A question about two common definitions Two definitions make me puzzled !  
1.
The definition of  $\textbf{Functions Differentiable at a Point}$:
A function $f$ defined in  a neighborhood $(x_{0}-\delta,x_{0}+\delta)$of a point $x_{0}$, if  $$\lim_{x\rightarrow {x}_{0}}\frac{f(x)-f({x}_{0})}{x-{x}_{0}}=A\in\mathbb{R}$$ then 
   We called the  function $f$ is differentiable at the point $x_{0}$!
2.
The defintion of  $\textbf{Functions Continuous at a Point}$:
Let $X \subset \mathbb{R},f(x):X\rightarrow \mathbb{R},x_{0}\in X$,
if
$$\forall \epsilon >0, \exists\delta>0;s.t.\forall x\in X,|x-x_{0}|<\delta \Rightarrow |f(x)-f(x_{0})|<\epsilon .$$ 
then We called the  function $f$ is Continuous at the point $x_{0}$!

My confusion:
$\textbf{A isolated point} \quad\hat{x}\in X,$ according to The defintion of  $\textbf{Functions Continuous at a Point}$,  $f$ is Continuous at the point $\hat{x}$.But $f$ has no vaules in  $\hat{x}$ neighborhood $(\hat{x}-\delta_{\hat{x}},\hat{x}+\delta_{\hat{x}})$ of the point $\hat{x}$,so on the basis of The definition of  $\textbf{Functions Differentiable at a Point}$, $f$ is not differentiable at the point $\hat{x}$.
But the conclusion drawn from the results described above is contradicted against $\textbf{f(x) differentiable at a point must be continuous at the point}.$

I need clartity to eliminate the confusion!Any of your help will be appreciated!
I am sorry I made some obvious mistakes! There is no contradiction!
 A: There is no contradiction.
As you noted, the function defined at an isolated point can be continuous at that point (vacuously satisfying the requirements). Also as you noted, the function is not differentiable at this point. You are correct on both counts.
This does contradict the idea that $$f(x) \text{ is continuous at } x_0 \qquad\Rightarrow\qquad f(x) \text{ is differentiable at } x_0$$
but that's okay because the above is a false statement. 

In general, the statement $A \Rightarrow B$ is not contradicted by $A$ being false and $B$ being true. If $A$ is true and $B$ is false, then you can say $A \not\Rightarrow B$. 
A: You are worried about a contradiction to the statement the differentiability $\implies$ continuity, but if you are not differentiable, you can't contradict this statement.
So, at an isolated point, you have that the function is not differentiable, but is continuous, and so there is no contradiction. You would only have a contradiction if the function were differentiable but not continuous.
