# Limit of generalized mean with exponent tend to zero

How to get the following limit: $$\lim_{p \to 0}\sqrt[p]{\frac{1}{n}\sum_{i=1}^{n}x_i^p}=\sqrt[n]{\prod_{i=1}^n x_i}$$ I.e. how to get geometric mean from definition of generalized means?

I attempted to apply exponentiation/derivation/logarithmization but all these gives nothing to me.

• For ease of notation, consider the $n=2$ case. You can write $\sqrt[p]{\frac{x_1^p + x_2^p}{2}}$ as $\exp\left(\frac{\ln\left(\frac{x_1^p + x_2^p}{2}\right)}{p}\right)$. You could then try using L'Hôpital's rule. – Minus One-Twelfth Feb 23 at 12:02
• An equivalent question here – Fabio Lucchini Feb 23 at 14:33

Take logarithm on the LHS to obtain $$\lim_{p\to 0}\frac{\log\left(\frac1{n}\sum_{i=1}^n x_i^p\right)}p=\lim_{p\to 0}\frac{\log\left(\frac1{n}\sum_{i=1}^n x_i^p\right)}{\frac1{n}\sum_{i=1}^n x_i^p-1}\cdot\lim_{p\to 0}\frac{\frac1{n}\sum_{i=1}^n (x_i^p-1)}{p}.$$ Note that the first term tends to $$1$$ since $$\lim_{t\to 1}\frac{\log t}{t-1}=(\log t)'|_{t=1}=1$$. The second term is equal to $$\frac1 n\sum_{i=1}^n\lim_{p\to 0}\frac{x_i^p-1}{p}=\frac1 n\sum_{i=1}^n\log(x_i)\lim_{p\to 0}\frac{e^{\log(x_i)\cdot p}-1}{\log(x_i)\cdot p}=\frac1 n\sum_{i=1}^n\log(x_i)$$ since $$\lim_{t\to 0}\frac{e^t-1}t=1$$. So, we have that the LHS converges to $$\exp\left(\frac1 n\sum_{i=1}^n\log(x_i)\right)=\left(\prod_{i=1}^n x_i\right)^{\frac1 n}$$ as wanted.