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### Limits of the Natural Logarithm

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

Hi,

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?

Thanks,
Andrew
```

```
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
infinite.

- Doctor Rob, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
High School Analysis
High School Calculus

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