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!
 A: Because $f(x)$ is continuously differentiable on $[0,1]$, tben by the mean value theorem:

$$f\left(\frac{i}{n+1}\right)-f\left(\frac{i}{n}\right) = f'(\xi_i)(\frac{i}{n+1} - \frac{i}{n}) \text{  where  } \xi_i \in \left[\frac{i}{n+1},\frac{i}{n}\right] \tag{1}$$ 

Then, by the given formula of $x_n$, we have
\begin{align*}
\ x_{n+1} - x_n &= \left[f\left(\frac{1}{n+1}\right)-f\left(\frac{1}{n}\right)\right]+\dots +\left[f\left(\frac{n-1}{n+1}\right)-f\left(\frac{n-1}{n}\right)\right]+\left[f\left(\frac{n}{n+1}\right)-f\left(1\right)\right] + f\left(1\right)
\\&=f(1) - \sum_{i=1}^{n}i\cdot f'(\xi_i)\left(\frac{1}{n(n+1)}\right)
\end{align*}  
Hence, by the defintion of definite integration, we have:
\begin{align*}
\\\lim_{n\rightarrow\infty}\left(x_{n+1}-x_n\right) &= f(1) - \lim_{n \rightarrow \infty}\sum_{i=1}^{n}i\cdot f'(\xi_i)\left(\frac{1}{n(n+1)}\right)
\\&=f(1) - \lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{i}{n+1}\cdot f'(\xi_i)\cdot\frac{1}{n} 
\\&=f(1) - \int_{0}^{1}xf'(x)dx \space\space\space\dots\text{See Note}
\\&=f(1) - \left[xf(x)|_{0}^{1}-\int_{0}^{1}f(x)dx\right]
\\ &=\int_{0}^{1}f(x)dx
\end{align*}


Note that:: 
  \begin{align*}
\\ \sum_{i=1}^{n}i\cdot f'(\xi_i)\left(\frac{1}{n(n+1)}\right) = \sum_{i=1}^{n}\frac{i}{n+1}\cdot f'(\xi_i)\cdot\frac{1}{n}
\end{align*}
  by (1), it is obvious that:
  $$\frac{i-1}{n} \lt \frac{i}{n+1} \lt \xi_i \lt \frac{i}{n}\space\space\space\text{(where } i \le n) \tag{2}$$
  which means $\xi_i, \frac{i}{n+1} \in \left[\frac{i-1}{n},\frac{i}{n}\right]$. Also by $(2)$, we know:
  $$\sum_{i=1}^{n}\frac{i}{n+1}\cdot f'(\xi_i)\cdot\frac{1}{n} \le \sum_{i=1}^{n}\xi_i\cdot f'(\xi_i)\cdot\frac{1}{n}$$
  Because $\frac{i}{n+1}$ and $\xi_i$ share the same subinterval, it implies that if we let $x_{i}^* = \xi_i$(sample points) and $\Delta x=\frac{1}{n}$, then:
  $$\lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{i}{n+1}\cdot f'(\xi_i)\cdot\frac{1}{n}=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}\xi_i\cdot f'(\xi_i)\cdot\frac{1}{n}=\int_{0}^{1}xf'(x)dx$$
Hence, it is ok to apply the definition of definite integration here.

A: Heuristically, $x_n \sim n \int_0^1 f$, so $x_{n+1}-x_n \sim \int_0^1 f$ and thus $$\lim_{n \rightarrow \infty} x_{n+1}-x_n= \int_0^1 f$$
A: Note that: 
$$\lim_{n\rightarrow\infty}x_n=\lim_{n\rightarrow\infty}\sum_{i=1}^{n-1}f \left(\frac in\right)=\lim_{n\rightarrow\infty}n\cdot\sum_{i=1}^{n-1}f\left(\frac in\right)\frac1n=\lim_{n\rightarrow\infty}n\cdot\sum_{i=1}^{n-1}f\left(x_i\right)\Delta x$$
where
$x_i=\frac in\in\left(0,1\right)$ and $\Delta x=\frac1n$.
