assume there is a limit, L. So we have that for all $\displaystyle {\epsilon}>0$ there exists a $\displaystyle {\sigma}>0$ such that $\displaystyle |x-0|=|x|<{\sigma} ->|sin(1/x)-L|<{\epsilon}$

Consider when x>_0 and let $\displaystyle {\epsilon}$=1/2.

Then |x|=x -> $\displaystyle 0<x<{\sigma}-> |sin(1/x)-L|<1/2$

I need a contradiction.

Thanks