Question

$\underset{n\to \infty }{lim}\left[\frac{1}{n^2}{\sec }^2\frac{1}{n^2}+\frac{2}{n^2}{\sec }^2\frac{4}{n^2}+....+\frac{1}{n}{\sec }^{2}1\right]$

• A. $\frac{1}{2}\sec 1$
• B. $\frac{1}{2}\csc 1$
• C. $\tan 1$
• D. $\frac{1}{2}\tan 1$

Answer 1

The correct option is D $\frac{1}{2}\tan 1$

$\underset{n\to \infty }{lim}\left[\frac{1}{n^2}{\sec }^2\frac{1}{n^2}+\frac{2}{n^2}{\sec }^2\frac{4}{n^2}+....+\frac{1}{n}{\sec }^{2}1\right]$

$=\underset{x\to 0}{lim}\frac{r}{n^2}se{c}^2\frac{r^2}{n^2}=\underset{x\to 0}{lim}\frac{r}{n}\times \frac{1}{n}se{c}^2\frac{r^2}{n^2}$

Given limit is equal to value of integral $\to$

${\int }_0^{1}xse{c}^2{x}^{2}dx=\frac{1}{2}{\int }_0^{1}2nse{c}^2{x}^{2}dx$

Put $x^2=t$

$\Rightarrow \frac{1}{2}{\int }_0^{1}se{c}^{2}tdt$

$=\frac{1}{2}|tan{|}_0^1$

$=\frac{1}{2}tan1$