n乗根の相加平均のn乗の極限が相乗平均になる話

\begin{eqnarray} \lim_{n\rightarrow\infty}\left(\sum_{k=1}^{m}\frac{\sqrt[n]{a_k}}{m}\right)^n&=&\lim_{x\rightarrow +0}\left(\sum_{k=1}^{m}\frac{a_k^x}{m}\right)^\frac{1}{x}\quad(x=\frac{1}{n}) \end{eqnarray}
$\lim$の中身の対数をとったものを考えると
\begin{eqnarray} \lim_{x\rightarrow +0}\log\left(\sum_{k=1}^{m}\frac{a_k^x}{m}\right)^\frac{1}{x}&=&\lim_{x\rightarrow +0}\frac{\log\sum_{k=1}^{m}\frac{a_k^x}{m}}{x}\\&=&\lim_{x\rightarrow +0}\frac{f(x)}{x}\quad (f(x)=\log\sum_{k=1}^{m}\frac{a_k^x}{m})\\&=&\lim_{x\rightarrow +0}\frac{f(x)-f(0)}{x-0}\quad(\because f(0)=\log1=0)\\&=&f^{\prime}(0)\quad (\because f(x)\text{は微分可能}) \end{eqnarray}
$$f^{\prime}(x)=\frac{\sum_{k=1}^{m}\frac{a_k^x}{m}\log a_k}{\sum_{k=1}^{m}\frac{a_k^x}{m}}=\frac{\sum_{k=1}^{m}a_k^x\log a_k}{\sum_{k=1}^{m}a_k^x}$$
より
$$f^{\prime}(0)=\frac{\sum_{k=1}^{m}\log a_k}{\sum_{k=1}^{m}1}=\frac{1}{m}\log\prod_{k=1}^{m}a_k=\log\sqrt[m]{\prod_{k=1}^{m}a_k}$$
つまり
$$\lim_{x\rightarrow +0}\log\left(\sum_{k=1}^{m}\frac{a_k^x}{m}\right)^\frac{1}{x}=\log\sqrt[m]{\prod_{k=1}^{m}a_k} $$
$\exp(x)$は連続関数だから
$$ \lim_{x\rightarrow+0}\left(\sum_{k=1}^{m}\frac{a_k^x}{m}\right)^\frac{1}{x}=\lim_{x\rightarrow+0}\exp\left(\log\left(\sum_{k=1}^{m}\frac{a_k^x}{m}\right)^\frac{1}{x}\right)=\sqrt[m]{\prod_{k=1}^{m}a_k}$$

つまり
$$\lim_{n\rightarrow\infty}\left(\sum_{k=1}^{m}\frac{\sqrt[n]{a_k}}{m}\right)^n=\sqrt[m]{\prod_{k=1}^{m}a_k}$$