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Let $f\in C^{1}[0,1]$,is this true?

\begin{align*} \lim_{n\to\infty}\frac{\int_{\frac{i}{n}}^{\frac{i+1}{n}}f(x)dx-\frac{1}{2n}[f(\frac{i}{n})+f(\frac{i+1}{n})]}{\frac{1}{n^2}}=0 \end{align*} ($0\leq i\leq n-1,i\in \mathbf{N}$)

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We have $$f(x) = f(i/n) + (x-i/n)f'(\xi_1) \,\,\,\,\,\,\,\,\, (\star)$$ $$f(x) = f((i+1)/n) + (x-(i+1)/n)f'(\xi_2) \,\,\,\,\,\,\,\,\, (\dagger)$$ where $\xi_1,\xi_2 \in [1/n,1/(n+1)]$. Hence, averaging $(\star)$ and $(\dagger)$, we get that $$f(x) = \dfrac{f(i/n) + f((i+1)/n)}2 + \dfrac12(x-i/n)f'(\xi_1) + \dfrac12(x-(i+1)/n)f'(\xi_2)$$ Hence, $$\int_{i/n}^{(i+1)/n} f(x) dx = \dfrac{f(i/n) + f((i+1)/n)}{2n} + \dfrac{f'(\xi_1)}{4n^2} - \dfrac{f'(\xi_2)}{4n^2}$$ Hence, we get that $$n^2 \left(\int_{i/n}^{(i+1)/n} f(x) dx - \dfrac{f(i/n) + f((i+1)/n)}{2n} \right) = \dfrac{f'(\xi_1)-f'(\xi_2)}{4}$$ Hence,$$\overbrace{\lim_{n \to \infty} n^2 \left(\int_{i/n}^{(i+1)/n} f(x) dx - \dfrac{f(i/n) + f((i+1)/n)}{2n} \right) = \lim_{\xi_1 \to \xi_2}\dfrac{f'(\xi_1)-f'(\xi_2)}{4}}^{\text{Because $\xi_1 , \xi_2 \in [i/n,(i+1)/n]$ and hence as $n \to \infty$, we get that $\xi_1 \to \xi_2$}}$$ But, $$\underbrace{\lim_{\xi_1 \to \xi_2}\dfrac{f'(\xi_1)-f'(\xi_2)}{4} = 0}_{\text{Because $f \in C^1$}}$$ Hence, we get what we want.

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  • $\begingroup$ I found one can also prove it by L'Hospital's law. $\endgroup$
    – Luqing Ye
    Jan 13 '13 at 8:00
  • $\begingroup$ Yes, You could also prove using L'Hospital rule. In general, though I prefer not to use L'Hospital rule when I can use other more direct methods. $\endgroup$
    – user17762
    Jan 13 '13 at 8:05

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