Question

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

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

Answer 1

Answer: (D)

$\underset{n\to \infty }{\lim }\left[\frac{1}{n}{\sec }^2\frac{1}{n^2}+\frac{2}{n^2}{\sec }^2\frac{4}{n^2}+\frac{3}{n^2}{\sec }^2\frac{9}{n^2}+....+\frac{1}{n}{\sec }^{2}1\right]$ is equal to
$\underset{n\to \infty }{\lim }\frac{r}{n^2}{\sec }^2\frac{r^2}{n^2}=\underset{n\to \infty }{\lim }\frac{1}{n}.\frac{r}{n}{\sec }^2\frac{r^2}{n^2}$
$\Rightarrow$ Given limit is equal to value of integral
$\underset{0}{\overset{1}{\int }}x{\sec }^2{x}^{2}dx$
or $\frac{1}{2}\underset{0}{\overset{1}{\int }}2x\sec x^{2}dx=\frac{1}{2}\underset{0}{\overset{1}{\int }}{\sec }^{2}tdt$ [put $x^2=t$ ]
$=\frac{1}{2}{\left(\tan t\right)}_0^1=\frac{1}{2}\tan 1$.