$\sin(1/x)$ is not uniformly continuous on $(0, \frac{\pi}{2}]$ 
Show that  $$ f(x)=\sin\frac1x $$ is not uniformly continuous on
  $(0,\frac\pi2]$.

It is looking easy to do this problem if it is asked for $(0, 1)$ but I am not getting for the given range of $(0, \frac{\pi}{2}]$ . Please help
 A: If it is not uniformly continuous on $A$, it is not uniformly continuous on $B\supset A$.
A: Consider $\delta > 0$. For $n$ big enough, 
$$0< y := \frac 1{\frac \pi 2 + (2n+1)\pi}< x := \frac 1{\frac \pi 2 + 2n\pi} <\delta$$ Now what is $f(x) - f(y)$?
A: Hint: Try finding two sequences $\langle x_n\rangle$ and $\langle y_n \rangle $ such that $x_n-y_n\to 0$ and $|f(x_n)-f(y_n)|>\epsilon_0(>0)$
Note: In $[a,\frac{\pi}{2}]$ for some $a>0$, $f(x)$ is continuous. That is, $f(x)$ is continuous on a closed and bounded interval and so is uniform continuous on the interval. The only possible point which is creating a problem is $x=0$. That's the reason why you need to choose $x_n-y_n\to 0$
Try $\langle x_n\rangle=\frac{1}{2n\pi}$ and $\langle y_n \rangle =\frac{1}{(4n+1)\pi /2} $, maybe.
A: For $n\in N$ let $x_n=1/\pi n$ and $y_n=1/(\pi n+\pi /2).$
We have $0<|x_n-y_n|<1/n^2.$
Let $f(x)=\sin 1/x$ for $x\ne 0.$ We have $f(x_n)=0$ and $f(y_n)=\pm 1$ so $|f(x_n)-f(y_n)|=1.$ 
There cannot exist $d>0$ such that $\forall x,y\in (0,\pi/2]\;(|x-y|<d\implies |f(x)-f(y)|<1).$ Because for any $d>0$ there exists $n\in N$ with $1/n^2<d,$ so $|x_n-y_n|<d$ but $|f(x_n)-f(y_n)|=1.$ 
Draw a graph of $f(x)$ for $x\in (0,r]$ for some (any) $r>0.$ 
