Maria Huey

2021-12-30

Does the following limit exist?
$\underset{x\to 0}{lim}x\mathrm{sin}\left(\frac{1}{x}\right)$
If yes, then I have to find it; if no, then I need to give reason why

Bernard Lacey

You can let
$z=\frac{1}{x}$
so then the limit becomes
$\underset{x\to \mathrm{\infty }}{lim}\frac{1}{z}\mathrm{sin}z$
which clearly pulls down to zero.

Edward Patten

First Approach
To begin with, notice that
$|x\mathrm{sin}\left(\frac{1}{x}\right)|\le |x|<ϵ$
Thus, for every $ϵ>0$, there corresponds a $\delta =ϵ$ s.t.
$0<|x-0|<\delta ⇒|x\mathrm{sin}\left(\frac{1}{x}\right)-0|<ϵ$
whence we conclude the given function converges to zero as x approaches zero.
Second Approach
The limit exists and converges to zero due to the sandwich theorem.
That is because $x\to 0$ and the sin function is bounded.

Vasquez

Yes, the limit there exist. Note that the funcion $f\left(x\right)=\mathrm{sin}\left(x\right)$ is a bounded function, with
$|\mathrm{sin}\left(x\right)|\le 1⇔-1\le \mathrm{sin}\left(x\right)\le 1$
Therefore,
$-1\le \mathrm{sin}\left(\frac{1}{x}\right)\le 1$
$\underset{x\to 0}{lim}-x\le \underset{x\to 0}{lim}x\mathrm{sin}\left(\frac{1}{x}\right)\le \underset{x\to 0}{lim}x$
$0\le \underset{x\to 0}{lim}x\mathrm{sin}\left(\frac{1}{x}\right)\le 0$
Finally, by the well-know Squeeze theorem we can conclude that
$\underset{x\to 0}{lim}x\mathrm{sin}\left(\frac{1}{x}\right)=0$

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