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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.

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  • $\begingroup$ Is it $a_1...a_n $ $\endgroup$ – Archis Welankar Sep 12 '16 at 16:59
  • $\begingroup$ take ln of the expression would help. $\endgroup$ – Qing Zhang Sep 12 '16 at 17:00
  • $\begingroup$ @ArchisWelankar That is not really helpful, is it? Answering "How to find the limit" by "here is the limit." $\endgroup$ – Clement C. Sep 12 '16 at 17:12
  • $\begingroup$ As you have given the answer so i didnt give it $\endgroup$ – Archis Welankar Sep 13 '16 at 5:21
  • $\begingroup$ You can given another short and nice answer... $\endgroup$ – neelkanth Sep 13 '16 at 5:42
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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 $$\begin{align} \left(\frac{a_{1}^{\frac{1}{x}}+a_{2}^{\frac{1}{x}}+\cdots+{a_{n}}^{\frac{1}{x}}}{n}\right)^{nx} &= \exp\left(nx \ln \frac{1}{n}\sum_{k=1}^n e^{\frac{1}{x}\ln a_{k}}\right)\\ &= \exp\left(nx \ln\left( 1+ \frac{1}{nx}\sum_{k=1}^n\ln a_{k}+ o\left(\frac{1}{x}\right)\right)\right) \\ &= \exp\left(nx \left( \frac{1}{nx}\sum_{k=1}^n\ln a_{k}+ o\left(\frac{1}{x}\right)\right)\right) \\ &= \exp\left(\sum_{k=1}^n\ln a_{k}+ o\left(1\right)\right) \\ &= \exp\left(\ln \prod_{k=1}^na_{k}+ o\left(1\right)\right) = e^{o(1)} \prod_{k=1}^na_{k} \\ & \xrightarrow[x\to\infty]{} \prod_{k=1}^na_{k}. \end{align}$$

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