when .
i proved that sin (1/x)<1/x
prove that sup{xsin (1/x)|x>0}=1
if we say that A={xsin (1/x)|x>0}
xsin (1/x)<x(1/x)=1
so one is upper bound
now i need to prove that there is no smaller upper bound so that 1 is the supremum
suppose that "t" is our smaller upper bound t<1 and epsilon=1-t
now i need to do some limit definition and |f(x)-1|<epsilon
|f(x)-1|<1-t
from that i need to get that t>1 so 1 is the only supremum
how to do that