Friday, 23 February 2018

calculus - The limit as n approaches infinity of n(a1/n1)



I need to know how to calculate this without using l'hospitals rule:



limit as x approaches infinity of: x(a1/x1)



I saw that the answer is log(a), but I want to know how they got it.

The book implies that I should be able to find it by just using algebraic manipulation and substitution.


Answer



METHOD 1:



lim






METHOD 2:



Another way to do this is to substitute y=a^x in (1). Then



\begin{align} \lim_{x\to \infty}x(a^{1/x}-1)&=\lim_{x\to 0^{+}}\frac{a^{x}-1}{x}\\\\ &=\lim_{y\to 1^{+}}\frac{y-1}{\log y/\log a}\\\\ &=\log a\,\lim_{y\to 1^{+}}\frac{y-1}{\log y} \end{align}



Noting that for y>1, \frac{y-1}{y}\le\log y\le y-1. Then,



1\le\frac{y-1}{\log y}\le y



and the squeeze theorem does the rest!


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