Q: Help finding the limit of $\lim_{n \to \infty}\prod_{k=1}^{n}\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right)$. 
Given that $f:[0,1]\to \Bbb{R}$ is a continuous function, I need to show that
$$\lim_{n \to \infty}\prod_{k=1}^{n}\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right)=e^{\int_0^1f(x)dx}$$

Writing $$y=\lim_{n \to \infty}\prod_{k=1}^{n}\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right)$$
and taking the logarithm of both sides and interchanging limit and logarithm(since $f$ is continuous), I get
$$\log y = \lim_{n\to \infty}\sum_{k=1}^{n}\log\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right)$$
Now I know that the limit of the Riemann sum $\lim_{n\to \infty}\sum_{k=1}^{n}\frac{b-a}{n}f\left(a+\frac{(b-a)}{n}k\right)=\int_a^bf(x)dx$. However if I want it to read $\int_0^1f(x)dx$ in the given problem, I would need the term inside the summation to read $\frac{1}{n}f\left(\frac{k}{n}\right)$. The way to do this it would seem is to take the series expansion of $\log(1+x)$ and ignoring the $x^2$ and higher order terms. Is it justifiable to do so?

A: $f$ is continuous on $[0,1]$, hence bounded.Lets assume that $n$ is large, so that the factors $1+\frac{1}{n}f(\frac{k}{n})>0$ hence taking logarithms makes sense. Now, $$\log(1+x) = x+O(x^2)$$ in particular, if we let $x = \frac{1}{n} \max_{[0,1]}(f)$, we see that
$$\left|\log\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right)-\frac{1}{n}f\left(\frac{k}{n}\right)\right|\leq \frac{C_f}{n^2} $$
for some constant $C_f$ which depends only on $f$.
It follows that
$$ \sum_{k=1}^n\left|\log\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right)-\frac{1}{n}f\left(\frac{k}{n}\right)\right|\leq \frac{C_f}{n}\to 0 \text{ as } n \to \infty$$
so the two series converge to the same limit.
