I need to know how to calculate this without using l'hospitals rule:
limit as x approaches infinity of: x(a1/x−1)
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!
No comments:
Post a Comment