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
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
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.