Limits and Integrals together.

Let's calculate the integral first:- $I_{n} = \int_{0}^{\dfrac{\pi}{4}} tan^{n}xdx$ $I_{n} = \int_{0}^{\dfrac{\pi}{4}} tan^{n-2}x tan^{2}xdx$ $I_{n} = \int_{0}^{\dfrac{\pi}{4}} tan^{n-2}x (sec^{2}x-1)dx$ $I_{n} = \int_{0}^{\dfrac{\pi}{4}}tan^{n-2}x sec^{2}x dx-\int_{0}^{\dfrac{\pi}{4}} tan^{n-2}xdx$ $I_{n} = \Bigg[ \dfrac{tan^{n-1}x}{n-1} \Bigg]_{0}^{\dfrac{\pi}{4}}-I_{n-2}$ $I_{n} + I_{n-2} = \dfrac{1}{n-1}$ As $n \to \infty, I_{n}=I_{n-2}=l(let)$ . Hence $l+l = \dfrac{1}{n-1}$ $l=\dfrac{1}{2(n-1)}$ Now we will calculate the limit. $\lim_{n \to \infty} \dfrac{n}{2(n-1)}$ Hence the limit is $\boxed{0.5}$ Please can anyone help me in solving the reduction formula so as to find the exact explicit formula of $I_{n}$ .

I have a query..is this last step correct? And if it isn't... doesn't the fact that the answer actually is 0.5 (see prakhar's solution) prove the last line to be correct?