$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{{{n^3} + 1}} + \frac{4}{{{n^3} + 1}} + \frac{9}{{{n^3} + 1}} + \dots + \frac{{{n^2}}}{{{n^3} + 1}}} \right] = $

  • A
    $1$
  • B
    $2/3$
  • C
    $1/3$
  • D
    $0$

Explore More

Similar Questions

$\lim _{x \rightarrow \infty} x^3 \left\{\sqrt{x^2+\sqrt{1+x^4}}-x \sqrt{2}\right\} = $

Let $[t]$ be the greatest integer less than or equal to $t$. Then the least value of $p \in N$ for which $\lim _{x}$ ${\rightarrow 0^{+}}\left(x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\ldots+\left[\frac{p}{x}\right]\right)-x^2\left(\left[\frac{1}{x^2}\right]+\left[\frac{2^2}{x^2}\right]+\ldots+\left[\frac{9^2}{x^2}\right]\right)\right) \geq 1$ is equal to . . . . . .

The limiting value of the function $f(x) = \frac{2\sqrt{2} - (\cos x + \sin x)^3}{1 - \sin 2x}$ as $x \to \frac{\pi}{4}$ is

$\mathop {\lim }\limits_{x \to {0^ + }} \left\{ {{{\left( {1 + x} \right)}^{\frac{2}{x}}}} \right\}$ is equal to (where $\{.\}$ denotes the fractional part of $x$)

Let $[x]$ denote the greatest integer less than or equal to $x$. Then,evaluate the limit: $\mathop {\lim }\limits_{x \to 0} \,\frac{{\tan \,(\pi \,{{\sin }^2}\,x) + \,{{(\left| x \right|\, - \,\sin \,(x\,[x]))}^2}}}{{{x^2}}}$

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