Hi guys,

please can someone here help me to prove that:

Assume that:

$\displaystyle \int_1^{\infty}c(x)^{1/2}<\infty$

this implies that:

$\displaystyle \lim_{x \to\infty}{x c(x)^{1/2}}=0$

thanks in advance

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- Dec 17th 2006, 10:33 AMmiss_lolittaUrgent and important??! integral inequality
Hi guys,

please can someone here help me to prove that:

Assume that:

$\displaystyle \int_1^{\infty}c(x)^{1/2}<\infty$

this implies that:

$\displaystyle \lim_{x \to\infty}{x c(x)^{1/2}}=0$

thanks in advance - Dec 17th 2006, 01:30 PMThePerfectHacker
- Dec 17th 2006, 01:57 PMmiss_lolitta
hi..ThePerfectHacker

right..c(x) is a decreasing function and continous:o

thanks:) so much - Dec 17th 2006, 03:05 PMThePerfectHacker
Then it is simple.

$\displaystyle c(x)\geq 0$

Otherwise the expression is undefined.

Now, if $\displaystyle c(x)$ is decreasing function and continous then,

$\displaystyle \sqrt{c(x)}$ is decreasing.

And we are told that,

$\displaystyle \int_1^{\infty} \sqrt{c(x)}dx$

Converges.

That means, by the integral test,

$\displaystyle \sum_{k=1}^{\infty} \sqrt{c(k)}$

Converges.

But that means that,

$\displaystyle \lim_{k\to \infty}\sqrt{c(k)}=0$.

Thus,

$\displaystyle \lim_{x\to \infty}x\cdot \sqrt{c(x)}=?$

We now that,

$\displaystyle \lim_{x\to \infty}\sqrt{c(x)}=0$*)

This is a L'Hopital rule problem.

(But then we need to know that the function is differenciable).

*)I am not sure about this. Perhaps this statement works only the other way.