Thread: \lim_{x\rightarrow 0} sin (\frac{1}{x})

Show that doesn't exist by using the squeeze theorem.

Thanks in advance.

Originally Posted by Biscaim

Show that 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.





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

and



There for the limit does not exists at 0