How do you solve this limit involving definite integration?

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 $L=\frac{2}{\pi}$.