# $f$ is differentiable in $\Bbb R \implies f$ is continuous in $\Bbb R$

Given a function $f:\Bbb R\to \Bbb R$, I need to show that if $f$ is differentiable then $f$ is continuous.

Here is my idea on how to show this: If $f$ is differentiable then $\forall x_0\in\Bbb R$, $f(x_0)$ is differentiable. Then this implies that $$\forall x_0\in\Bbb R, \lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=f'(x_0)\iff \lim_{x\to x_0}[f(x)-f(x_0)]=\lim_{x\to x_0}[x-x_0]\frac{f(x)-f(x_0)}{x-x_0}.$$ Since $\lim_{x\to x_0}[x-x_0] = 0$, then from this we have, $$\lim_{x\to x_0}[f(x)-f(x_0)] = 0.$$ Moreover, since these limits exist independently, then we have $$\lim_{x\to x_0}f(x)=f(x_0)\implies f\text{ is continuous}$$ Does this seem correct or is there some error in the logic? Thanks.

• This is correct. – copper.hat Feb 21 '14 at 23:29
• It is correct, see Ayman's answer here for another way:math.stackexchange.com/questions/269666/… – user10444 Feb 21 '14 at 23:34
• Thanks. I am aware of the $\epsilon$-$\delta$ style of proof, but it was something that I was trying to avoid in this case to make it more concise. – homegrown Feb 21 '14 at 23:37

You seem like you have the right approach, but I am utterly confused by your symbols. You write an equality which means "$f$ is differentiable at $x_0$" then claim this is $\iff$ a statement about limits which is in fact necessarily true purely from the cancellation of $[x - x_0]$ from the top and bottom of that fraction. One could argue about what it even means if the limit does not exist, but certainly if $f$ is merely continuous it does, so the RHS of your $\iff$ is more often true than the LHS.
I am posting this as an answer because it would be too much for a comment. Anyways, what I was saying is this: $$\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=f'(x_0)$$*is logically equivalent to* $$\lim_{x\to x_0}[f(x)-f(x_0)]=\lim_{x\to x_0}[x-x_0]\lim_{x\to x_0}f'(x_0), \text{where}\lim_{x-x_0}\frac{f(x)-f(x_0)}{x-x_0}=f'(x_0), \text{etc...}$$ I was just cutting out some additional steps
• However, I stand by my answer. A continuous function satisfies the right hand side of your $\iff$ but not the left hand side, so I just don't believe that the $\Leftarrow$ direction is true. – Ben Millwood Mar 1 '14 at 19:46