Differntiable and continuous Is it true that a function which is not continuous at a point will not be differentiable at that point? Graphically it seems so, but can we prove this formally?
Also, if the above statement is incorrect, can we still disprove it?
Note that we know "All differentiable functions are continuous, but not all continuous functions are differentiable". But I am asking about the negation of this in some sense
 A: By definition, a function is continuous if
$$f(a)=\lim_{x\to a}f(x)$$
(naturally this entails $f(a)$ exists and the limit exists, I'm just rolling all three parts into one)
and differentiable at $a$ if
$$\lim_{x\to a} {f(x)-f(a)\over x-a}$$
exists, and we call this limit $f'(a)$.
If it's not continuous, then either
($i$) $f(a)$ is not defined and so the derivative limit does not exist and we fail to be differentiable.
($ii$)$f(x)$ doesn't approach a limit as $x\to a$ where again we fail to have the derivative limit exist.
($iii$) The limit will be "finite, non-zero number/ $0$" because the $f(x)$ limit is not equal to $f(a)$. This makes the derivative limit undefined, so by definition the function is not differentiable at $a$.
A: If a function is differentiable at $x$ then it is continuous at $x$.
Hence if it is not continuous at $x$ it cannot be differentiable at $x$.
A: Suppose that $f(x)$ is differentiable at $a$, with derivative $f'(a)$. Then
$$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=f'(a).\tag{1}$$
It follows in particular that $f(x)$ is defined in a neighbourhood of $a$. For $x\ne a$ in that neighbourhood, let 
$$\frac{f(x)-f(a)}{x-a}=f'(a)+g(x,a).\tag{2}$$
It follows from (1) that $\lim_{x\to a} g(x,a)=0$.
Rewrite (2) as 
$$f(x)=f(a)+(x-a)f'(a)+(x-a)g(x,a).\tag{3}.$$
Now let $x\to a$. The right-hand side of (3) has limit $f(a)$. It follows that $\lim_{x\to a}f(x)=f(a)$, meaning that $f$ is continuous at $a$.
