What is $\lim_{x\to\infty}\left(\frac{a_1^{1/x}+a_2^{1/x}+\dots+a_n^{1/x}}{b_1^{1/x}+b_2^{1/x}+\dots+b_n^{1/x}}\right)^x$?

Question

$$\lim_{x\to\infty}\left(\frac{a_1^{1/x}+a_2^{1/x}+\dots+a_n^{1/x}}{b_1^{1/x}+b_2^{1/x}+\dots+b_n^{1/x}}\right)^x$$ Can you try to do this limit without L'Hôpital's rule or Taylor expansion? Thank you for trying.

Answer 1

By the definition of the derivative, we have $$ \begin{align} \lim_{1/x\to0}\frac{a^{1/x}-1}{1/x} &=\left.\frac{\mathrm{d}}{\mathrm{d}t}a^t\,\right|_{t=0}\\ &=\left.\frac{\mathrm{d}}{\mathrm{d}t}e^{t\log(a)}\,\right|_{t=0}\\ &=\left.\vphantom{\frac{\mathrm{d}}{\mathrm{d}t}}\log(a)\,e^{t\log(a)}\,\right|_{t=0}\\[3pt] &=\log(a) \end{align} $$ Therefore, we have $a_k^{1/x}=1+\log(a_k)/x+o\!\left(1/x\right)$ as $x\to\infty$, $$ \begin{align} \lim_{x\to\infty}\left(\frac{a_1^{1/x}+a_2^{1/x}+\cdots+a_n^{1/x}}{b_1^{1/x}+b_2^{1/x}+\cdots+b_n^{1/x}}\right)^x &=\lim_{x\to\infty}\left(\frac{1+\frac1n\log(a_1a_2\cdots a_n)/x+o\!\left(1/x\right)}{1+\frac1n\log(b_1b_2\cdots b_n)/x+o\!\left(1/x\right)}\right)^{\large x}\\[3pt] &=\left(\frac{a_1a_2\cdots a_n}{b_1b_2\cdots b_n}\right)^{1/n} \end{align} $$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\lim_{x \to \infty}\pars{a_{1}^{1/x} + \cdots a_{n}^{1/x} \over b_{1}^{1/x} + \cdots b_{n}^{1/x}}^{x} = \exp\pars{\lim_{x \to 0^{+}}\bracks{\ln\pars{a_{1}^{x} + \cdots + a_{n}^{x}} - \ln\pars{b_{1}^{x} + \cdots + b_{n}^{x}}\over x}} \\[5mm] = &\ \exp\pars{\lim_{x \to 0^{+}}\bracks{% {a_{1}^{x}\ln\pars{a_{1}} + \cdots + a_{n}^{x}\ln\pars{a_{n}} \over a_{1}^{x} + \cdots + a_{n}^{x}} - {b_{1}^{x}\ln\pars{b_{1}} + \cdots + b_{n}^{x}\ln\pars{b_{n}} \over b_{1}^{x} + \cdots + b_{n}^{x}}}} \\[5mm] = &\ \exp\pars{\bracks{% {\ln\pars{a_{1}} + \cdots + \ln\pars{a_{n}} \over n} - {\ln\pars{b_{1}} + \cdots + \ln\pars{b_{n}} \over n}}} = \exp\pars{{1 \over n}\ln\pars{a_{1} \ldots a_{n} \over b_{1} \ldots b_{n}}} \\[5mm] = &\ \bbox[#ffe,10px,border:1px dotted navy]{\ds{% \pars{a_{1} \cdots a_{n} \over b_{1} \cdots b_{n}}^{1/n}}} \end{align}