lim , _ n [ 1 n^2 ^2 1 n^2 + 2 n^2 ^2 4 n^2 + ..... + 1 n ^2 1 ] … - Vidyadip

MCQ

$\mathop {{\rm{lim}}\,}\limits_{n \to \infty } \left[ {\frac{1}{{{n^2}}}{{\sec }^2}\frac{1}{{{n^2}}} + \frac{2}{{{n^2}}}{{\sec }^2}\frac{4}{{{n^2}}} + ..... + \frac{1}{n}{{\sec }^2}1} \right]$ equals

A
$\tan 1$
✓
$\frac{1}{2}\tan 1$
C
$\frac{1}{2}\sec 1$
D
$\frac{1}{2}{\rm{cosec}}1$

✓

Answer

Correct option: B.

$\frac{1}{2}\tan 1$

b
(b) $\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{{{n^2}}}{{\sec }^2}\frac{1}{{{n^2}}} + \frac{2}{{{n^2}}}{{\sec }^2}\frac{4}{{{n^2}}} + \frac{3}{{{n^2}}}{{\sec }^2}\frac{9}{{{n^2}}} + ..... + \frac{1}{n}{{\sec }^2}1} \right]$ is equal to

$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {\frac{r}{{{n^2}}}{{\sec }^2}\frac{{{r^2}}}{{{n^2}}}} $

$= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r = 1}^n {\frac{r}{n}{{\sec }^2}\frac{{{r^2}}}{{{n^2}}}} $

Given limit is equal to the value of integral $\int_0^1 {x{{\sec }^2}{x^2}dx} $

=$\frac{1}{2}\int_0^1 {2x{{\sec }^2}{x^2}dx} = \frac{1}{2}\int_0^1 {{{\sec }^2}t\;dt} $, [Put ${x^2} = t$]

$ = \frac{1}{2}[\tan \;t]_0^1 = \frac{1}{2}\tan 1$.

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