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Let $f:[1,\infty)$ a continuious and non-negative function s.t. $\int_1^\infty f(x)$ converges.

I don't understand why the existence of $lim_{x\rightarrow\infty} f($x) neccessairly means that $lim_{x\rightarrow\infty} f($x) = 0.(why?)

Plus, Does $lim_{x\rightarrow\infty} f(x) = 0$ even if the assumption of the existence of is not given?(why?)

Lastly, Is the function $f$ bounded over the line $[1,\infty)$?(why?)

Thanks in adnvance!

user21312
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1 Answers1

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Suppose that $l:= \lim_{x\rightarrow\infty} f(x)>0$. Then there is $c_1>1$ such that $f(x) \ge l/2$ for $x \ge c_1$. For $u$ with $u>c_1$ we then have

$\int_u^{u+1}f(x) dx \ge l/2$.

If $ 0< \epsilon < l/2$ we have by Cauchy: there is $c_2 \ge c_1$ such that

$\int_u^{u+1}f(x) dx < \epsilon$ for $u>c_2$, a contradiction.

Yes, $f$ is bounded: there is $r>1$ such that $f(x) \le 2l$ for $x>r$. Since $f$ is continuous, $f$ is also bounded on $[1,r]$

Fred
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  • In the part Plus, Does limx→∞f(x)=0 even if the assumption of the existence of is not given?(why?). Can I infer that this is already given by the fact that f(x) is convergent? – user21312 Apr 28 '17 at 13:59