$\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n (k^2 x)$ का मान है

  • A
    $x$
  • B
    $\frac{x}{2}$
  • C
    $\frac{x}{3}$
  • D
    $\frac{x}{4}$

Explore More

Similar Questions

$a \in R, |a| > 1$ के लिए,मान लीजिए $\lim _{n \rightarrow \infty} \left( \frac{1+\sqrt[3]{2}+\ldots+\sqrt[3]{n}}{n^{7/3} \left( \frac{1}{(an+1)^2} + \frac{1}{(an+2)^2} + \ldots + \frac{1}{(an+n)^2} \right)} \right) = 54$. तो $a$ का/के संभावित मान है/हैं:
$(1) 8$ $(2) -9$ $(3) -6$ $(4) 7$

धनात्मक पूर्णांक $n$ के लिए,$f(n) = n + \sum_{r=1}^n \frac{16r + (9-4r)n - 3n^2}{4rn + 3n^2}$ परिभाषित करें। तो,$\lim_{n \rightarrow \infty} f(n)$ का मान किसके बराबर है?

$\lim _{n \rightarrow \infty} \frac{1}{n^2} \sum_{k=1}^{2n} k e^{k/n} = $

$\lim _{n \rightarrow \infty}\left[\frac{1}{n^2} \sec ^2 \frac{1}{n^2}+\frac{2}{n^2} \sec ^2 \frac{4}{n^2}+\ldots+\frac{1}{n} \sec ^2 1\right]=$

$\lim _{n \rightarrow \infty}\left[\frac{n^{3 / 2}}{n^{5 / 2}}-\frac{n^{1 / 2}}{n^{3 / 2}}+\frac{n^{3 / 2}}{(n+2)^{5 / 2}}-\frac{n^{1 / 2}}{(n+3)^{3 / 2}}+\ldots+\frac{n^{3 / 2}}{(n+2(n-1))^{5 / 2}}-\frac{n^{1 / 2}}{(n+3(n-1))^{3 / 2}}\right]=$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo