Show $\displaystyle f(x)=\frac{1}{x^2}$ is not uniformly continuous on $\displaystyle I=(0,\infty)$ .
Basically, it can't be uniformly continuous in that interval because the slope goes to (negative) infinity as x goes to zero. To prove this rigorously, remember the definition of uniform continuity: for any $\displaystyle \epsilon>0$, there is a fixed $\displaystyle \delta$ such that $\displaystyle |x-y|<\delta$ implies $\displaystyle |\frac{1}{x^2}-\frac{1}{y^2}|< \epsilon$.
To show that it's not uniformly continuous, choose any epsilon, and show that for any fixed delta, $\displaystyle |\frac{1}{x^2}-\frac{1}{(x+\delta)^2}|$ goes to infinity as x goes to 0.
There is a theorem concerning uniform continuity that says:
f is uniform continuous on its domain ,and in our case (0,$\displaystyle \infty$) iff for any pair of sequences on (0,$\displaystyle \infty$ {$\displaystyle x_{n}$},{$\displaystyle y_{n}$} we have:
$\displaystyle lim_{n\to\infty} |x_{n}-y_{n}| = 0\Longrightarrow$ $\displaystyle lim_{n\to\infty}|f(x_{n})-f(y_{n})| =0$
Hence to prove that :$\displaystyle \frac{1}{x^2}$ is not uniform continuous on (0,$\displaystyle \infty$) we must find a pair of sequences that:
$\displaystyle lim_{n\to\infty} |x_{n}-y_{n}| = 0$ ,and
$\displaystyle lim_{n\to\infty}|f(x_{n})-f(y_{n})|\neq 0$
So if we put :
$\displaystyle x_{n}=\frac{1}{\sqrt{n}}$ and
$\displaystyle y_{n} =\frac{2}{\sqrt{n}}$,then
$\displaystyle lim_{n\to\infty} |x_{n}-y_{n}| = 0$ and
$\displaystyle lim_{n\to\infty}|f(x_{n})-f(y_{n})|\neq 0$