derivative of that function with only one function: $f'(x)=\frac{x}{|x|}$, the problem is that this new function is discontinuous at $f(0)$. I am starting to guess that for a function to be differentiable at a point, all it's derivatives should also be continuous at the same point. Is my guess correct?
 A: No, a function doesn't even need to be differentiable to be continuous, like in this case: clearly $f(x)=|x|$ is continuous in the whole real line, but it isn't differentiable in the origin.
Your incremental ratio is wrong, by the way: the limit operator is missing (or else, that is not the derivative), and $\sqrt{(x+h)^2}=x+h$ and $\sqrt{x^2}=x$ only for $x,x+h\geq 0$.
A: If you write $f(x)=|x|$ as a piecewise function, then 
$f(x)= \begin{cases} 
      -x & x\leq 0 \\
       x & x> 0 \\
   \end{cases}
$
A function is differentiable at a point $a$ if the derivative exists at this point.
Let our point $a=0$
$\lim\limits_{x\rightarrow a^-} \frac{f(x)-f(a)}{x-a}=\frac{-x-(-(0))}{x-0}=-1$
and
$\lim\limits_{x\rightarrow a^+} \frac{f(x)-f(a)}{x-a}=\frac{x-((0))}{x-0}=1$
Therefore $f(x)=|x|$ is not differentiable at $x=0$.  
With respect to continuity, if a function is differentiable at $x=a$ then it must be continuous at $x=a$;however, continuity does not imply differentiability.  
If, however, a function is not continuous at $x=a$, we can say for certain that it is not differentiable at $x=a$.  
