Let $x_1,x_2,\ldots,x_n$ be positive real numbers, and let $w_1,w_2,\ldots,w_n$ be positive real numbers such that $w_1+w_2+\cdots+w_n=1$ . For $r\neq 0$ , the $r$ -th weighted power mean of $x_1,x_2,\ldots,x_n$ is $$ M_w^r(x_1,x_2,\ldots,x_n)=(w_1x_1^r+w_2x_2^r+\cdots+w_nx_n^r)^{1/r}. $$ Using the Taylor series expansion $e^t=1+t+{\mathcal O}(t^2)$ , where ${\mathcal O}(t^2)$ is Landau notation for terms of order $t^2$ and higher, we can write $x_i^r$ as $$ x_i^r=e^{r\log x_i}=1+r\log x_i+{\mathcal O}(r^2). $$ By substituting this into the definition of $M_w^r$ , we get \begin{eqnarray*} M_w^r(x_1,x_2,\ldots,x_n)&=&\left[w_1(1+r\log x_1) +\cdots+w_n(1+r\log x_n)+{\mathcal O}(r^2)\right]^{1/r}\\ &=&\left[1+r(w_1\log x_1+\cdots+w_n\log x_n)+{\mathcal O}(r^2)\right]^{1/r}\\ &=&\left[1+r\log(x_1^{w_1}x_2^{w_2}\cdots x_n^{w_n})+{\mathcal O}(r^2)\right]^{1/r}\\ &=&\exp\left\{\frac{1}{r}\log\left[1+r\log(x_1^{w_1}x_2^{w_2}\cdots x_n^{w_n})+{\mathcal O}(r^2)\right]\right\}. \end{eqnarray*}Again using a Taylor series, this time $\log (1+t)=t+{\mathcal O}(t^2)$ , we get \begin{eqnarray*} M_w^r(x_1,x_2,\ldots,x_n)&=&\exp\left\{\frac{1}{r} \left[r\log(x_1^{w_1}x_2^{w_2}\cdots x_n^{w_n})+{\mathcal O}(r^2)\right]\right\}\\ &=&\exp\left[\log(x_1^{w_1}x_2^{w_2}\cdots x_n^{w_n})+{\mathcal O}(r)\right]. \end{eqnarray*}Taking the limit $r\to 0$ , we find \begin{eqnarray*} M_w^0(x_1,x_2,\ldots,x_n)&=&\exp\left[\log (x_1^{w_1}x_2^{w_2}\cdots x_n^{w_n})\right]\\ &=&x_1^{w_1}x_2^{w_2}\cdots x_n^{w_n}. \end{eqnarray*}In particular, if we choose all the weights to be $\frac{1}{n}$ , $$ M^0(x_1,x_2,\ldots,x_n)=\sqrt[n]{x_1x_2\cdots x_n}, $$ the geometric mean of $x_1,x_2,\ldots,x_n$ .