TITLE: whats the proof for $\lim_{x → 0} [(a_1^x + a_2^x + .....+ a_n^x)/n]^{1/x} = (a_1.a_2....a_n)^{1/n}$
QUESTION [9 upvotes]: This equation is directly given in my book and I am don't know anything about its proof.I tried L'Hospital rule by differentiating the both numerator as well as denominator(division rule), but the result is still coming in indeterminate forms.I am a beginner , and haven't practiced limits that much. This formula is really confusing me.

REPLY [4 votes]: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
&\color{#f00}{%
\lim_{x \to 0}\bracks{%
\pars{a_{1}^{x} + a_{2}^{x} + \cdots + a_{n}^{x} \over n}^{1/x}}} =
\exp\pars{\lim_{x \to 0}\bracks{{1 \over x}\,
\ln\pars{a_{1}^{x} + a_{2}^{x} + \cdots + a_{n}^{x} \over n}}}
\\[3mm] = &\
\exp\pars{\lim_{x \to 0}{\bracks{%
a_{1}^{x}\ln\pars{a_{1}} + a_{2}^{x}\ln\pars{a_{2}} + \cdots + a_{n}^{x}\ln\pars{a_{n}}}/n \over
\bracks{a_{1}^{x} + a_{2}^{x} + \cdots + a_{n}^{x}}/n}}\qquad
\pars{~\mbox{L'H}\mathrm{\hat{o}\mbox{pital Rule}}~}
\\[3mm] = &\
\exp\pars{\ln\pars{a_{1}} + \ln\pars{a_{2}} + \cdots +\ln\pars{a_{n}} \over n} =
\color{#f00}{\pars{a_{1}a_{2}\ldots a_{n}}^{1/n}}
\end{align}