Question

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

• A. $\frac{1}{2}cosec1$
• B. $\frac{1}{2}\sec 1$
• C. $\frac{1}{2}\tan 1$
• D. $\tan 1$

Answer 1

Answer: (C)

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

$\underset{n\to \infty }{lim}\frac{1}{n^2}{\sec }^2\frac{1}{n^2}+\frac{2}{n^2}{\sec }^2\left(\frac{4}{n^2}\right)+...+\frac{1}{n}{\sec }^{2}1$
$=\underset{n\to \infty }{lim}\frac{1}{n^2}{\sec }^2\frac{1}{n^2}+\frac{2}{n^2}{\sec }^2\left(\frac{4}{n^2}\right)+...+\frac{n}{n^2}{\sec }^2\left(\frac{n^2}{n^2}\right)$
$=\underset{n\to \infty }{lim}\sum _{r=1}^{r=n}\left(\frac{r}{n^2}\right){\sec }^2{\left(\frac{r}{n}\right)}^2=\underset{n\to \infty }{lim}\sum _{r=1}^{r=n}\frac{1}{n}\left(\frac{r}{n}\right){\sec }^2{\left(\frac{r}{n}\right)}^2$
$={\int }_0^{1}x{\sec }^2\left(x^2\right)dx=\frac{1}{2}\tan 1$
Hence, option 'C' is correct.