F(x) integral converge, f(x) uniformly continuous ,prove that f(x) limit = 0

[ThankYou]

Homework Statement

\int_{a}^\infty\ f(x) dx <--- converge
f(x) uniformly continuous in [a,\infty]
prove that lim_{x\rightarrow \infty} f(x) = 0

Homework Equations

The Attempt at a Solution

I know that if f(X) has a limit in \infty it has to be 0
I think that the solution has to be conected to the fact that if f(x) uniformly continuous ,there is a M that |f'(x)|<M,
I think I can prove that if f(x) does not have limit it's Derivative has to change infinite times form + to - , so f(x) has to go up and down infinite times..
and when it go up there is a limit to how low her max can be,
I think I have to put it all together with Cauchy test
But I can't seem to do it.
Thank you

 

[Dick]

Try a proof by contradiction. If f(x) does NOT converge to zero, then there is an epsilon>0 such that for all N>0 there is an x>N such that |f(x)|>epsilon. Do you agree with that statement? If you draw a picture, then that means that the graph of f(x) must have an infinite number of points where |f(x)|>epsilon stretching off to infinity. If f(x) is also uniformly continuous, then there is a corresponding delta such that |f(x)-f(y)|<epsilon/2 for |x-y|<delta. That means all of those x's in your picture correspond to a bump in the graph of f(x) with area greater than delta*epsilon/2. Can you show this contradicts the convergence of the integral?

 

[ThankYou]

Thank you very much
That was very helpful