$\displaystyle \lim_{n\to \infty} \int_{0}^{\frac{\pi}{2}} \frac{1}{tg^nx+ctg^nx} dx$
Hi,
Let $\displaystyle u=\tan^n x$ ($\displaystyle \implies$ $\displaystyle x=\arctan(\sqrt[n]{u})$ and $\displaystyle \mathrm{d}x=\frac{1}{n}\cdot \frac{u^{\frac{1-n}{n}}}{1+u^2}\,\mathrm{d}u$)
$\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{1}{\tan^nx +\mathrm{cotan}^n x} \,\mathrm{d}x=\frac{1}{n} \int_0^\infty \frac{1}{u+\frac{1}{u}}\cdot \frac{u^{\frac{1-n}{n}}}{1+u^2}\,\mathrm{d}u=\frac{1}{n} \int_0^\infty \frac{\sqrt[n]{u}}{(1+u^2)^2}\,\mathrm{d}u $
Now it must be clear what the limit is...
finding a good upper bound for the integral looks more interesting to me than the limit, which is 0:
let $\displaystyle \tan^{2n}x=t.$ then: $\displaystyle 0 \leq I_n=\int_0^{\frac{\pi}{2}} \frac{dx}{\tan^n x + \cot^n x}=\frac{1}{2n} \int_0^{\infty} \frac{t^{\frac{1}{2n} - \frac{1}{2}}}{(1+t)(1+\sqrt[n]{t})} \ dt \leq \frac{1}{2n} \int_0^{\infty} \frac{t^{\frac{1}{2n} - \frac{1}{2}}}{1+t} \ dt.$ but: $\displaystyle \int_0^{\infty} \frac{t^{\frac{1}{2n} - \frac{1}{2}}}{1+t} \ dt=\frac{\pi}{\cos \left(\frac{\pi}{2n} \right)}.$ thus: $\displaystyle 0 \leq I_n \leq \frac{\pi}{2n\cos \left(\frac{\pi}{2n} \right)},$ and hence: $\displaystyle \lim_{n\to\infty}I_n = 0.$
I posted an answer to this question here
http://www.mathhelpforum.com/math-he...ral-limit.html
This back when I was less careful, so you may want to wait for NCA or another senior member to verify it.
The basic gist is that since $\displaystyle \left|\frac{1}{\tan^n(x)+\cot^n(x)}\right|\leqslan t 1$ on $\displaystyle \left[0,\frac{\pi}{2}\right]$ and $\displaystyle \frac{1}{\tan^n(x)+\cot^n(x)}\to 0$ almost everywhere (pointwise) then by Lebesgue's Dominated Convergence Theorem $\displaystyle \lim_{n\to\infty}\int_0^{\frac{\pi}{2}}\frac{dx}{\ tan^n(x)+\cot^n(x)}=\int_0^{\frac{\pi}{2}}\lim_{n\ to\infty}\frac{dx}{\tan^n(x)+\cot^n(x)}dx$ and using the fact that $\displaystyle \lim_{n\to\infty}\frac{dx}{\tan^n(x)+\cot^n(x)}=\l eft\{\begin{array}{rcl} \frac{1}{2}& \mbox{if} & x=\frac{\pi}{4}\\ 0 & \mbox{if} & \text{otherwise} \end{array}\right.$.
NOTE: Once again the original post should not have probably been done because I was using machinery a bit beyond my capabilities of yet so you should wait for confirmation...but I thought it may give another perspective to this problem.