# whats the proof for $\lim_{x → 0} [(a_1^x + a_2^x + …+ a_n^x)/n]^{1/x} = (a_1.a_2…a_n)^{1/n}$

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.

• Take a logarithm, then l'hopital it. – Ivan Neretin Jul 5 '16 at 12:53
• You need the $a_i$ positive for this to work. Also, not only the limit holds, but it approaches from the right; that is the expression on the left (without the limit) is always bigger than the expression on the right (the geometric mean) for all $x>0$ – Ant Jul 6 '16 at 21:38

You can compute the limit of the logarithm of your expression: $$\lim_{x\to0}\frac{\log(a_1^x+\dots+a_n^x)-\log n}{x}$$ which is the derivative at $0$ of $$f(x)=\log(a_1^x+\dots+a_n^x)$$ Since $$f'(x)=\frac{a_1^x\log a_1+\dots+a_n^x\log a_n}{a_1^x+\dots+a_n^x}$$ we have $$f'(0)=\frac{\log a_1+\dots+\log a_n}{n}= \log\bigl((a_1\dotsm a_n)^{1/n}\bigr)$$ and therefore your limit is $$\lim_{x\to0}\left(\frac{a_1^x + a_2^x +\dots+ a_n^x}{n}\right)^{\!1/x} =e^{f'(0)}=(a_1\dotsm a_n)^{1/n}$$
I suppose that all the $a_k$ are positive. Let $u_n(x)$ be your expression. Put $\displaystyle v_n(x)=\log u_n(x)=\frac{\log(\frac{a_1^x+\cdots+a_n^x}{n})}{x}$ and $\displaystyle F(x)= \log(\frac{a_1^x+\cdots+a_n^x}{n})=\log G(x)$. Then we have $\displaystyle v_n(x)=\frac{F(x)-F(0)}{x}$. Now $G$ has a derivative at $x=0$, and $F$ also. So $v_n(x)$ has for limit the derivative of $F$ at $0$. As $\displaystyle F^{\prime}(x)=\frac{G^{\prime}(x)}{G(x)}$ and $G(0)=1$, this is $G^{\prime}(0)$. Now $a_k^x=\exp(x\log a_k)$, and it is easy to finish.
The simplest route is to take logarithms. Let $$f(x) = \left(\frac{a_{1}^{x} + a_{2}^{x} + \cdots + a_{n}^{x}}{n}\right)^{1/x}$$ and we need to calculate the limit of $f(x)$ as $x \to 0$. Let $L$ be this desired limit then we have \begin{align} \log L &= \log\left(\lim_{x \to 0}f(x)\right)\notag\\ &= \lim_{x \to 0}\log f(x)\text{ (via continuity of log)}\notag\\ &= \lim_{x \to 0}\frac{1}{x}\log\frac{a_{1}^{x} + a_{2}^{x} + \cdots + a_{n}^{x}}{n}\notag\\ &= \lim_{x \to 0}\frac{1}{x}\cdot\left(\dfrac{a_{1}^{x} + a_{2}^{x} + \cdots + a_{n}^{x}}{n} - 1\right)\cdot\dfrac{\log\left(1 + \dfrac{a_{1}^{x} + a_{2}^{x} + \cdots + a_{n}^{x}}{n} - 1\right)}{\dfrac{a_{1}^{x} + a_{2}^{x} + \cdots + a_{n}^{x}}{n} - 1}\notag\\ &= \frac{1}{n}\lim_{x \to 0}\frac{a_{1}^{x} + a_{2}^{x} + \cdots + a_{n}^{x} - n}{x}\notag\\ &= \frac{1}{n}\lim_{x \to 0}\sum_{k = 1}^{n}\frac{a_{k}^{x} - 1}{x}\notag\\ &= \frac{1}{n}\sum_{k = 1}^{n}\lim_{x \to 0}\frac{a_{k}^{x} - 1}{x}\notag\\ &= \frac{1}{n}\sum_{k = 1}^{n}\log a_{k}\notag\\ &= \log(a_{1}a_{2}\cdots a_{n})^{1/n}\notag \end{align} It now follows that $$L = (a_{1}a_{2}\cdots a_{n})^{1/n}$$ We have used the following two standard limits in the above derivation $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1,\,\lim_{x \to 0}\frac{a^{x} - 1}{x} = \log a$$