How to prove $\sin(1/x)$ is not uniformly continuous How do I go about proving $f(x)=\sin(1/x)$ is not uniformly continuous?
(Or: different question, but same intention* how do I prove that $x\sin(x)$ is not uniformly continuous)
*I'm trying to grasp how one would prove $f$ is not uniformly continuous for functions other than the simple $x^n$. I have seen one technique being to set an $\epsilon$ and set $x, y$ in the form of $\delta$ (e.g. $\delta/2$, etc.) then subsequently proving that $f(x)-f(y)\ge\epsilon$
 A: Ultimately a very brief solution could be given to this problem, but I decided to write in some detail how you might approach it.
You want to negate the following: $$\forall \varepsilon>0,\exists\delta>0,\forall x,y, |x-y|<\delta\implies|f(x)-f(y)|<\varepsilon.$$
You can write out what that negation is rather mechanically, swapping universal and existential quantifiers, until you finally negate the implication by ensuring that $|x-y|<\delta$ and $|f(x)-f(y)|\geq \varepsilon$.  See here for a discussion of dissecting analysis problems like this by Tim Gowers.
That is, you want to prove:
$$\exists \varepsilon>0,\forall\delta>0,\exists x,y,\text{ such that } |x-y|<\delta\text{ and }|f(x)-f(y)|\geq\varepsilon.$$
Before giving the final argument, it is a good idea to experiment in a "backwards" fashion; think about where you want to end up and how you can get there.  Roughly, the conclusion "$|x-y|<\delta\text{ and }|f(x)-f(y)|\geq\varepsilon$" will be saying that $x$ and $y$ will be close while $f(x)$ and $f(y)$ will stay a distance $\varepsilon$ away.  The property of $\sin(1/x)$ that allows this to happen is that it oscillates like crazy between $1$ and $-1$ over smaller and smaller intervals of $x$ values.  So there will be "nearby" $x$ and $y$ such that $f(x)=-1$ and $f(y)=1$.  The distance between the function values here is $2$, while the distance between the input values can be arbitrarily small.  This leads to the conclusion that $\varepsilon = 2$ will be a sufficient choice.  
Next, with $\varepsilon$ fixed at $2$, and $\delta>0$ arbitrary but fixed, you need to show that there are $x$ and $y$ with $|x-y|<\delta$ and $|f(x)-f(y)|\geq 2$.  As indicated above, the last part can be achieved by ensuring that $f(x)=-1$ and $f(y)=1$.  For what $x$ and $y$ is it true that $\sin(1/x)=-1$ and $\sin(1/y)=1$?  Use what you know about the sine function to answer this question (I'll leave this to you).  Notice that the choices of such $x$ and $y$ get arbitrarily close to $0$, and in particular you can choose such $x$ and $y$ with $0<x,y<\delta$, which implies that $|x-y|<\delta$.

Largely the same approach applies to $x\sin(x)$, except that its reason for not being uniformly continuous changes.  Now the problem is where $x$ gets very large, and the rate of oscillation doesn't change, but the amplitude does.  A hint is to consider the function values at $x=2\pi n$ and $y=2\pi n + c$, where $c>0$ is "small", as $n$ goes to infinity.
A: Choose two sequences $T_n = \frac{1}{n}$ and $S_n = \frac{1}{n+\pi}$. Their difference goes to zero as $n$ goes to infinity. But $|f(S_n)-f(T_n)|=\left|2 \sin\left(\frac{n+\pi-n}{2}\right)\cos\left(\frac{n+\pi+n}{2}\right)\right|=\left|2 \cos\left(n+\frac{\pi}{2}\right)\right|=\left|2\sin(n)\right|>0$ for $n \in \mathbb{N}$, so the function is not uniformly continuous.
