5
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ This limit does not exist. Wolfram Alpha says so too in parentheses. $\endgroup$ – bames Feb 11 '18 at 6:36
  • 2
    $\begingroup$ Thanks for all the answers. I understand what's going on now. $\endgroup$ – Stephen Feb 11 '18 at 6:48
7
$\begingroup$

For every $a\in[-1,1]$, there is some sequence $(x_{n})$ such that $x_{n}\rightarrow 0$ and $\sin(1/x_{n})\rightarrow a$.

$\endgroup$
  • $\begingroup$ 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
  • 3
    $\begingroup$ 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
6
$\begingroup$

$$\Box \ \nexists \lim_{x\to 0}\bigg(\sin\frac1x\bigg).$$ 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. $\qquad \qquad\qquad\qquad\quad\,\,\,\,$

$\endgroup$
6
$\begingroup$

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 $0$.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.