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Math Help - \lim_{x\rightarrow 0} sin (\frac{1}{x})

  1. #1
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    \lim_{x\rightarrow 0} sin (\frac{1}{x})

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

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by Biscaim View Post
    Show that \lim_{x\rightarrow 0} sin (\frac{1}{x}) doesn't exist by using the squeeze theorem.

    Thanks in advance.
    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.

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

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

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

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

    \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
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