# limit of an integrand whose integral is convergent

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...

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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

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 $-a$.

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