Limits question Watch

Edamame
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#1
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#1
How do I find the limit of

ln(lnx)/sqrt(x)

as x tends to infinity??

Thanks
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Chewwy
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#2
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ln(ln(x)) < x^0.25 for large x, by quite some way.

now, sandwich..
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Edamame
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(Original post by Chewwy)
ln(ln(x)) < x^0.25 for large x, by quite some way.

now, sandwich..
Hmmm?? I can't use qualitative description, I think. I tried L'Hopital's rule, but that led me nowhere.
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Edamame
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How?? On the whole expression??
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Chewwy
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(Original post by Edamame)
Hmmm?? I can't use qualitative description, I think. I tried L'Hopital's rule, but that led me nowhere.
huh? i'm pretty sure i set you on track to a perfectly rigourous and simple answer..
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Edamame
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Sorry what do you mean by now, sandwich??
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Zhen Lin
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Sandwich theorem. If a function is between two other functions for all values, then, the limit of that function is in between the limit of those two functions.
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Edamame
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Ok I've never learned sandwich theorem before. Then I must use Maclaurin's series somehow...
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Edamame
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#9
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I find this question impossible. But it's not, because I've plotted the graph, and it tends to a positive value.

Please help.

I tried expanding it by Maclaurin's series, but I'm still getting no where!
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DFranklin
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Take k > 0. First prove that k ln x < x for sufficiently large x.

Deduce ln(ln x) < ln x for sufficiently large x and ln(x) < x^{1/k} for sufficiently large x.

This makes it easy to show ln(ln(x))/sqrt(x) < x^a for some a < 0.
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Chewwy
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(Original post by Edamame)
Ok I've never learned sandwich theorem before. Then I must use Maclaurin's series somehow...
perfect logic..
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Edamame
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(Original post by Chewwy)
perfect logic..
It's exam logic... What isn't taught won't be examined. Ok I still don't get it. Please forgive, because I don't do maths, but I am forced to do it cause I do science (at uni). Argh... I don't hate maths, but I am just so slow sometimes.

I still interested in the Maclaurin series. I think it's the key to solving this question, but how to use it properly? Do I have to manipulate the limit somehow?
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DFranklin
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(Original post by Edamame)
It's exam logic... What isn't taught won't be examined.
I think it's the "I must use Maclaurin's series somehow..." bit Chewwy was questioning.

In fact, this question has nothing to do with Maclaurin, but because you seem to be fixated on that, you are essentially ignoring all the suggestions people have made.

If you think that's good exam logic, you're spectacularly wrong...
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Edamame
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#14
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#14
Sorry for being so ignorant and stupid. Can somebody please explain to me how to do this question step by step?
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Totally Tom
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(Original post by DFranklin)
Take k > 0. First prove that k ln x < x for sufficiently large x.

Deduce ln(ln x) < ln x for sufficiently large x and ln(x) < x^{1/k} for sufficiently large x.

This makes it easy to show ln(ln(x))/sqrt(x) < x^a for some a < 0.
read again.
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Maths Buster
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First show that ln x < x for x >1. Call this result *.
Then put x = y^4 so that y tends to infinity as x tends to infinity.
Your expression becomes
{ln[4lny]}/[y^2] = [ln4]/[y^2] + {ln[lny]}/[y^2]
< [ln4]/[y^2] + {lny]/[y^2], using *.
< [ln4]/[y^2] + 1/y, using *.
And this tends to zero as y tends to infinity.
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Edamame
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(Original post by Maths Buster)
First show that ln x < x for x >1. Call this result *.
Then put x = y^4 so that y tends to infinity as x tends to infinity.
Your expression becomes
{ln[4lny]}/[y^2] = [ln4]/[y^2] + {ln[lny]}/[y^2]
< [ln4]/[y^2] + {lny]/[y^2], using *.
< [ln4]/[y^2] + 1/y, using *.
And this tends to zero as y tends to infinity.
But it doesn't tend to zero! It tends to a positive value... try plot a graph..
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Maths Buster
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x = 50 gives you 0.193.
x = 1000 gives you 0.0611.
x = one million gives you 0.0026.
As x increases to infinity, your expression tends to zero.
Convinced now?
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Edamame
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#19
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(Original post by Maths Buster)
x = 50 gives you 0.193.
x = 1000 gives you 0.0611.
x = one million gives you 0.0026.
As x increases to infinity, your expression tends to zero.
Convinced now?
Yes. This function is a bit tricky - it tends to zero so slowly...
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Edamame
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#20
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(Original post by Maths Buster)
First show that ln x < x for x >1. Call this result *.
Then put x = y^4 so that y tends to infinity as x tends to infinity.
Your expression becomes
{ln[4lny]}/[y^2] = [ln4]/[y^2] + {ln[lny]}/[y^2]
< [ln4]/[y^2] + {lny]/[y^2], using *.
< [ln4]/[y^2] + 1/y, using *.
And this tends to zero as y tends to infinity.
How to prove that lnx < x for x>1 ?

Is this good enough:

x < e^x for x > 1 (just because - I am not good at proofs)

so

ln x < x
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