Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'd appreciate a hint to my problem:

If $\int_{1}^{\infty} f(x) \, dx$ converges where $f$ is continuous on $[0,\infty)$ with $\lim_{x\rightarrow \infty} f(x) < \infty$ then show that $\lim_{x\rightarrow \infty} f(x) = 0$.

My attempt: suppose that $\lim_{x\rightarrow \infty} f(x) = a$. Then set $\epsilon >0$. By definition $\exists \, M > 0 \,\, \text{s.t. } |f(x)-a|<\epsilon \text{ whenever } x>M.$

So $a - \epsilon < f(x) < a+ \epsilon \Rightarrow \int_{M}^{\infty} (a - \epsilon) \, dx < \int_{M}^{\infty} f(x) \, dx <\infty$ as $f$ is convergent. But for $\int_{M}^{\infty} (a - \epsilon) \, dx < \infty, \,\, a = \epsilon.$

Now I'm stuck...

share|improve this question
    
Assume $a\ne0$. What can you then say about $\int_M^\infty f(x)\,dx$ for $M$ big? –  David Mitra Feb 5 '13 at 17:32

1 Answer 1

You are on the right track, can continue:

If $a-\epsilon>0$ then the integral from $M$ to $\infty$ would be infinity. Hence it already follows that $a\le\epsilon\ $ for all $\epsilon>0$. So, if $a\ge 0$ was assumed we are ready.

For the case $a<0$, we can use this for $-f$ and $-a$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.