Limit of (log x / sqrt x)
Mark Thakkar
x) converges to 0 at x tends to infinity.
So far, I have the following:
1. For x >= 8, the function is monotone decreasing;
2. For x >= 1, the function is non-negative.
So I know that the limit *exists*, but I don't know how to show that it is
zero.
I have been trying to show that, given any positive number r, we can find
an x-value which will give f(x) = r, thereby showing that the limit must
be zero.
However, I can't think of a way to solve this equation for x:
(log x) / (sqrt x) = r.
Any ideas?
Thanks,
Mark.
Mark Thakkar
> I'm having enormous difficulty proving that (log x) / (sqrt
> x) converges to 0 at x tends to infinity.
>
>
> I have been trying to show that, given any positive number r, we can
find
> an x-value which will give f(x) = r, thereby showing that the limit must
> be zero.
>
> However, I can't think of a way to solve this equation for x:
>
> (log x) / (sqrt x) = r.
Another thought: we can rewrite (log x) / (sqrt x) as log (x ^ 1/sqrt x).
Then we get exp(r) = x ^ 1/sqrt x.
So, to assign a value of x to each small positive r, we need to show that
(x ^ 1/sqrt x) tends to 1 as x => infinity.
So, my question slightly changes focus: prove that (x ^ 1/sqrt x) tends to
1 as x tends to infinity. (I have checked this result using Maple, and it
is correct. I need the /algebra/, though...)
Thanks,
Mark.
Mark Thakkar
> So, my question slightly changes focus: prove that (x ^ 1/sqrt x) tends
to
> 1 as x tends to infinity. (I have checked this result using Maple, and
it
> is correct. I need the /algebra/, though...)
I have shown by differentiation that this function is monotone decreasing
after x=7.4, and by reasoning that f(x) >= 1 for x >= 1.
So there is a limit. Why is this limit 1?
Anyone?
Mark.
Wizard of Id
infinity/infinity so you can take the derivative top and bottom and when you
flip the fraction and multiply you will get
1/infinity which has a value of 0.
--
--
The Wizard of Id
pig...@vif.com
Mark Thakkar <mark.t...@balliol.ox.ac.uk> wrote in message
news:85i4nv$7t9$1...@news.ox.ac.uk...
| Hello, all. I'm having enormous difficulty proving that (log x) / (sqrt
| x) converges to 0 at x tends to infinity.
|
| So far, I have the following:
|
| 1. For x >= 8, the function is monotone decreasing;
|
| 2. For x >= 1, the function is non-negative.
|
| So I know that the limit *exists*, but I don't know how to show that it is
| zero.
|
| I have been trying to show that, given any positive number r, we can find
| an x-value which will give f(x) = r, thereby showing that the limit must
| be zero.
|
| However, I can't think of a way to solve this equation for x:
|
| (log x) / (sqrt x) = r.
|
| Any ideas?
|
| Thanks,
|
| Mark.
|
|
yuc...@my-deja.com
> Use l'Hopital's rule. log x / sqrt x as x approaches infinity becomes
> infinity/infinity so you can take the derivative top and bottom and
> when you flip the fraction and multiply you will get
>
> 1/infinity which has a value of 0.
But L'Hopital's rule only applies when both the numerator and the
denominator tend to /zero/ as x tends to some constant a.
Here, our numerator and denominator are both infinite, and we wish x to
approach infinity.
Does that sound right? Any other ideas?
Mark.
Sent via Deja.com http://www.deja.com/
Before you buy.
ne...@arn.net
Mark,
I think you need calculus here... Let L be the limit of x^(1/sqrt x)
as x approaches infinity. Then take the natural log of both sides.
ln(L) = ln(x^(1/sqrt x))
ln(L) = (1/sqrt x)(ln x)
ln(L) = (ln x)/(sqrt x)
Then apply l'Hopital's Rule. I.e. the limit of an indeterminant fraction (0/0
or inf/inf) is equal to the limit of the derivative of the numerator over the
derivative of the denominator.
Now,
ln(L) = (1/x)/((1/2)x^(-1/2))
ln(L) = 2/x^(1/2)
As x goes to inf, ln(L) = 0.
So, L = e^0 = 1.
Good luck! =)
--Denise
ne...@arn.net
Mark Thakkar
If anyone's interested, the method is:
=
Let x = e^y so that (log x)/sqrt(x) = y /exp(y/2). You want to show that
as y -> infinity then y/exp(y/2) -> 0. But for positive y, exp(y/2) > 1 +
y/2 + y^2/8 > y^2/8. (From the series). Hence
y/exp(y/2) < 8/y. This is enough.
=
Mark.
CR
>
> But L'Hopital's rule only applies when both the numerator and the
> denominator tend to /zero/ as x tends to some constant a.
>
> Here, our numerator and denominator are both infinite, and we wish x to
> approach infinity.
>
> Does that sound right? Any other ideas?
L'Hopital's rule can be applied provided that both the numerator and
denominator in the given function approach either 0 or infinity when the
variable approaches anything (infinity, zero, any other constant, from
the right, left, or both). You can't use it if the numerator and/or
denominator approaches something other than 0 or infinity, or if the
numerator approaches zero while the denominator approaches infinity or
vice versa.
This problem can be readily handled by L'Hopital's rule.
Bill Bliss
>But L'Hopital's rule only applies when both the numerator and the
>denominator tend to /zero/ as x tends to some constant a.
>Mark.
It also applies when numerator and denominator tend to infinity.
-Bill-
Sebastian Bleasdale
>But L'Hopital's rule only applies when both the numerator and the
>denominator tend to /zero/ as x tends to some constant a.
>
>approach infinity.
>
>Does that sound right? Any other ideas?
Take the reciprical of the both the numerator and the denominator. Then
they'll both tend to zero :). Or don't bother - it can be proven that
it works at infinity as well.
--
Sebastian Bureaucrat, n.: A person who cuts red tape sideways.