Let $f$ be positive and continuous on $[0,1]$. Find $\lim_{p \to 0} ( \int_{0}^{1} f(x)^p dx)^{1/p}$
My approach: $f$ is continuous on $[0,1] \implies$ $\exists \min_{x\in[0,1]}{f(x)}$ (call it $m$) and $\max_{x\in[0,1]}f(x)$ (call it $M$). so we have $0<m\leq f(x) \leq M$ for all $x\in[0,1]$ therefore $m\leq (\int_{0}^{1} f(x)^p dx)^{1/p} \leq M$ so by Intermediate value theorem, $\exists c\in[0,1]$ such that $f(c) = (\int_{0}^{1} f(x)^p dx)^{1/p}$ Then $\lim_{p \to 0} (\int_{0}^{1} f(x)^p dx)^{1/p} = \lim_{p\to 0} f(c)$
Can't proceed further
