# Limits, Riemann Sum and Definite Integral

$\large \large \lim_{n\to\infty } \left \lfloor \frac{\Sigma_{r=0}^{n-1} \frac1n f(\frac{r}n) + \Sigma_{r=1}^{n} \frac1n f(\frac{r}n)}{2 \int_0^1f(x) \, dx}\right \rfloor$

Does the above limit exist? If so, what is its value: 0 or 1?

Note by Shubhamkar Ayare
1 year, 6 months ago

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## Comments

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Top Newest

Trapezoidal Rule - Wikipedia or Trapezium Rule - Brilliant. Nothing can be said about the limit, unless sgn(f''(x)) is given.

- 1 year, 6 months ago

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$\displaystyle\lim_{n \to \infty} \sum_{r=0}^{n-1} \dfrac{1}{n}f\left(\dfrac{r}{n}\right) = \dfrac{1}{n}f\left(0\right)+\int_{\left(\small\displaystyle\lim_{n \to \infty}\frac{1}{n}\right)}^{\left(\small\displaystyle\lim_{n \to \infty}\frac{n-1}{n}\right)} f(x)dx = \int_{0}^{1} f(x) dx +\dfrac{1}{n}f\left(0\right)$

$$\text{Similarly , }$$

$\displaystyle\lim_{n \to \infty} \sum_{r=1}^{n} \dfrac{1}{n}f\left(\dfrac{r}{n}\right) = \dfrac{1}{n}f\left(\dfrac{n}{n}\right)+\int_{\left(\small\displaystyle\lim_{n \to \infty}\frac{1}{n}\right)}^{\left(\small\displaystyle\lim_{n \to \infty}\frac{n-1}{n}\right)} f(x)dx = \int_{0}^{1} f(x) dx+\dfrac{1}{n}f\left(1\right)$

$$\text{Our limit becomes}$$

$\displaystyle\lim_{n \to \infty} \left\lfloor \dfrac{\dfrac{1}{n}\left[ f\left( 1 \right)+f\left(0\right) \right]+2 \displaystyle\int_{0}^{1} f(x) dx}{2 \displaystyle\int_{0}^{1} f(x) dx} \right\rfloor \\ \displaystyle\lim_{n \to \infty} \left\lfloor \dfrac{\dfrac{1}{n}\left[ f\left( 1 \right)+f\left(0\right) \right]}{2 \displaystyle\int_{0}^{1} f(x) dx} + 1 \right\rfloor = 1$

- 1 year, 6 months ago

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May I consider that as: unless f(x) is given, the limit does not exist?

If a geometrical approach is used (definite integration is area under the curve; the sum is the total area bars of width 1/n and height f(r/n)), is there any way to arrive at a conclusion?

What if f(x) is given as increasing or decreasing? What if n does not tend to infinity? (I had a question in class test: n did not tend to infinity and f(x) was decreasing. Our sir, with whom I disagree in this matter, said that if n tends to infinity, the answer would be 1, from 'Definite Integration as a Limit of Sum' concept. )

- 1 year, 6 months ago

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