Prove: $\lim_{n\rightarrow\infty}\prod_{k=1}^{n}\left(1+\frac{1}{n}f(\frac{k}{n})\right)=e^{\int_0^1f(x)dx}$

Question

Here, $f:[0,1]\rightarrow\mathbb{R}$ is a continious function. I tried by Riemann Integration till the step:
$$\int_0^1f(x)dx=\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\left(\frac{1}{n}f(\frac{k}{n})\right)$$
$$\implies e^{\int_0^1f(x)dx}=\lim_{n\rightarrow\infty}\prod_{k=1}^{n}\left(e^{\frac{1}{n}f(\frac{k}{n})}\right)$$ But, I can't seem to proceed further!

Answer 1

Note that $$x-\frac{x^2}{2}\le\ln(1+x)\le x$$

Let $x=\frac1nf\left(\frac kn\right)$

$$\frac1nf\left(\frac kn\right)-\frac1{2n^2}f^2\left(\frac kn\right)\le\ln\left(1+\frac1nf\left(\frac kn\right)\right)\le \frac1nf\left(\frac kn\right)$$

and

$$\prod_{k=1}^{n}\left(1+\frac{1}{n}f(\frac{k}{n})\right)=e^{\displaystyle \sum_{k=1}^{n}\ln\left(1+\frac{1}{n}f(\frac{k}{n})\right)}$$

So we get

$$ e^{\displaystyle \sum_{k=1}^{n}\frac{1}{n}f(\frac{k}{n})-\frac1{2n^2}f^2\left(\frac kn\right)}\le\prod_{k=1}^{n}\left(1+\frac{1}{n}f(\frac{k}{n})\right)\le e^{\displaystyle \sum_{k=1}^{n}\frac{1}{n}f(\frac{k}{n})}\tag{1}$$

Note that

$$\int_0^1 f^2(x)dx<M$$

is bounded, hence

$$\lim_{n\to\infty}\sum_{k=1}^{n}\frac1{2n^2}f^2\left(\frac kn\right)=0$$

By squeeze theorem, (1) gives

$$\lim_{n\rightarrow\infty}\prod_{k=1}^{n}\left(1+\frac{1}{n}f(\frac{k}{n})\right)=e^{\int_0^1 f(x)dx}$$