Evaluate the limit: mathop {lim}limits_{n to | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given limit is of the form $\mathop {\lim}\limits_{n 	o \infty } \frac{1}{n} \sum\limits_{r=0}^{n-1} f\left(\frac{r}{n}ight) = \int\limits_

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) The given limit is of the form $\mathop {\lim}\limits_{n 	o \infty } \frac{1}{n} \sum\limits_{r=0}^{n-1} f\left(\frac{r}{n}ight) = \int\limits_0^1 f(x) \, dx$. Here,the expression can be written as $\frac{\pi}{2} \cdot \frac{1}{n} \sum\limits_{r=0}^{n-1} \cos\left(\frac{\pi}{2} \cdot \frac{r}{n}ight)$. Comparing this with the definite integral definition,we have $f(x) = \cos\left(\frac{\pi x}{2}ight)$. Thus,the limit is equal to $\frac{\pi}{2} \int\limits_0^1 \cos\left(\frac{\pi x}{2}ight) \, dx$. Evaluating the integral: $\frac{\pi}{2} \left[ \frac{\sin(\frac{\pi x}{2})}{\frac{\pi}{2}} ight]_0^1 = \left[ \sin\left(\frac{\pi x}{2}ight) ight]_0^1 = \sin\left(\frac{\pi}{2}ight) - \sin(0) = 1 - 0 = 1$.

Evaluate the limit: $\mathop {\lim}\limits_{n \to \infty } \frac{\pi }{2n} \left( 1 + \cos \frac{\pi }{2n} + \cos \frac{2\pi }{2n} + \dots + \cos \frac{(n - 1)\pi }{2n} \right)$

Similar Questions

If $f(n) = \frac{1}{n} [(n+1)(n+2)(n+3) \ldots (2n)]^{\frac{1}{n}}$,then $\lim_{n \rightarrow \infty} f(n) =$

Evaluate the following definite integral as the limit of a sum: $\int_{-1}^{1} e^{x} dx$

The value of $\lim _{n \rightarrow \infty} \left\{ \frac{\sqrt{n+1}+\sqrt{n+2}+\ldots+\sqrt{2n-1}}{n^{3/2}} \right\}$ is

Let $[ \cdot ]$ denote the greatest integer function and $f(x) = \lim_{n \to \infty} \frac{1}{n^3} \sum_{k=1}^n \left[ \frac{k^2}{3^x} \right]$. Then $12 \sum_{j=1}^{\infty} f(j)$ is equal to ........... .

$\lim _{n \rightarrow \infty}\left[\frac{n}{(n+1) \sqrt{2n+1}}+\frac{n}{(n+2) \sqrt{2(2n+2)}}+\frac{n}{(n+3) \sqrt{3(2n+3)}}+\ldots n \text{ terms}\right]=\int_0^1 f(x) d x$,then $f(x)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module