limn→∞∑k=1nf(2k−12n)12n\lim_{n\to\infty}\sum _{k=1}^{n}f\left ( \frac {2k- 1}{2n}\right ) \frac 1{2n} .
limn→∞∑k=1nf(2k−12n)1n.\lim_{n\to\infty}\sum_{k=1}^nf\left(\frac{2k-1}{2n}\right)\frac1n.
limx→0∑k=12nf(k2n)⋅2n.\lim_{x\to_0}\sum_{k=1}^{2n}f\left(\frac{k}{2n}\right)\cdot\frac{2}{n}.
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