Limit exists v. Differentiable Does existence of a limit at a point not necessarily mean that it's differentiable at that point?
Take this function: 
$$f(x) = \frac{(x - 1)^{2}}{x - 1}$$
The function is not defined at x = 1, but as x approaches 1, f(x) goes to 0; i.e., a "removable discontinuity." But the function is not differentiable at 1, right?
 A: You're right. $f(x)$ as written has a discontinuity at $x=1$, so it is not considered to be differentiable.  However, if you modify $f(x)$ by inserting the removable discontinuity, $f(x)$ becomes a differentiable function.  That is
$$
g(x)=x-1 =
\begin{cases}
\frac{(x-1)^2}{x-1} & x\neq 1\\
0 & x = 1
\end{cases}
$$
is a differentiable function
As you said, the existence of a limit at a point (even along with differentiability in the neighborhood of that point) does not guarantee that the function is differentiable at that point.  In order for a function to be differentiable at a point, it must first be continuous at that point.
Related: differentiability implies continuity.
A: Continuity is necessary for the existence of a $f'$, that means if a function $f$ has discontinuity at $x=a$ then $f$ is not differentiable at $x=a$. Your function is not continuous at $x=1$, therefore it's not differentiable at $x=1$. Moreover, continuity is not a sufficient condition.For example $|x|$ is continuous over $\mathbb{R}$ but it's not differentiable at $x=0$.
A: For the example that you have given, the function is not differentiable at that point because of the discontinuity that exists at that point.
Remember that the definition of a derivative at point uses a limit e.g. 
$$f' (a) = \lim_{h\rightarrow 0}\frac{f(a+h) - f(a)}{h}$$
Thus, it is only differentiable when a limit exists so yes, you are right when you say that if a limit exists, the function is differentiable and when it doesn't exist, the function is not differentiable.
A: By the very definitions of these concepts, a function is continuous only at points $a$ where it is defined ($\lim_{x\to a} f(x)=\color{red}{f(a)}$) and it is differentiable only at points $a$ where it is defined ($\lim_{x\to a}\frac{f(x)-\color{red}{f(a)}}{x-a}$ exists). We also find as a simple theorem that functions are continuous where they are differentiable, but for the problem at hand it suffices to note that $f$ is not even defined at $x=1$, hence not differentiable there. 
One must distinguish between the given function $f$ and the function obtained by removing the removable singukarity (which is of course continuous and in this case also differentiable at th epoint in question)
A: I think it depends on the domain of the function.


*

*if the domain is $\mathbb{R}$, then $f(x)$ is not defined at $x=1$, so it's not continuous and not differentiable;

*if the domain is $\mathbb{R}$ but $x\ne 1$, then it's continuous and differentiable in its domain.
