calculus - If $a_1,a_2,dotsc,a_n>0 $, then $limlimits_{x to infty} left[frac {a_1^{1/x}+a_2^{1/x}+dotsb+a_n^{1/x}}{n}right]^{nx}=a_1 a_2 dotsb a_n$

If $a_1,a_2,\dotsc,a_n $ are positive real numbers, then prove that

$$\lim_{x \to \infty} \left[\frac {a_1^{1/x}+a_2^{1/x}+.....+a_n^{1/x}}{n}\right]^{nx}=a_1 a_2 \dotsb a_n.$$

My Attempt:

Let $P=\lim_{x \to \infty} \left[\dfrac {a_1^{\frac{1}{x}}+a_2^{\frac {1}{x}}+.....+a_n^{\frac {1}{x}}}{n}\right]^{nx} \implies \ln P=\lim_{x \to \infty} \ln \left[\frac {a_1^{\frac{1}{x}}+a_2^{\frac {1}{x}}+.....+a_n^{\frac {1}{x}}}{n}\right]^{nx} =\lim_{x \to \infty} nx \ln \left[\frac {a_1^{\frac{1}{x}}+a_2^{\frac {1}{x}}+.....+a_n^{\frac {1}{x}}}{n}\right]= \lim_{x \to \infty} n \left[\frac {\ln (a_1^{1/x}+a_2^{1/x}+...+a_n^{1/x})-\ln n}{1/x}\right]$

and this is $0/0$ form and so I have to apply L'Hospital's rule. Now things get a bit complicated during derivative.

Can someone point me in the right direction? Thanks in advance for your time.

Answer

Let $$\begin{align}P &=\lim_{x \to \infty} \Big[\dfrac {a_1^{\dfrac{1}{x}}+a_2^{\dfrac {1}{x}}+.....+a_n^{\dfrac {1}{x}}}{n}\Big]^{nx}\\ \implies
\ln P &=\lim_{x \to \infty} \ln \Big[\dfrac {a_1^{\dfrac{1}{x}}+a_2^{\dfrac {1}{x}}+.....+a_n^{\dfrac {1}{x}}}{n}\Big]^{nx} \\&=\lim_{x \to \infty} nx \ln \Big[\dfrac {a_1^{\dfrac{1}{x}}+a_2^{\dfrac {1}{x}}+.....+a_n^{\dfrac {1}{x}}}{n}\Big]\\&= \lim_{x \to \infty} n \Big[\dfrac {\ln (a_1^{1/x}+a_2^{1/x}+...+a_n^{1/x})-\ln n}{1/x}\Big]\\&=\lim_{z \to 0} n \Big[\dfrac {\ln (a_1^{z}+a_2^{z}+...+a_n^{z})-\ln n}{z}\Big]\end{align}$$

and this is $0/0$ form and so I have to apply L'Hospital's rule. So,$$\begin{align}\ln P &=n \lim_{z \to 0}\dfrac {1}{ (a_1^{z}+a_2^{z}+...+a_n^{z})} \times \{a_1^z \ln a_1+a_2^z \ln a_2+.....+a_n^z \ln a_n\} \\&=n \times \dfrac {1}{n}\{\ln a_1+\ln a_2+....+\ln a_n\}\\&=\ln (a_1.a_2...a_n)\\\implies P&=a_1.a_2...a_n \end{align}$$.

This completes the proof.