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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 $f(x)-f(y)\ge\epsilon$

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    @Jonas Meyer: I do not mean to compare both functions when they are small. I mean the two functions are "similar". So an analogous proof should be able to constructed if the author notice $Sin[x]$ has a maximum difference of 2 near the point $\frac{1}{x}$ is not well-defined. In fact the author's statement is not clear, because by stating "is not uniformly continuous" one is assuming the function is in some underlying domain already. If it is an close interval with no singular points we may just apply Cantor's theorem.2011-11-07

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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, 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.

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    I don't know what you are trying to do. If you are trying to write a proof of something in analysis, like showing that $1/x$ is not uniformly continuous, it is typically insufficient to consider finitely many calculator values, or the shape of the graph, without giving precise, valid, logical reasons for your assertions. E.g., $1/x$ is not uniformly continuous because if I take $\varepsilon=1$, then for all \delta>0, there exist $x$ and $y$ with |x-y|<\delta but |1/x - 1/y|>1. You can even find formulas for $x$ and $y$ that will work, depending on $\delta$, if you want.2013-07-30
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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 $|f(S_n)-f(T_n)|=2|\cos n|$. Thus, can't find sigma for any epsilon greater than zero (from definition).