Vtyjkyuk

What is
$$lim_xrightarrow0left( sinfrac1xright) $$?

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?

For every $ain[-1,1]$, there is some sequence $(x_n)$ such that $x_nrightarrow 0$ and $sin(1/x_n)rightarrow a$.

The limit $$lim_xrightarrow0left( sinfrac1xright)$$

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 $0$.

When $frac1xtoinfty$ 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.

Take $a_n=dfrac1npi$. Clearly, $limlimits_ntoinfty a_n=0$ and $b_n=dfrac1frac12(4npi +pi)$. Clearly, $limlimits_ntoinftyb_n=0$. Then, $limlimits_ntoinftysinleft(dfrac1a_nright)=sin(npi)=0$ and $limlimits_ntoinftysinleft(dfrac1b_nright)=sinleft(frac12(4npi +pi)piright)=1$. Thus, the limit doesnt' exist.

$$Box nexists lim_xto 0bigg(sinfrac1xbigg).$$ Proof: Let $u = dfrac1x$ then $$lim_xto 0frac1x = infty$$ since $$lim_xtoinftyfrac1x = 0.$$ Therefore, we get $$lim_xto 0bigg(sinfrac 1xbigg) = lim_utoinfty(sin u)$$ but this cannot exist because sine is a periodic fluctuating function. $qquad qquadqquadqquadquad,,,,bigcirc$