Proving Uniform Continuity of f on [1, $\infty$]

[talolard]

Hello,

Homework Statement

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

The Attempt at a Solution

We will mark $$lim_{x->\infty}f(x) = L$$. So we know that there exists $$x_{0}$$ such that for every $$x>x_{0} |L-f(x)|<\epsilon$$ so f is uniformly continuos in $$(x_{0},\infty)$$
We shall look at the segment $$[1,x_{0}+1]$$. 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 continuous function ina closed segment has a minimum and maximal value there. Then we can look at the closed segment $$[1,x_{0}+1]$$ and mark $$m=MIN(f(x)), M= MAX(f(x))$$. But since f converges to L from $$x=,x_{0}$$ we have taken into account all of the possible values of f.
we will mark $$d= | |M| - |m| |$$ 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

 

[torquil]

I think that I would try to use your "route 1", if you know that the mapping is uniformly continuous in both domains, and they overlap on [x0,x0+1]. Let's call them domains 1 and 2.

Uniform continuity in a domain demands that for any constant epsilon>0, there should be a constant delta>0 such that |f(y)-f(x)| < epsilon for all y with |y-x| < delta.

For a given epsilon, you can find such a delta for domain 1, and one for domain 2, just choose the smallest one of those, and I think is it OK, right? I guess delta also has to be less than the overlap of the two domains. You can always choose such a delta.

It's been a while since I worked on such problems, so please be very critical towards my statements!

Torquil