Question
$$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{{{n^2}}}{{\sec }^2}\frac{1}{{{n^2}}} + \frac{2}{{{n^2}}}{{\sec }^2}\frac{4}{{{n^2}}} + ..... + \frac{1}{n}{{\sec }^2}1} \right]$$ equals-
A.
$$\frac{1}{2}\sec \,1$$
B.
$$\frac{1}{2}{\text{cosec}}\,1$$
C.
$$\tan 1$$
D.
$$\frac{1}{2}\tan 1$$
Answer :
$$\frac{1}{2}\tan 1$$
Solution :
$$\eqalign{
& \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{{{n^2}}}{{\sec }^2}\frac{1}{{{n^2}}} + \frac{2}{{{n^2}}}{{\sec }^2}\frac{4}{{{n^2}}} + ..... + \frac{1}{n}{{\sec }^2}1} \right]{\text{is equal to}} \cr
& \mathop {\lim }\limits_{n \to \infty } \frac{r}{{{n^2}}}{\sec ^2}\frac{{{r^2}}}{{{n^2}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}.\frac{r}{n}{\sec ^2}\frac{{{r^2}}}{{{n^2}}} \cr
& \Rightarrow {\text{Given limit is equal to value of integral}} \cr
& \int_0^1 {x\,{{\sec }^2}{x^2}dx = \frac{1}{2} \times } \int_0^1 {2x\,{{\sec }^2}{x^2}dx} \cr
& {\text{Put }}{x^2} = t, \cr
& {\text{The integral becomes}} = \frac{1}{2}\int_0^1 {{{\sec }^2}t\,dt = \frac{1}{2}\left( {\tan \,t} \right)_0^1 = \frac{1}{2}\tan \,1} \cr} $$