Proving $1\over x^2$ is not uniformly continuous I need to show that $1 \over x^2$ is not uniformly continuous on the interval $(0,2]$ using the definition of uniform continuity.
Definition of Uniform Continuity on a set A:
Let $A \subseteq \Bbb R$, let $f:A \rightarrow \Bbb R$.
$f$ is uniformly continuous if $ \forall \epsilon > 0$ $\exists \delta > 0$ such that if for any $x \in A$, $x$ satisfies the inequality:
$|x - u| < \delta \implies |f(x) - f(u)| < \epsilon$
To prove this function is not uniformly continuous by the given definition, I negated the definition such that:
$f$ is not uniformly continuous on the interval $(0,2]$ whenever
$\exists \epsilon_0 > 0$ $ \forall \delta > 0$ $\exists$ $x_\delta \in A$ satisfies $|x_\delta-u_\delta| < \delta $ but $|f(x_\delta) - f(u_\delta)| > \epsilon_0$ 
I chose to let $\epsilon_0 = 2$
For any $ 0< \delta < 2$ choose $x_\delta = \delta$ and $u_\delta$ = $\delta  \over 2$ 
Then $|x_\delta - u_\delta|$ = $|\delta -$ $\delta \over 2|$ = $\delta \over 2$
But $|f(x_\delta) - f(u_\delta)|$ = |$1 \over \delta^2$ - $\frac 1 1 \over \delta^2$ | = |$(\frac 1\delta)^2 - \frac 2\delta^2)|$ = $3 \over \delta^2$ $> 2 = \epsilon_0$
But on the interval $(0,2]$ this does not always hold true.
Can anyone help me with this? I am thinking that my choice of $\delta$ is wrong but I'm not sure.

Here is another way:
If $f$ was uniformly continuous, then for some $\delta>0$ we would have $f(x)-f(x+\delta) < 1$ for all $x$ such that $x,x+\delta \in (0,2]$.
Let $f(x) = {1 \over x^2}$, and
let $\phi(x) = f(x)-f(x+\delta)$ for $x \in (0,2-\delta)$.
Since $f$ is bounded on any compact subinterval of $(0,2]$, we see that $\lim_{x \downarrow 0} \phi(x) = \infty$. Hence $f$ is not uniformly continuous.
