Evaluate the limit by Riemann sum or otherwise Let $f:[0,1]\rightarrow [0,\infty)$ and let $g:[0,1]\rightarrow [0,\infty)$ be two continuous functions.
Evaluate:
$\lim_{n\rightarrow \infty}\sqrt[n]{f\left(\frac{1}{n}\right)g\left(\frac{n}{n}\right)+f\left(\frac{2}{n}\right)g\left(\frac{n-1}{n}\right)+...+f\left(\frac{n}{n}\right)g\left(\frac{1}{n}\right)}$
My attempt:Let $L=\lim_{n\rightarrow \infty}\sqrt[n]{f\left(\frac{1}{n}\right)g\left(\frac{n}{n}\right)+f\left(\frac{2}{n}\right)g\left(\frac{n-1}{n}\right)+...+f\left(\frac{n}{n}\right)g\left(\frac{1}{n}\right)}$
$\log_{e}L=\lim_{n\rightarrow \infty}\frac{1}{n}\log_{e}\left(\sum_{r=1}^{n}f\left(\frac{r}{n}\right)g\left(\frac{n-r+1}{n}\right)\right)=\int_{0}^1\log_{e}\left(f(x)g(1-x)\right)dx$
$\log_{e}L=\int_{0}^1\log_{e}\left(f(1-x)g(x)\right)dx$
$2\log_{e}L=\int_{0}^1\log_{e}\left(f(x)f(1-x)g(x)g(1-x)\right)dx$
$2\log_{e}L=\int_{0}^1\log_{e}\left(f(x)\right)dx+\int_{0}^1\log_{e}\left(f(1-x)\right)dx+\int_{0}^1\log_{e}\left(g(x)\right)dx+\int_{0}^1\log_{e}\left(g(1-x)\right)dx$
$2\log_{e}L=\int_{0}^1\log_{e}\left(f(x)\right)dx+\int_{0}^1\log_{e}\left(f(x)\right)dx+\int_{0}^1\log_{e}\left(g(x)\right)dx+\int_{0}^1\log_{e}\left(g(x)\right)dx$
$\log_{e}L=\int_{0}^1\log_{e}\left(f(x)\right)dx+\int_{0}^1\log_{e}\left(g(x)\right)dx$
$\log_{e}L=\int_{0}^1\log_{e}\left(f(x)g(x)\right)dx$
 A: Let 
$$x_n =\sqrt[n]{f\left(\frac{1}{n}\right)g\left(\frac{n}{n}\right)+f\left(\frac{2}{n}\right)g\left(\frac{n-1}{n}\right)+...+f\left(\frac{n}{n}\right)g\left(\frac{1}{n}\right)}.$$
Then
$$\log x_n = \frac{1}{n}\log\left(\sum_{k=1}^nf(k/n)g(1-k/n+1/n) \right) \\ = \frac{1}{n}\log\left(n\frac1{n}\sum_{k=1}^nf(k/n)g(1-k/n+1/n) \right)\\ = \frac{\log n}{n} + \frac{1}{n}\log\left(\frac1{n}\sum_{k=1}^nf(k/n)g(1-k/n+1/n) \right).$$
Hence 
$$\lim_{n \to \infty} \log x_n = \lim_{n \to \infty}\frac{\log n}{n} + \left(\lim_{n \to \infty}\frac{1}{n} \right) \log\left(\lim_{n \to \infty}\frac1{n}\sum_{k=1}^nf(k/n)g(1-k/n+1/n) \right) \\ = 0 + 0 \cdot \log\left(\int_0^1f(x)g(1-x) \, dx\right) \\ = 0
,$$
and
$$\lim_{n \to \infty}x_n = \exp\left( \lim_{n \to \infty} \log x_n\right) = 1.$$
Note that the sum does not appear to be a conventional Riemann sum since $f(x)$ and $g(1-x)$ are not evaluated at the same intermediate points.  Nevertheless it converges to the integral as it can be written as
$$\frac1{n}\sum_{k=1}^nf(k/n)g(1-k/n+1/n) \\ = \frac1{n}\sum_{k=1}^nf(k/n)g(1-k/n) + \frac1{n}\sum_{k=1}^nf(k/n)[g(1-k/n+1/n)-g(1 - k/n)].$$
Since $g$ is uniformly continuous, for any $\epsilon > 0$ we have $|g(1-k/n+1/n)-g(1 - k/n)| < \epsilon$ for $n$ sufficiently large and
$$\lim_{n \to \infty} \left| \frac1{n}\sum_{k=1}^nf(k/n)[g(1-k/n+1/n)-g(1 - k/n)] \right| \\ \leqslant \epsilon  \int_0^1|f(x)| \, dx.$$  
