My Math Forum Limit involving Logarithm

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March 31st, 2016, 10:24 AM   #1
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Limit involving Logarithm

Hello,
Would you mind help me please for solve it?
Or told me can i solve it with which logarithm feature?
Thanks
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Last edited by life24; March 31st, 2016 at 10:33 AM.

 March 31st, 2016, 10:46 AM #2 Senior Member   Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics I'm not sure what your question is. Is there any equation containing these expressions that will allow you to solve for n?
March 31st, 2016, 11:03 AM   #3
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Quote:
 Originally Posted by 123qwerty I'm not sure what your question is. Is there any equation containing these expressions that will allow you to solve for n?
I want compare pair with each other when limitation go to infinity.
For example:
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 March 31st, 2016, 11:15 AM #4 Senior Member   Joined: Feb 2014 Posts: 112 Thanks: 1 in other word Which greater than each other? Thanks
March 31st, 2016, 11:18 AM   #5
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Quote:
 Originally Posted by life24 Thank you for your reply, I want compare pair with each other when limitation go to infinity. For example:
That's equivalent to $\displaystyle \lim_{n \rightarrow \infty} \frac{\sqrt n }{2 \log n}$, so I guess it's infinity.

 March 31st, 2016, 12:24 PM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,729 Thanks: 705 The first, $\displaystyle log(n^{\sqrt{n}})$, is the same as $\displaystyle \sqrt{n}log(n)$. The second , $\displaystyle log(\sqrt{n}^n)$, is the same as $\displaystyle nlog(n^{1/2})= \frac{1}{2}n log(n)$. $\displaystyle \sqrt{n}< \frac{1}{2} n$ as long as $\displaystyle n< n^2/4$, $\displaystyle n^2- 4n= n(n- 4)> 0$, n> 4. Thanks from life24
March 31st, 2016, 03:09 PM   #7
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Very good,
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March 31st, 2016, 05:09 PM   #8
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Quote:
The second one is $\displaystyle n^2 \log(\log n)^{n^2}$. I tried to solve the limit by hand but it was quite complicated (though a rather clear win for the denominator). Graphing the two functions, it's pretty clear the denominator wins out, so the limit is 0.

March 31st, 2016, 05:20 PM   #9
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Quote:
 Originally Posted by 123qwerty That's equivalent to $\displaystyle \lim_{n \rightarrow \infty} \frac{\sqrt n }{2 \log n}$, so I guess it's infinity.
To clarify,
$\displaystyle \lim_{n \rightarrow \infty} \frac{\log (n)^{\sqrt n}}{\log{n^2}^{\log n} } = \lim_{n \rightarrow \infty} \frac{\sqrt{n}\log (n)}{2\log n \log{n}}= \lim_{n \rightarrow \infty} \frac{\sqrt{n}}{2\log{n}}= \lim_{n \rightarrow \infty} \frac{0.5n^{-0.5}}{\frac{2}{n \ln {10}}}= \lim_{n \rightarrow \infty} \frac{\ln{10} \sqrt{n}}{4} = \infty$

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