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March 31st, 2016, 11:24 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?
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Last edited by life24; March 31st, 2016 at 11:33 AM.
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March 31st, 2016, 11:46 AM   #2
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I'm not sure what your question is. Is there any equation containing these expressions that will allow you to solve for n?
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March 31st, 2016, 12:03 PM   #3
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Quote:
Originally Posted by 123qwerty View Post
I'm not sure what your question is. Is there any equation containing these expressions that will allow you to solve for n?
Thank you for your reply,
I want compare pair with each other when limitation go to infinity.
For example:
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March 31st, 2016, 12:15 PM   #4
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in other word Which greater than each other? Thanks
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March 31st, 2016, 12:18 PM   #5
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Quote:
Originally Posted by life24 View Post
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.
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March 31st, 2016, 01:24 PM   #6
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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.
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March 31st, 2016, 04:09 PM   #7
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Very good,
Please help me for it.
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March 31st, 2016, 06:09 PM   #8
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Originally Posted by life24 View Post
Very good,
Please help me for it.
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.
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March 31st, 2016, 06:20 PM   #9
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Originally Posted by 123qwerty View Post
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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