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What is $$\lim\limits_{x\rightarrow0}{\left( \sin{\frac{1}{x}}\right)} $$?

Wolfram says "-1 to 1", but I don't know what that means.

In fact, I thought this limit didn't exist, so what does "-1 to 1" mean in this context?

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    $\begingroup$ This limit does not exist. Wolfram Alpha says so too in parentheses. $\endgroup$ – bames Feb 11 '18 at 6:36
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    $\begingroup$ Thanks for all the answers. I understand what's going on now. $\endgroup$ – Stephen Feb 11 '18 at 6:48
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For every $a\in[-1,1]$, there is some sequence $(x_{n})$ such that $x_{n}\rightarrow 0$ and $\sin(1/x_{n})\rightarrow a$.

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  • $\begingroup$ Of course, so the limit doesn't exist as expected. It must mean what RRL said. That is, it "oscillates" between -1 and 1. $\endgroup$ – Stephen Feb 11 '18 at 6:40
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    $\begingroup$ Yes, the function assumes every point in $[-1,1]$ as adherent point. $\endgroup$ – user284331 Feb 11 '18 at 6:41
  • $\begingroup$ In other words, $$-1\leqslant \sin x \leqslant 1,$$ for all $x\in\mathbb{R}$. $\endgroup$ – Mr Pie Feb 20 '18 at 10:29
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$$\Box \ \nexists \lim_{x\to 0}\bigg(\sin\frac1x\bigg).$$ Proof: Let $u = \dfrac{1}{x}$ then $$\lim_{x\to 0}\frac{1}{x} = \infty$$ since $$\lim_{x\to\infty}\frac{1}{x} = 0.$$ Therefore, we get $$\lim_{x\to 0}\bigg(\sin\frac 1x\bigg) = \lim_{u\to\infty}(\sin u)$$ but this cannot exist because sine is a periodic fluctuating function. $\qquad \qquad\qquad\qquad\quad\,\,\,\,$

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The limit $$\lim_{x\rightarrow0}{\left( \sin{\frac{1}{x}}\right)}$$

does not exist.

Note that as x approaches $0$, $\sin( 1/x )$ covers the closed interval $[-1, 1]$ infinitely many times.

For example at $ x= \frac {2}{(2n+1)\pi } $ we have $\sin(1/x)=\pm 1.$

Thus there is no limit for $\sin(1/x)$ as $x$ approaches $0$.

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When $\frac1x\to\infty$ it can be any value resulting from -1 to 1 because $-1<sin($any real number$)<1$ and the exact angle of $\infty$ cannot be determined.

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Take $a_n=\dfrac{1}{n\pi}$. Clearly, $\lim\limits_{n\to\infty} a_n=0$ and $b_n=\dfrac{1}{\frac{1}{2}(4n\pi +\pi)}$. Clearly, $\lim\limits_{n\to\infty}b_n=0$. Then, $\lim\limits_{n\to\infty}\sin\left(\dfrac{1}{a_n}\right)=\sin(n\pi)=0$ and $\lim\limits_{n\to\infty}\sin\left(\dfrac{1}{b_n}\right)=\sin\left(\frac{1}{2}(4n\pi +\pi)\pi\right)=1$. Thus, the limit doesnt' exist.

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