Suppose that $f(x)$ is continuous and positive on $[0,1]$ Prove that $$\lim_{p \to 0^+}(\int_{0}^{1}f^p(x)\text dx)^{1/p}=e^{\int_{0}^{1}\ln f(x)\text dx} $$.
This is my former question.Actually I have get a way by consider the limitation: $$ \lim_{x \to 0}(1+x)^{\frac{1}{x}}=e $$ So I changed the form :$$(\int_{0}^{1}f^p(x)\text dx)^{1/p}=(1+\int_{0}^{1}f^p(x)\text dx-1)^{\frac{1}{\int_{0}^{1}f^p(x)\text dx-1}\cdot\frac{\int_{0}^{1}f^p(x)\text dx-1}{p}}$$
Then in this situation I think the problem is equal to prove that: $$\lim_{p\to0^+}\frac{\int_{0}^{1}f^p(x)\text dx-1}{p}=\int_{0}^{1}\ln f(x)\text dx$$
I have considered mean value theorem and L'Hospital rule. But doesn't work.
So I really want to know is there any way to prove the last equation. Or I had thinking in a wrong way at first.
PS:My first and second question has deleted and this is my third question.And I am in a danger of being blocked to ask question. I really have searched before asking.If the question is also duplicate and too stupid.Please comment and let know.