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
 A: Your negation is correct; you should specify though that what you're defining is continuity at the point $x_0$, which is distinct from continuity in the whole domain $S$.
Your choice of $\epsilon=1/2$ is fine. However you need to do some more work to show that $f$ can't be continuous. Suppose we try to make $f$ into a continuous function by assigning $f(0)=y_0$. Take any $\delta>0$.
Case 1: Suppose $y_0<0$. Let $x=1/(\pi/2+2\pi N)$ where $N$ is chosen large enough so $|x|<\delta$. Then
$|f(x)-f(x_0)|=|1-y_0|\geq1>\epsilon$ which proves discontinuity.
Case 2: Suppose $y_0\geq0$. Let $x=1/(-\pi/2+2\pi N)$ where $N$ is chosen large enough so $|x|<\delta$. Then
$|f(x)-f(x_0)|=|-1-y_0|\geq1>\epsilon$ which again proves discontinuity.
Thus we conclude there's no choice of $y_0=f(0)$ which makes $f$ continuous at zero.
A: The negation is:

there exists $\epsilon >0$ such that for any $\delta>0$ we can find an $x$ such that $|x-x_0|<\delta$  and $|f(x)-f(x_0)| > \epsilon$.

And you have just proved this.
A: 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 $|f(x_\delta)−f(x_0)|≥ϵ$
