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

If yes, then I have to find it; if no, then I need to give reason why

Bernard Lacey

Beginner2021-12-31Added 30 answers

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.

so then the limit becomes

which clearly pulls down to zero.

Edward Patten

Beginner2022-01-01Added 38 answers

First Approach

To begin with, notice that

$\left|x\mathrm{sin}\left(\frac{1}{x}\right)\right|\le \left|x\right|<\u03f5$

Thus, for every$\u03f5>0$ , there corresponds a $\delta =\u03f5$ s.t.

$0<|x-0|<\delta \Rightarrow |x\mathrm{sin}\left(\frac{1}{x}\right)-0|<\u03f5$

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.

To begin with, notice that

Thus, for every

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

Vasquez

Expert2022-01-08Added 669 answers

Yes, the limit there exist. Note that the funcion

Therefore,

Finally, by the well-know Squeeze theorem we can conclude that

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