Does the following limit exist? limx→0xsin⁡(1x) If yes, then I have to find it; if no, then I need to give reason why

Question

Does the following limit exist?  limx→0xsin⁡(1x)  If yes, then I have to find it; if no, then I need to give reason why

Bernard Lacey · Accepted Answer

You can let   z=1x   so then the limit becomes  limx→∞1zsin⁡z  which clearly pulls down to zero.

Edward Patten · Accepted Answer

First Approach    To begin with, notice that   |xsin⁡(1x)|≤|x|&amp;lt;ϵ  Thus, for every ϵ&amp;gt;0, there corresponds a δ=ϵ s.t.   0&amp;lt;|x−0|&amp;lt;δ⇒|xsin⁡(1x)−0|&amp;lt;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→0 and the sin function is bounded.

Vasquez · Accepted Answer

Yes, the limit there exist. Note that the funcion f(x)=sin⁡(x) is a bounded function, with |sin⁡(x)|≤1⇔−1≤sin⁡(x)≤1 Therefore, −1≤sin⁡(1x)≤1 limx→0−x≤limx→0xsin⁡(1x)≤limx→0x 0≤limx→0xsin⁡(1x)≤0 Finally, by the well-know Squeeze theorem we can conclude that limx→0xsin⁡(1x)=0

Does the following limit exist? \lim_{x \to 0} x \sin(\frac{1}{x}) If yes,

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