$\lim _{n \rightarrow \infty}\left[\frac{1}{n^2} \sec ^2 \frac{1}{n^2}+\frac{2}{n^2} \sec ^2 \frac{4}{n^2}+\frac{3}{n^2} \sec ^2 \frac{9}{n^2}+\ldots+\frac{n}{n^2} \sec ^2 \frac{n^2}{n^2}\right]=$

  • A
    $\tan^{-1} 1$
  • B
    $\frac{1}{2} \tan^{-1} 1$
  • C
    $\frac{1}{2} \tan 1$
  • D
    $\frac{1}{2} \sec 1$

Explore More

Similar Questions

The value of $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^3}{r^4+n^4}$ is

If $\mathop {\lim }\limits_{n \to \infty } \frac{{{1^a} + {2^a} + \dots + {n^a}}}{{{{\left( {n + 1} \right)}^{a - 1}}\left[ {\left( {na + 1} \right) + \dots + \left( {na + n} \right)} \right]}} = \frac{1}{{60}}$ for some positive real number $a$,then $a$ is equal to

$\lim _{n \rightarrow \infty}\left(\frac{1}{1+n}+\frac{1}{2+n}+\frac{1}{3+n}+\ldots+\frac{1}{2 n}\right)$ is equal to :-

Given that $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n p} f\left(\frac{r}{n}\right)=\int_0^p f(x) d x$. If $f: R \rightarrow R$ is defined by $f(x)=x^2+2$, then $\lim _{n \rightarrow \infty} \frac{3}{n}\left[f\left(\frac{7}{n}\right)+f\left(\frac{14}{n}\right)+f\left(\frac{21}{n}\right)+\ldots+f(7)\right]=$

By the definition of the definite integral, the value of $\lim _{n \rightarrow \infty}\left(\frac{1^4}{1^5+n^5}+\frac{2^4}{2^5+n^5}+\frac{3^4}{3^5+n^5}+\ldots+\frac{n^4}{n^5+n^5}\right)$ is

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