March 31st, 2016, 10:24 AM  #1 
Senior Member Joined: Feb 2014 Posts: 114 Thanks: 1  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 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  
Senior Member Joined: Feb 2014 Posts: 114 Thanks: 1  Quote:
I want compare pair with each other when limitation go to infinity. For example:  
March 31st, 2016, 11:15 AM  #4 
Senior Member Joined: Feb 2014 Posts: 114 Thanks: 1 
in other word Which greater than each other? Thanks

March 31st, 2016, 11:18 AM  #5 
Senior Member Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics  
March 31st, 2016, 12:24 PM  #6 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,236 Thanks: 884 
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.

March 31st, 2016, 03:09 PM  #7 
Senior Member Joined: Feb 2014 Posts: 114 Thanks: 1 
Very good, Please help me for it. 
March 31st, 2016, 05:09 PM  #8 
Senior Member Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics  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  
Senior Member Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics  Quote:
$\displaystyle \lim_{n \rightarrow \infty} \frac{\log (n)^{\sqrt n}}{\log{n^2}^{\log n} } = \lim_{n \rightarrow \infty} \frac{\sqrt{n}\log (n)}{2\log n \log{n}}= \lim_{n \rightarrow \infty} \frac{\sqrt{n}}{2\log{n}}= \lim_{n \rightarrow \infty} \frac{0.5n^{0.5}}{\frac{2}{n \ln {10}}}= \lim_{n \rightarrow \infty} \frac{\ln{10} \sqrt{n}}{4} = \infty$  

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