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