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Limits of the Natural Logarithm

Date: 07/22/98 at 04:53:59
From: Andrew C.
Subject: Limits 


I have a question about limits as x tends to infinity. We came across 
a question that asked us to evaluate the limit of x (ln x)^n as x tends 
to infinity. We were supposed to answer whether or not this limit tends 
to 0. We considered n = 1, ie x (ln x) and my teacher said that this 
did tend to 0 as x-> infinity but wouldn't prove it. Can you help with 
this? And can you also help with whether or not the limit x (ln x)^n 
as x-> infinity? 


Date: 07/22/98 at 11:28:14
From: Doctor Rob
Subject: Re: Limits 

The way to evaluate this type of limit is to use the fact that the 
natural logarithm function and the exponential function are both 
continuous for all arguments > 0. Note that:

   lim x*(ln[x])^n = exp(lim ln[x*(ln[x])^n])
                   = exp(lim ln[x] + n*ln[ln(x)])

Now if n >= 0, you know that lim ln[x] = infinity, and also lim 
ln[ln(x)] = infinity, so the limit of the sum will be infinity, and 
your limit will also be infinity. 

If n < 0, things become a little more complicated, and now you should 
use L'Hopital's Rule:

   lim x*(ln[x])^n = lim (ln[x])^n/(1/x)
                   = lim (n*[ln(x)]^[n-1]*[1/x])/(-1/x^2)
                   = (-n)*lim x*(ln[x])^(n-1)

This relates the limiting values for differing values of n. We want to 
use it backwards, since n < 0:

   lim x*(ln[x])^n = (1/[-n-1])*lim x*(ln[x])^(n+1)
                   = (1/[-n-1])*(1/[-n-2])*lim x*(ln[x])^(n+2)

and so on. Ultimately, using this formula, you will reduce the limit
computation to finding the limit when n >= 0, which, as we saw above, 
is infinite.

Our conclusion is that for any real value of n, this limit is 

- Doctor Rob, The Math Forum
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Associated Topics:
High School Analysis
High School Calculus

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