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

In summary, the conversation discusses proving that a function f(x) with a limit at infinity must converge to 0, given that f(x) is uniformly continuous in the interval [a, infinity]. The suggested approach is to use a proof by contradiction, showing that if f(x) does not converge to zero, then there must be an infinite number of points where the function is greater than a certain epsilon, which contradicts the convergence of the integral.
  • #1
ThankYou
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Homework Statement


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


Homework Equations





The Attempt at a Solution



I know that if f(X) has a limit in [tex]\infty[/tex] 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
 
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  • #2
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?
 
  • #3
Thank you very much
That was very helpful
 

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

1. What does it mean for an F(x) integral to converge?

When an F(x) integral converges, it means that the limit of the integral approaches a finite value as the upper and lower limits of integration approach infinity. In other words, the area under the curve approaches a finite value and does not continue to increase indefinitely.

2. How is uniform continuity related to the convergence of f(x)?

Uniform continuity is a property of a function that guarantees that the function does not have any sudden jumps or discontinuities. When a function is uniformly continuous, it is easier to determine if the integral of the function will converge or not. If a function is not uniformly continuous, it may still have a convergent integral, but it is harder to prove.

3. How can one prove that the limit of f(x) is equal to 0?

In order to prove that the limit of f(x) is equal to 0, one must use the definition of a limit. This involves showing that for any small value of epsilon, there exists a corresponding delta such that if the distance between x and the limit point is less than delta, then the distance between f(x) and 0 is less than epsilon. In other words, as x gets closer and closer to the limit point, f(x) gets closer and closer to 0.

4. Is the converse true? If f(x) limit = 0, does that mean that the integral of f(x) converges?

No, the converse is not necessarily true. Just because the limit of f(x) is equal to 0 does not automatically mean that the integral of f(x) will converge. The convergence of an integral depends on various factors, such as the behavior of the function near its limits of integration and its continuity properties.

5. What are some common methods used to prove the convergence of an integral?

The most commonly used methods to prove the convergence of an integral include the comparison test, limit comparison test, ratio test, and integral test. These methods involve comparing the given integral to a known, simpler integral or series, and using various convergence tests to determine if the integral converges or diverges.

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