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derivation of geometric mean as the limit of the power mean

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Fix $x_1, x_2, \ldots, x_n \in \mathbb{R}^+$.  Then let 
\mu(r) := \left(\frac{x_1^r+\cdots+x_n^r}{n}\right)^{1/r}.

For $r\neq 0$, by definition $\mu(r)$ is the $r$th power mean of the $x_i$.  It is also clear that $\mu(r)$ is a differentiable function for $r\neq 0$.  What is $\lim_{r\to 0} \mu(r)$?  

We will first calculate $\lim_{r\to 0} \log\mu(r)$ using \PMlinkname{l'H\^opital's rule}{LHpitalsRule}.
\lim_{r\to 0} \log\mu(r) & = \lim_{r\to 0} \frac{\log\left(\frac{x_1^r+\cdots +x_n^r}{n}\right)}{r}\\
& = \lim_{r\to 0} \frac{\left(\frac{x_1^r\log x_1+\cdots+x_n^r\log x_n}{n}\right)}{\left(\frac{x_1^r+\cdots+x_n^r}{n}\right)}\\
& = \lim_{r\to 0} \frac{x_1^r\log x_1+\cdots+x_n^r\log x_n}{x_1^r+\cdots+x_n^r}\\
& = \frac{\log x_1+\cdots+\log x_n}{n}\\
& = \log \sqrt[n]{x_1\cdots x_n}.

It follows immediately that 
\lim_{r\to 0} \left(\frac{x_1^r+\cdots+x_n^r}{n}\right)^{1/r} = \sqrt[n]{x_1\cdots x_n}.