# Any idea on this problem $\lim \limits_{x\to\infty}f(x)=0$ [closed]

This is a question in real analysis. I think it needs integral definition to prove, but not sure. Any idea is welcome.

Let $f$ be a real valued uniform continuous function on $[0,\infty]$ such that $$\lim \limits_{x\to\infty}\int_{0}^{x}f(t)dt$$ exists. Prove that $$\lim \limits_{x\to\infty}f(x)=0$$

## closed as off-topic by user223391, user147263, user91500, Claude Leibovici, N. F. TaussigMay 23 '15 at 9:32

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• try mean value. – chenbai May 23 '15 at 3:19
• This might be one of the questions in analysis which is most often asked on the site. – Did May 24 '15 at 8:57

Hint: Let $F(x) = \int_0^xf.$ So we have $F'(x) + F(x) \to L$ at $\infty.$ Consider the function $e^xF(x).$
maybe this can help: Let $F(x)$ be the integral of $f(x)$, so $E(x) = f(x)+F(x)-F(0)=exp(-x)*d[e^x*F(x)]/dx-F(0)$and $lim_x E(x)=k$, for some $k$ constant, but this is equivalent to say that $lim$ of $exp(-x)*d[e^x*F(x)]dx=k+F(0)=C$; follows by intermediate theorem of Cauchy that exist a,b with $a<x<b$ such; $d[e^x*F(x)]dx*exp[-x]=(e^a*F(a)-e^b*F(b))/(e^a-e^b)$, applying the limits with b tending to infinity and making some manipulations, we got $e^b*(F(b)-C)=e^a(F(a)-C)$ (this is a constant, lets say M); $e^b(F(b)-C)=M$,so $F(b)-C=e^(-b)*M=0$ because b tends to infinity, we have $F(b)=C$, and this is equivalent to $f(b)=0$.