How do you solve this limit involving definite integration? $$ \lim \limits_{r \to \infty} \frac {r^C \int_0^{\frac{\pi}{2}} x^r \sin(x)\, dx}{\int_0^{\frac{\pi}{2}} x^r \cos(x)\, dx} = L$$
Find the value of $\pi L - C$, given that $C\in\mathbb{R}$ and $L>0$.
My approach:
I tried to apply integration by parts to both the numerator and denominator to get a recurring relation, hoping to cancel something off, but to no avail. I'm not getting any other method to solve it, so any help will be appreciated. 
 A: I probably have a simple solution that many missed, through a probabilistic/distributional approach. It is quite trivial that:
$$ \lim_{r\to +\infty}\frac{\int_{0}^{\pi/2}x^{r+1}\sin(x)\,dx}{\int_{0}^{\pi/2}x^r\sin(x)\,dx} = \frac{\pi}{2} $$
since the integrand functions in the numerator/denominator get more and more concentrated around the right endpoint as $r$ increases, and their ratio at $x=\frac{\pi}{2}$ is exactly $\frac{\pi}{2}$. By using integration by parts, we have:
$$ \lim_{r\to +\infty}\frac{(r+1)\int_{0}^{\pi/2}x^{r}\cos(x)\,dx}{\int_{0}^{\pi/2}x^r\sin(x)\,dx} = \frac{\pi}{2}$$
hence the given limit is finite iff $C=-1$ and in such a case $L=\frac{2}{\pi}$.
A: This is 2011 putnam A3 problem,you can see some solution :http://www.artofproblemsolving.com/community/c7h449984p2531777
A: One may solve the problem go the OP by an application of the following result

Proposition: Suppose $(\Omega,\mathscr{F},\mu)$ is a finite measure space and $f\in L_\infty$. Define $\alpha_p =\int_X |f|^p\,d\mu$. Then
$$\frac{\alpha_{p+1}}{\alpha_p}\xrightarrow{p\rightarrow\infty}\|f\|_\infty$$

to the case $([0,\pi/2],\mathscr{B}([0,\pi/2]),\mu)$, where $\mu(dx)=\mathbb{1}_{[0,\pi/2]}(x)\,\sin(x)\,dx$, and  $f(x)=x$.  In this case, after an application of  integration by parts, we obtain
$$ \lim_{r\to \infty}\frac{\int_{0}^{\pi/2}x^{r+1}\sin x\,dx}{\int_{0}^{\pi/2}x^r\sin x\,dx} = \lim_{r\to\infty} \frac{(r+1)\int^{\pi/2}_0 x^r\cos x\,dx}{\int^{\pi/2}_0 x^r\sin x\,dx}=\|f\|_\infty=\frac{\pi}{2} $$
This gives  possible values $L=\frac{2}{\pi}$ and $C=-1$.

Note:

*

*A proof of the Proposition above can be found here. The proof is based on Holder's inequality along with the well known limit $\lim_{p\rightarrow\infty}\|f\|_p=\|f\|_\infty$.

