Show that $\displaystyle \lim_{x\rightarrow 0} sin (\frac{1}{x})$ doesn't exist by using the squeeze theorem.

Thanks in advance.

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- Apr 12th 2009, 04:49 PMBiscaim\lim_{x\rightarrow 0} sin (\frac{1}{x})
Show that $\displaystyle \lim_{x\rightarrow 0} sin (\frac{1}{x})$ doesn't exist by using the squeeze theorem.

Thanks in advance. - Apr 12th 2009, 05:10 PMTheEmptySet
I don't see how the squeeze theorem will help but here is a method.

By using the sequental char of limits all we need to do is find two different sequences that converge to 0 that have different function limits.

$\displaystyle x_n=\frac{1}{n\pi}$

$\displaystyle y_n=\frac{1}{(\frac{\pi}{2}+2n\pi)}$

note that both of these go to zero and n goes to infinity, but

$\displaystyle \sin\left(\frac{1}{x_n}\right)=\sin(n\pi)=0$ and

$\displaystyle \sin\left( \frac{1}{y_n}\right)=\sin\left( \frac{\pi}{2}+2\pi\right)=1$

There for the limit does not exists at 0