Question

$\underset{n\to \infty }{lim}[\frac{1}{n^2}{\sec }^2\frac{1}{n^2}+\frac{2}{n^2}{\sec }^2\frac{4}{n^2}+\cdots +\frac{n}{n^2}{\sec }^{2}1]$ is equal to

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

Answer 1

Answer: (A)

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

Let $A=\underset{n\to \infty }{lim}(\frac{1}{n^2}{\sec }^2\frac{1}{n^2}+\frac{2}{n^2}{\sec }^2\frac{4}{n^2}+\cdots +\frac{n}{n^2}{\sec }^{2}1)$
$=\underset{n\to \infty }{lim}\frac{1}{n}[\frac{1}{n}{\sec }^2{\left(\frac{1}{n}\right)}^2+\frac{2}{n}{\sec }^2{\left(\frac{2}{n}\right)}^2+\cdots +\frac{n}{n}{\sec }^2{\left(\frac{n}{n}\right)}^2]$

$=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^n\left(\frac{r}{n}\right){\sec }^2{\left(\frac{r}{n}\right)}^2$
$\Rightarrow A={\int }_0^{1}x{\sec }^2{\left(x\right)}^{2}dx$
On put $x^2=t$, we get
$\Rightarrow 2xdx=dt$

$\Rightarrow xdx=\frac{1}{2}dt$
Therefore, $A=\frac{1}{2}{\int }_0^1{\sec }^{2}tdt$
$=\frac{1}{2}{\left[\tan t\right]}_0^1$
$=\frac{1}{2}\tan 1$