$${I_n} = \int\limits_0^{\frac{\pi }{4}} {{{\tan }^n}x\,dx} $$    then $$\mathop {\lim }\limits_{n \to \infty } n\left[ {{I_n} + {I_{n + 2}}} \right]$$    equals-

Solution :
$$\eqalign{ & {I_n} + {I_{n + 2}} = \int\limits_0^{\frac{\pi }{4}} {{{\tan }^n}x\left( {1 + {{\tan }^2}x} \right)\,dx} \cr & = \int\limits_0^{\frac{\pi }{4}} {{{\tan }^n}x\,{{\sec }^2}\,x\,dx = \left[ {\frac{{{{\tan }^{n + 1}}x}}{{n + 1}}} \right]_0^{\frac{\pi }{4}}} \cr & = \frac{{1 - 0}}{{n + 1}} \cr & = \frac{1}{{n + 1}} \cr & \therefore {I_n} + {I_{n + 2}} = \frac{1}{{n + 1}} \Rightarrow \mathop {\lim }\limits_{n \to \infty } n\left[ {{I_n} + {I_{n + 2}}} \right] \cr & = \mathop {\lim }\limits_{n \to \infty } \,n.\frac{1}{{n + 1}} \cr & = \mathop {\lim }\limits_{n \to \infty } \frac{n}{{n + 1}} \cr & = \mathop {\lim }\limits_{n \to \infty } \frac{n}{{n\left( {1 + \frac{1}{n}} \right)}} = 1 \cr} $$