Originally Posted by
johnsomeone Is this the statement?
$\displaystyle \text{Prop: f is continuous at x } \Longleftrightarrow \lim_{h \to 0+} \left( \underset{|t| < h}{sup} \ \lvert f(x+t) - f(x) \rvert \right) = 0.$
If so, then proof is an $\displaystyle \epsilon - \delta$ proof. To get you started:
$\displaystyle \text{f is continuous at x means } \lim_{y \to x} f(y) = f(x), $
$\displaystyle \text{or equivalently } \lim_{y \to x} (f(y) - f(x)) = 0,$
$\displaystyle \text{or equivalently } \lim_{t \to 0} (f(x+t) - f(x)) = 0,$
$\displaystyle \text{or equivalently } \lim_{t \to 0} |f(x+t) - f(x)| = 0.$
Each of those requires an argument, but is very straightforward. So you can see that the final form there is almost identical to what you're trying to prove, except for that sup (supremum = least upper bound) part.
To prove the proposition, you'll need to prove two things, the $\displaystyle \Rightarrow$ direction, and the $\displaystyle \Leftarrow$ direction.
When you're into the guts of the proof, you'll want to choose (in one direction, though this works for both) the $\displaystyle \delta > 0$ so that
it makes the limit you know $\displaystyle \frac{\epsilon}{2}.$
----------------------------------------------
This won't make any sense until you try to prove it yourself (but perhaps then it will be helpful):
If everything in a set is less than $\displaystyle \frac{\epsilon}{2}$, then the supremum of that set will be less than or equal to $\displaystyle \frac{\epsilon}{2}$, which is less than $\displaystyle \epsilon.$
That will be your decisive observation when proving it in one direction.
In the other direction, you'll observe that if the supremeum of a set is less than $\displaystyle \epsilon$ (or you could use $\displaystyle \frac{\epsilon}{2}$ there too),
then any individual one of those things in that set is less than $\displaystyle \epsilon.$