Improper integral convergence and implications of infinite limits

[Szichedelic]

Homework Statement

Let f be a continuous function on [1,∞) such that \lim_{x\rightarrow ∞}f(x)=α. Show that if the integral \int^{∞}_{1} f(x)dx converges, then α must be 0.

Homework Equations

Definition of an Improper Integral
Let f be a continuous function on an interval [a,∞). then we define \int^{∞}_{a} f(x)dx to be \lim_{b \rightarrow ∞}\int^{b}_{a} f(x)dx if this limit exists. In this case, the integral is said to be convergent.

The Attempt at a Solution

My attempt at this solution was first to start with the epsilon-delta version of a limit of a function. That is, since \lim_{x\rightarrow ∞}f(x)=α, then \forall \epsilon>0 with x_{0}\in [1,∞) \exists~ \delta>0 s.t. |f(x) - α | whenever |x-x_{0}|<\delta. Then, I assumed that α was not equal to zero. From there I split up my inequality so that

-ε<f(x) - α< ε \Rightarrow α - ε<f(x)< ε + α \Rightarrow \int^{∞}_{1}(α - ε)dx < \int^{∞}_{1} f(x)dx < \int^{∞}_{1} (α + ε)dx

From here, I am getting stuck in my logic and I think that perhaps this is not the route I want to go. I would appreciate it if I could be offered any hints on where to go from here, or on whether or not I should start over. Thanks!

 

[tiny-tim]

Hi Szichedelic!

My attempt at this solution was first to start with the epsilon-delta version of a limit of a function. That is, since lim_{x\rightarrow∞}f(x)=α, then \forall \epsilon>0 with x_{o} \in [1,∞) \exists \delta>0 s.t. |f(x) - α | whenever |x-x_{o}|<\delta.

No, for ∞, your δ is a number "near ∞" …

we usually use N instead, and we look for an N such that x > N => |f(x) - α| < ε