Show that the geometric mean is the limit of the $t$th power mean as $t \to 0$

Question

Let $a_1,a_2,\dots,a_n$ be any $n$ positive real numbers. Show that $$\lim_{t \to 0^+}\left[\frac1n \sum_{i=1}^{n}a_i^t\right]^{1/t}$$ is the geometric mean of $a_1,a_2,\dots,a_n$.

I know $\text{GM}= \left[\displaystyle\prod_{i=1}^{n} a_i\right]^{1/n}$

$\displaystyle\lim_{t\to 0^+} a_i^t\approx 1 ~\forall~ i$ so AM=GM

So, $$\lim_{t \to 0^+}\left[\left(\displaystyle\prod_{i=1}^{n} a_i^t\right)^{1/n}\right]^{1/t}=\lim_{t \to 0^+} \left[\displaystyle\prod_{i=1}^{n} a_i\right]^{\displaystyle\frac tn\times \frac 1t}=\text{GM}$$

Am I correct? I can't give a solid argument. Can this problem solve in any other fashion.

Answer 1

The logarithm of our expression is equal to $$\frac{\log(\frac{1}{n}\sum a_i^t)}{t}.$$ Using L'Hospital's Rule we find that the limit is $$\lim_{t\to 0^+} \frac{\frac{1}{n} \sum(\log a_i) a_i^t }{\frac{1}{n}\sum a_i^t}.$$ This limit is $\frac{1}{n}\sum \log a_i$. Exponentiate.

Remark: The AM-GM argument is nice. However, some estimates would be needed to verify the intuition that we are "nearly" in the equality case.