A) \[\frac{1}{2}\]
B) 1
C) \[\infty \]
D) zero
Correct Answer: B
Solution :
Since, \[{{l}_{n}}=\int_{0}^{\pi /4}{{{\tan }^{n}}x\,dx}\] \[\therefore \] \[{{l}_{n+2}}=\int_{0}^{\pi /4}{{{\tan }^{n+2}}xdx}\] Now, \[{{l}_{n}}+{{l}_{n+2}}=\int_{0}^{\pi /4}{{{\tan }^{n}}xdx+\int_{0}^{\pi /4}{{{\tan }^{n+2}}xdx}}\] \[=\int_{0}^{\pi /4}{{{\tan }^{n}}x(1+{{\tan }^{2}}x)dx}\] \[=\int_{0}^{\pi /4}{{{\sec }^{2}}x{{\tan }^{n}}xdx}\] Put \[\tan x=t\Rightarrow {{\sec }^{2}}xdx=dt\] \[\therefore {{l}_{n}}+{{l}_{n+2}}=\int_{0}^{1}{{{t}^{n}}\,dt=\left[ \frac{{{t}^{n+1}}}{n+1} \right]_{0}^{1}=\frac{1}{n+1}}\] \[\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\,n[{{l}_{n}}+{{l}_{n}}]=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n}{n+1}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{1+\frac{1}{n}}=1\]
You need to login to perform this action.
You will be redirected in
3 sec
