upper bound turning into supremum

nhrock3

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

╔(σ_σ)╝

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.

micromass

To elaborate on what ╔(σ_σ)╝ said. You could try to find [tex]\lim_{x\rightarrow +\infty}{x\sin(1/x)}[/tex]. With this limit, it is easy to see that the supremum is 1...