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1. The problem statement, all variables and given/known data

Given that f is continuous in [tex][1,\infty)[/tex] and [tex] lim_{x->\infty}f(x) [/tex] exists and is finite, prove that f is uniformly continuous in [1,[tex]\infty)[/tex]

3. The attempt at a solution

We will mark [tex] lim_{x->\infty}f(x) = L [/tex]. So we know that there exists [tex]x_{0}[/tex] such that for every [tex] x>x_{0} |L-f(x)|<\epsilon [/tex] so f is uniformly continuos in [tex] (x_{0},\infty) [/tex]

We shall look at the segment [tex] [1,x_{0}+1] [/tex]. We know that a continuous function ina closed segment is uniformly continuous.

From here I am puzled. I have two routes.

The easy route sais that since we showed f is uniformly continuous in those two segments and they overlap then f is uniformly continuous on the whole segment. I havea feeling this is not enough.

Route 2:

We know that a continous function ina closed segment has a minimum and maximal value there. Then we can look at the closed segment [tex] [1,x_{0}+1] [/tex] and mark [tex] m=MIN(f(x)), M= MAX(f(x)) [/tex]. But since f converges to L from [tex] x=,x_{0} [/tex] we have taken into account all of the possible values of f.

we will mark [tex] d= | |M| - |m| | [/tex] the biggest difference in value f gets.

and from here I am stuck. I am lost as to how to tie up the proof. Gudiance would be greatly apreciated.

Thanks

Tal

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# Homework Help: Uniform continuity

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