Taylor polynomial approximation (Little $o$) Suppose $(a_1,...,a_n) \in {\mathbb{R}_{*}^{+}}^n$, how can I prove that using the little $o$ (Taylor polynomial approximation):
$\displaystyle\lim _{ x\to +\infty  }{ \left( \cfrac { \sum _{ i=1 }^{ n }{ { a_{ i } }^{ \frac { 1 }{ x }  } }  }{ n }  \right) ^{ x } } =\sqrt [ n ]{ \prod _{ i=1 }^{ n }{ a_{ i } }  } $
 A: For each $i$ we have
$$
a_i^{1/x} = \exp\left(\frac{\log a_i}{x}\right) = 1 + \frac{\log a_i}{x} + o\left(\frac{1}{x}\right)
$$
as $x \to \infty$ by using the Taylor series for the exponential function, so that
$$
\begin{align}
\frac{1}{n}\sum_{i=1}^{n} a_i^{1/x} &= \frac{1}{n}\left[n + \frac{1}{x} \sum_{i=1}^{n} \log a_i + o\left(\frac{1}{x}\right)\right] \\
&= 1 + \frac{1}{x} \log\sqrt[n]{\prod_{i=1}^{n} a_i} + o\left(\frac{1}{x}\right)
\end{align}
$$
as $x \to \infty$.  Thus
$$
\begin{align}
\left(\frac{1}{n}\sum_{i=1}^{n} a_i^{1/x}\right)^x &= \exp\left\{x\log\left[\frac{1}{n}\sum_{i=1}^{n} a_i^{1/x}\right]\right\} \\
&= \exp\left\{ x \log\left[ 1 + \frac{1}{x} \log\sqrt[n]{\prod_{i=1}^{n} a_i} + o\left(\frac{1}{x}\right) \right] \right\} \\
&= \exp\left\{ x \left[ \frac{1}{x} \log\sqrt[n]{\prod_{i=1}^{n} a_i} + o\left(\frac{1}{x}\right) \right] \right\} \\
&= \exp\left\{ \log\sqrt[n]{\prod_{i=1}^{n} a_i} + o(1) \right\}
\end{align}
$$
as $x \to \infty$.  Note that in the third line we used the fact that $\log(1+y) = y + o(y)$ as $y \to 0$ in the third line.  From this the result follows.
