## upper bound turning into supremum

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

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 I would have to think about how to show it directly but you can apply some theorems to show that 1 is the suprema. x_nsin(1/(x_n)) is a bounded sequence hence it has a monotone subsequence. The monotone subsequence either converges to the infimum of the sequence or the suprema ( you may have to prove this.) Your job would be to try to find such a sequence. I am not sure how to show the suprema is 1 directly without using sequences or other sequence approach.
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus To elaborate on what ╔(σ_σ)╝ said. You could try to find $$\lim_{x\rightarrow +\infty}{x\sin(1/x)}$$. With this limit, it is easy to see that the supremum is 1...