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Prove that the function defined by

$$f(x)=\sin\left(\frac{1}{x}\right)$$

is not uniformly continuous on the interval $(0,1)$.

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  • $\begingroup$ Perhaps they wanted you to show your work. Perhaps someone didn't know what uniformly continuous means. $\endgroup$ – Omnomnomnom Dec 8 '15 at 17:17
  • $\begingroup$ The information about Quora is superfluous. As to this site, it would be better next time if you showed what you tried, where you're stuck and such details, which helps people answer your question better. $\endgroup$ – Pedro Tamaroff Dec 8 '15 at 17:21
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We need to show:

$$(\exists \epsilon >0) \space \space (\forall \delta >0) \space \space (\exists \space x, y \in \mathbb{R})$$ with $$\left |\space x-y\space \right|<\delta,\space\space\space\space \left|\space f(x)-f(y) \space\right|\geq\epsilon$$

Let $\epsilon=1$. Let $\delta>0$ be arbitrary. By the corollary of the Archimedean property $\exists n_0 \in \mathbb{N}$ such that $\frac{1}{n_0}<2\pi\delta$. Let $x_n = \frac{1}{2n\pi}$ and $y_n = \frac{1}{(2n+1/2)\pi}$.$\space$Then for every $n \in \mathbb{N}, n>n_0$ we have

$$\left|\space x_n-y_n\space \right|=\left|\space \frac{1}{2n\pi}-\frac{2}{(4n+1)\pi}\space \right|= \frac{1}{2n\pi(4n+1)}<\frac{1}{2n\pi}<\frac{1}{2n_0\pi}<\delta$$ but $$\left|\space f(x_n)-f(y_n)\space \right|=\left| \space sin(2n\pi)-sin((2n+\frac{1}{2})\pi) \space \right|=1\geq\epsilon.$$

$$\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \blacksquare$$

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A uniformly continuous function sends cofinal sequences to cofinal sequences, that is, if $|x_n-y_n|\to 0$ and $f$ is uniformly continuous, then $|f(x_n)-f(y_n)|\to 0$. You should be able to prove this using the definition of uniformly continuous function. This doesn't happen with $\sin(x^{-1})$ over $(0,1)$, for the sequences defined by $1/x_n=2\pi n$ and $1/y_n = 2\pi n +1/2$ are cofinal in $(0,1)$ yet their image sequences are not.

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