mathop {Lim}limits_{n to infty } ,,sumlimits_ | vedclass

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

Question

(C) We are given the limit $L = \mathop {Lim}\limits_{n 	o \infty } \sum\limits_{k = 1}^n \frac{n}{n^2 + k^2 x^2}$.Divide the numerator and denominat

Vedclass · Accepted Answer

$\frac{	an^{-1}(x)}{x}$

Vedclass · Answer

(C) We are given the limit $L = \mathop {Lim}\limits_{n 	o \infty } \sum\limits_{k = 1}^n \frac{n}{n^2 + k^2 x^2}$.Divide the numerator and denominator by $n^2$:$L = \mathop {Lim}\limits_{n 	o \infty } \frac{1}{n} \sum\limits_{k = 1}^n \frac{1}{1 + (k/n)^2 x^2}$.This is a Riemann sum of the form $\int_0^1 f(t) dt$ where $t = k/n$ and $dt = 1/n$ as $n 	o \infty$.Thus,$L = \int_0^1 \frac{1}{1 + t^2 x^2} dt$.Let $u = tx$,then $du = x dt$,so $dt = du/x$. When $t=0, u=0$ and when $t=1, u=x$.$L = \int_0^x \frac{1}{1 + u^2} \cdot \frac{du}{x} = \frac{1}{x} [	an^{-1}(u)]_0^x$.$L = \frac{1}{x} (	an^{-1}(x) - 	an^{-1}(0)) = \frac{	an^{-1}(x)}{x}$.

$\mathop {Lim}\limits_{n \to \infty } \,\,\sum\limits_{k = 1}^n {\frac{n}{{{n^2} + {k^2}{x^2}}}} $,$x > 0$ is equal to

Similar Questions

$\lim _{n \rightarrow \infty} \frac{(2n(2n-1) \dots (n+1))^{1/n}}{n} = $

$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {\frac{1}{n}{e^{\frac{r}{n}}}} $ is

$\lim _{n \rightarrow \infty}\left[\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\ldots+\frac{n}{n^2+n^2}\right]=$

The value of $\lim _{n \rightarrow \infty} \prod_{r=1}^n\left(1+\frac{r^2}{n^2}\right)^{\frac{2 r}{n^2}}$ is equal to

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 ........... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module