$1/x^2$ is not uniformly continuous on $(0,1)$ The following is what I tried. I just start studying analysis. I am not sure if I am right. Please let me know if you see any mistake.
Prove $f(x)=1/x^2$ is not uniformly continuous on (0,1).
Definition : $f(x)=1/x^2$ is uniformly continuous on (0,1) if, given $\epsilon>0$, there exists a $\delta>0$ such that for all $x,y\in(0,1)$ with $|x-y|<\delta$, we have $|\frac{1}{x^2}-\frac{1}{y^2}|<\epsilon$.
So, my goal is to find a epsilon that contradicts the definition so that I prove $f(x)$ is not uniformly continuous.
Let $x\in(0,1)$. 
Since $0<x<1$, we have $1<\frac{1}{x}$. So, $1<\frac{1}{x^2} \Leftrightarrow 3<\frac{3}{x^2}$.
And, Suppose $y=\frac{x}{2}$.
Then, $|f(x)-f(y)|=|\frac{1}{x^2}-\frac{1}{y^2}|=|\frac{1}{x^2}-\frac{4}{x^2}|=|\frac{3}{x^2}|>3$.
So, if we choose $\epsilon =2$ and $y=x/2$, $|f(x)-f(y)|$ is never less than $\epsilon$.
Thus, it is not uniformly continuous.
(I feel like I should use $\delta$ somewhere , but not sure where to use it...please give me some advise.
 A: You are correct. With regard your uncertainty regarding $\delta$...
You are saying for your chosen $\epsilon$, no $\delta$ will work. Suppose such a $\delta$ existed for your $\epsilon$. Then, if $|x-y|=|x-\frac{x}{2}|=|\frac{x}{2}|<\delta$, we have $|f(x)-f(y)|>3>\epsilon$. Most importantly, we need the existence of such an $x$ in our domain and for $\frac{x}{2}$ to also be in the domain. This works as $(0,1)$ is our domain and not say $(1,\infty)$. 
In short, you implictly mention it by having $\frac{x}{2}$.
A: Assume that your function is uniformly continuous on $(0,1)$. Then for a given $\varepsilon >0$ it must exists a $\delta=\delta(\varepsilon)>0$ such that for every $x,y \in (0,1)$ such that $|x-y|<\delta$ it is $|f(x)-f(y)|<\varepsilon$. Now, let be $x=2/n$ and $y=1/n$ with $n$ positive integer greater than $2$. You will get
$$
\left| {x - y} \right| < \delta  \Rightarrow \left| {\frac{1}
{{x^2 }} - \frac{1}
{{y^2 }}} \right| < \varepsilon 
$$
or 
$$
\frac{{\left| {y - x} \right|\left| {y + x} \right|}}
{{x^2 y^2 }} < \varepsilon 
$$
Since $|x-y|=1/n$, if $n>1/\delta$ you would have that for such values of $n$ it must be
$$
\frac{{\left| {\frac{1}
{n} - \frac{2}
{n}} \right|\left| {\frac{1}
{n} + \frac{2}
{n}} \right|}}
{{\frac{1}
{{n^2 }} \cdot \frac{4}
{{n^2 }}}} < \varepsilon 
$$
or $3n^2/4<\varepsilon$ which is absurd.
