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