Show that $\displaystyle \lim_{x\rightarrow 0} sin (\frac{1}{x})$ 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.
$\displaystyle x_n=\frac{1}{n\pi}$
$\displaystyle y_n=\frac{1}{(\frac{\pi}{2}+2n\pi)}$
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