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Limits

Posted December 1st, 2009 by alpha
in

$ \mathop{\lim}\limits_{x \to \infty}(a^\frac{1}{x}-1)x $

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answer

On December 1st, 2009 Structure says:

Take

$$y=\frac{1}{x}$$
$$ \mathop{\lim}\limits_{x \to \infty}(a^\frac{1}{x}-1)x =\lim_{y\to 0}\frac{a^y-1}{y}=\lim_{y\to 0}\frac{e^{\ln a^y}-1}{y}=\lim_{y\to 0}\frac{e^{y\ln a}-1}{y}=\ln a$$

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