Negation of definition of continuity This should be a very easy question but it might just be that I'm confusing myself. So we have the definition of a function $f$ on $S$ being continuous at $x_0$:

For any $\epsilon$>0, there exists $\delta>0$ such that: whenever $|x-x_0|<\delta$, we have $|f(x)-f(x_0)|<\epsilon$

And I assume the negation is

There exists $\epsilon$>0 such that for all $\delta>0$, $|x-x_0|<\delta$ yet $|f(x)-f(x_0)|\ge \epsilon$.

Now I want to show that the function $f(x)=\sin(\frac{1}{x})$ together with $f(0)=0$ cannot be made into a continuous function at $x=0$. So I need to show that there exists $\epsilon>0$ such that for all $\delta>0$, $|x|<\delta$ yet $|f(x)|\ge\epsilon$.
Let $\epsilon = \frac{1}{2}$. Then no matter what $\delta$ we choose, let $|x|<\frac{1}{2}$. It is certainly possible that $|f(x)|\ge \frac{1}{2}$, because, well, $\frac{1}{x}$ can really take on arbitrarily large value as $x$ is small.
Now, what confuses me is that, as $x$ gets small, $f(x)$ can certainly be greater than $\frac{1}{2}$ for infinitely many times, but it will be less than that infinitely many times, too. But I suppose it doesn't really matter. So I think there's something wrong with my negation but I couldn't figure out where.
Update: The correct version can be found here. Watch for Lemma 4.6

The negation is:
There exists $ϵ>0$ such that for all $δ>0$, there is an $x_\delta$ such that $ |x_\delta−x_0|<δ$ yet $|f(x_\delta)−f(x_0)|≥ϵ$
