$f(x) = \frac{2}{x}$ 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. 

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  $(0,1]$.
