Limits of the Natural Logarithm
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/
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