differentiability - basic understanding From this discussion :
Why is the absolute value function not differentiable at $x=0$?
I understand the idea of computing the left (n-) and right (n+) limits and if they are a different number, then we consider that the general limit of of x when x->n does not exist.
This was explained for the |x| and it makes sense.
However I tried to do the same for $x^2$ which is given as an example of a differentiable function (especially in the point where x=0), and i end up to 2 different left and right limits which would mean that $x^2$ would not be differentiable in 0 .... which I know (or at least i think i know? is false - but not so sure now...).
this is how i did, using the derivative's "limit definition" as in the link provided
I imagine that i made a mistake of course... but i really don't see where, as I strictly repeated the same procedure as for |x|.
Could someone explain if possible in detail why?
 A: $0^+$ and $0^-$ are considered to be equal for the purposes of deciding whether the limits from $+$ and $-$ are the "same" value.  Note that when you try the same thing with $|x$ you get $+1$ and $-1$ and these are of course not equal.
A: I think we need to take this back to what is the definition of a limit.
$\lim_\limits{h\to 0} \frac {h^2}{h} = 0$
means
For any $\epsilon>0$ there exists a $\delta > 0$ such that $| \frac {h^2}{h}|<\epsilon$ when $|h|<\delta$
For every neighborhood around $0$ in the image of $f(h) = \frac {h^2}{h}$ (sometimes called an epsilon ball) that we define, we can find an a neighborhood around $0$ in the domain (called the delta ball) such that everything in the delta ball maps inside the epsilon ball.
This is not a particularly easy concept, and may take some time to "grok in fullness."
left hand and right hand limits.
$\lim_\limits{x\to 0^+} \frac {h^2}{h} = 0$ means
$\forall \epsilon>0,\exists \delta >0: 0<h<\delta \implies |\frac {h^2}{h}-0|<\epsilon$
and $\lim_\limits{h\to 0^-} \frac {h^2}{h} = 0$ means
$\forall \epsilon>0,\exists \delta >0: 0>h>-\delta \implies |\frac {h^2}{h}-0|<\epsilon$
These limits do not equal $0^+$ and $0^-$ respectively, they equal exactly $0.$
A: There is no number called "$0^+$". 
And there is no number called "$0^-$". 
There is only a number called "$0$". 
When you write 
$$\lim_{h \to 0^+} h = 0^+
$$ 
it does not mean that the limit is equal to "the number $0^+$" because, as just explained, there is no number called "$0^+$".
Instead, what it means is that as $h$ approaches $0$ from above, the limit of $h$ is equal to the number $0$ and, furthermore, it approaches that number from above. 
