prove: $\lim\limits_{t \to 0} (\frac{1}{n}\sum_{k=1}^{n}x_{k}^{t})^{\frac{1}{t}} = (x_{1}\cdots x_{n})^{\frac{1}{n}} $

Question

prove that for all: $x_{1},...,x_{n} > 0$

$\lim\limits_{t \to 0} (\frac{1}{n}\sum_{k=1}^{n}x_{k}^{t})^{\frac{1}{t}} = (x_{1}\cdots x_{n})^{\frac{1}{n}} $

i tried to some how use: Inequality of arithmetic and geometric means but with no success

Answer 1

Let $ a_{1},\cdots ,a_{n} $ be positive reals, $ n\in\mathbb{N}^{*} \cdot $

\begin{aligned} \displaystyle\lim_{x\to 0}{\left(\displaystyle\frac{1}{n}\displaystyle\sum_{k=1}^{n}{a_{k}^{x}}\right)^{\frac{1}{x}}}&=\displaystyle\lim_{x\to 0}{\mathrm{e}^{\frac{1}{x}\ln{\left(\frac{1}{n}\sum\limits_{k=1}^{n}{a_{k}^{x}}\right)}}} \\ &=\displaystyle\lim_{x\to 0}{\exp{\left(\displaystyle\frac{\frac{1}{n}\sum\limits_{k=1}^{n}{\left(a_{k}^{x}-1\right)}}{x}\times\displaystyle\frac{\ln{\left(\frac{1}{n}\sum\limits_{k=1}^{n}{a_{k}^{x}}\right)}}{\frac{1}{n}\sum\limits_{k=1}^{n}{a_{k}^{x}}-1}\right)}} \\ &=\displaystyle\lim_{x\to 0}{\exp{\left(\displaystyle\frac{1}{n}\displaystyle\sum_{k=1}^{n}{\ln{a_{k}}\displaystyle\frac{\mathrm{e}^{x\ln{a_{k}}}-1}{x\ln{a_{k}}}}\times\displaystyle\frac{\ln{\left(\frac{1}{n}\sum\limits_{k=1}^{n}{a_{k}^{x}}\right)}}{\frac{1}{n}\sum\limits_{k=1}^{n}{a_{k}^{x}}-1}\right)}} \\ &=\exp{\left(\displaystyle\frac{1}{n}\displaystyle\sum_{k=1}^{n}{\ln{a_{k}}}\right)} \\ \displaystyle\lim_{x\to 0}{\left(\displaystyle\frac{1}{n}\displaystyle\sum_{k=1}^{n}{a_{k}^{x}}\right)^{\frac{1}{x}}}&=\left(\displaystyle\prod_{k=1}^{n}{a_{k}}\right)^{\frac{1}{n}} \end{aligned}