Evaluating $\lim_{n\rightarrow\infty}x_{n+1}-x_n$ 
Let $f(x)$ be continuously differentiable on $[0,1]$ and
$$x_n = f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+f\left(\frac{3}{n}\right)+\ldots+f\left(\frac{n-1}{n}\right)$$  
Find $$\lim_{n\rightarrow\infty}\left(x_{n+1}-x_n\right)$$



Confusion: I just found a subtle problem in my solution.By the definition of definite integration, like $\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(x_i^*)\Delta x$, the sample point $x_i^*$ should be included in its $ith$ subinterval $[x_{i-1},x_i]$. In this case, I am not sure whether the subinterval $[\frac{i-1}{n(n+1)},\frac{i}{n(n+1)}]$ includes the sample point $i$ and $\xi_i$ or not.I can not deduce it from the given condition,  


$$\xi_i \in [\frac{i}{n+1},\frac{i}{n}] \implies \frac{i-1}{n(n+1)}\lt \frac{i}{n+1}\lt\xi_i\lt?\lt\frac{i}{n}$$

If those sample points is not always included in the corresponding subinterval, I might not apply the definite integration here. Hope somebody can take a look at this solution.

Update: I rewrite part of my solution and fixed the problem I have before. Thanks very much for all the help!

Heuristically, $x_n \sim n \int_0^1 f$, so $x_{n+1}-x_n \sim \int_0^1 f$ and thus $$\lim_{n \rightarrow \infty} x_{n+1}-x_n= \int_0^1 f$$
