real analysis - Finding $lim_{x to infty} left(frac{a_{1}^{1/x}+a_{2}^{1/x}+cdotcdotcdot{a_{n}}^{1/x}}{n}right)^{nx}$

For non zero positive reals $a_{1},a_{2}\cdot\cdot\cdot a_{n}$ how to find
$$\lim_{x \to \infty} \left(\frac{a_1^{1/x} +a_2^{1/x}+\ldots +a_n^{1/x}}{n}\right)^{nx}?$$

It becomes indeterminant form $1^{\infty}.$ But difficult to solve by L'Hospital's Rule. By using A.M.-G.M. inequality it comes that limit is $\geq a_{1}a_{2}\cdot\cdot\cdot a_{n}.$ I also tried by using Squeeze. Please help.Thanks.

Answer

tl;dr

For $a_1,\dots, a_n$ positive numbers, and $x>0$, $$\lim_{x\to\infty}\left(\frac{a_{1}^{\frac{1}{x}}+a_{2}^{\frac{1}{x}}+\cdots+{a_{n}}^{\frac{1}{x}}}{n}\right)^{nx} = \prod_{k=1}^na_{k}$$

Proof. Detailed approach, where we use the (low-order) Taylor expansions $e^u=1+u+o(u)$ and $\ln(1+u)=u+o(u)$ when $u\to 0$.

You can rewrite
$$\begin{align} \left(\frac{a_{1}^{\frac{1}{x}}+a_{2}^{\frac{1}{x}}+\cdots+{a_{n}}^{\frac{1}{x}}}{n}\right)^{nx} &= \exp\left(nx \ln \frac{a_{1}^{\frac{1}{x}}+a_{2}^{\frac{1}{x}}+\cdots+{a_{n}}^{\frac{1}{x}}}{n}\right) = \exp\left(nx \ln \frac{1}{n}\sum_{k=1}^na_{k}^{\frac{1}{x}}\right)\\ &= \exp\left(nx \ln \frac{1}{n}\sum_{k=1}^n e^{\frac{1}{x}\ln a_{k}}\right) \end{align}$$
Since $n$ is a constant and $\frac{1}{x}\ln a_{k}\xrightarrow[x\to\infty]{} 0$ for each $k$, we have
$$e^{\frac{1}{x}\ln a_{k}} = 1 + \frac{1}{x}\ln a_{k} + o\left(\frac{1}{x}\right)$$
for each $k$ as $x\to\infty$, and therefore
$$\begin{align} \left(\frac{a_{1}^{\frac{1}{x}}+a_{2}^{\frac{1}{x}}+\cdots+{a_{n}}^{\frac{1}{x}}}{n}\right)^{nx} &= \exp\left(nx \ln \frac{1}{n}\sum_{k=1}^n e^{\frac{1}{x}\ln a_{k}}\right)\\ &= \exp\left(nx \ln\left( 1+ \frac{1}{nx}\sum_{k=1}^n\ln a_{k}+ o\left(\frac{1}{x}\right)\right)\right) \\ &= \exp\left(nx \left( \frac{1}{nx}\sum_{k=1}^n\ln a_{k}+ o\left(\frac{1}{x}\right)\right)\right) \\ &= \exp\left(\sum_{k=1}^n\ln a_{k}+ o\left(1\right)\right) \\ &= \exp\left(\ln \prod_{k=1}^na_{k}+ o\left(1\right)\right) = e^{o(1)} \prod_{k=1}^na_{k} \\ & \xrightarrow[x\to\infty]{} \prod_{k=1}^na_{k}. \end{align}$$