$f(x) = \frac{2}{x}$ is not uniformly continuous on $(0,1]$ Show that $f(x) = \frac{2}{x}$ is not uniformly continuous on $(0,1]$
WLOG Suppose, $0< \delta \leq 1.$ Let, $\epsilon = 1$ and $x = \frac{\delta}{2}, y = x + \frac{\delta}{3},  x,y \in (0,1]$
Skipping all the details: $\mid f(x) - f(y)\mid =...= \mid \frac{24}{5\delta}\mid \geq 1 = \epsilon$, since $0<\delta\leq 1$.
I have seen people restricting the $\delta$ like I did here, but I have never done this before and I thought the only condition applies to $\delta$ is $\delta >0$(though restricting the $\delta$ does make things easier). So, since this is one of my HW question, am I allowed to do it like this (just to make sure I don't make any stupid mistake)? Do you think my approach is correct? Please let me know. Thanks. 
 A: You should choose $\varepsilon$ before $\delta$. Also $|x-y|< \delta$ must hold. So, it is better to say:
Assume for a contradiction that $f$ is uniformly continuous on $(0,1]$ and let $\varepsilon =1$. Then one can find $\delta' > 0$ such that the uniform continuity definition hold. 
Let $\delta = \min \{ \delta' , 1 \}$
Now let $$x= \frac{\delta}{2} \quad \text{ and  } \quad y= \frac{\delta}{3}$$
so that $$|x-y| < \delta$$
Hence,
$$\left| f(x) - f(y)\right| = \left| \frac{4}{\delta} - \frac{6}{\delta} \right| \geq 2 > \varepsilon$$
Contradiction.
A: Let us see what you do. You assume a function is uniformly continuous. You take $\epsilon = 1$. Then you have some $\delta$ such that for all $x,y$ with $|x-y| < \delta$ something  should hold. But if this something holds for all $|x-y| < \delta$ then to say it holds for all $|x-y| < \delta'$ for some $\delta' < \delta$ is a weaker assumption. 
Thus you can always assume a smaller delta and thus in particular you can assume  $\delta$ is less than $1$ if it helps.    
You could write: 
Assume for a contradiction that $f$ is uniformly continuous, and chose $\epsilon = 1$. There exists a $\delta > 0 $ such that for all $x,y$ with $|x-y|< \delta $ one has $|f(x)-f(y)| < 1$. WLOG we can assume $\delta <  1$.  Now let $x= \delta/2$ ...
A: I am really unsure that this is right, I think it's wrong but that's what I tried :Consider $x_{n}=\frac{1}{n}$, $y_{n}=\frac{1}{2n}$. $x_{n}-y_{n}=\frac{1}{2n}$. $|f(x_{n})-f(y_{n})|=2n$. Let $\epsilon =1$. Let $\delta>0$. Pick $n$ such that $\frac{1}{n}<\delta $. $|x_{n}-y_{n}|<\frac{1}{n}<\delta$ and $|f(x_{n})-f(y_{n}|>1$
A: Use  this  theorem  : A real  valued  continuous  function  is  uniformly  continuous  in  an  interval $(a,b)$   or $(a,b]$  or $[a,b)$  iff  it  can  be  extended  continuously  on $[a,b]$ . 
Now $2\over x$  is continuous  on $(0,1]$  but  not  continuous  at  the  point  $0$  as  $1\over x$  does  not  take  any  value  at  $0$ since  it cannot  even  be  defined  at  $0$. So  the  continuous  extension not  being  possible  it  is  not  uniformly  continuous  on  $(0,1]$.
