Difference between derivative of a function at a point and limit of the differentiated function at that point $f'(x_0)\;=\lim\limits_{x \to x_0}\frac{f(x)-f(x_0)}{x-x_0}$
While solving a question:

Prove that the function {$f(x)=\frac{|x|}{x}\; \; \forall x\neq0$ and $f(x) =0 \;\;\forall x=0$} is not differentiable  at 0.

I have done it by evaluating derivative from the left and right individually which gives $\infty$. So, I have concluded that $f$ is not differentiable at 0.
But I  noticed that $f'(x)=0\;\;\;\forall x\neq0$. So, shouldn't the limit of $f'(x)$  be $0$? I know that I have made a false argument. Please tell me where is contradiction in my argument.
Edit:  In short, what is the difference between $f'(0)$ and $\lim\limits_{x \to 0}f'(x)$?
 A: You have the correct computations: the limit is $\infty$, so the derivative does not exist.
Now, when everything is nice, the derivative exists and it is continuous, so the derivative coincides with the limit of the derivatives at nearby points. 
My guess is that you are asking for an intuition of what fails when things are not nice. Intuitions are subjective, so I am going to give you mine: geometry. Do not think of the derivative as a number, but as a mechanism to build the tangent line, which is a (linear) approximation to the function. The number you obtain as the derivative is the slope of this line, but the line is also characterized by the point it is attached to.
In this case, you should plot the function. You would see you could place a horizontal line approximating the function at both sides of $0$, but these lines are placed on different points. Since the approximation does not work the same way from both sides, it cannot exist.
So, if I propose this exercise in a test, your answer would be correct, but it is not the answer I am expecting and it would be somehow considered as lacking intuition (and wasting time with the limit). The "correct" answer (in my opinion) is: the function is not continuous at 0, so the derivative cannot exist.
A: $\frac{f(x)-f(0)}{x-0}=\frac{1}{|x|} \to \infty$ as $x \to 0.$
Hence, $f$ is not differentiable at $0$.
