Putnam 2011, Problem A-3

Find a real number $c$ and a positive number $L$ for which
$$\lim_{r \to \infty} \frac{\int_0^{\pi/2} x^r \sin x dx}{\int_0^{\pi/2} x^r \cos x dx} = L.$$

Solution:
We claim that \((c,L) = (-1, 2/\pi)\) works. Write \(f(r) = \int_0^{\pi/2} x^r \sin x \, dx\). Then  
\[
f(r) < \int_0^{\pi/2} x^r \, dx = \frac{(\pi/2)^{r+1}}{r + 1}
\]  
while since \(\sin x \geq 2x/\pi\) for \(x \leq \pi/2\),  
\[
f(r) > \int_0^{\pi/2} \frac{2x^{r+1}}{\pi} \, dx = \frac{(\pi/2)^{r+1}}{r + 2}.
\]  
It follows that  
\[
\lim_{r \to \infty} r \left( \frac{2}{\pi} \right)^{r+1} f(r) = 1,
\]  
whence  
\[
\lim_{r \to \infty} \frac{f(r)}{f(r + 1)} = \lim_{r \to \infty} \frac{r(2/\pi)^{r+1} f(r)}{(r + 1)(2/\pi)^{r+2} f(r + 1)} \cdot \frac{2(r + 1)}{\pi r} = \frac{2}{\pi}.
\]  
Now by integration by parts, we have  
\[
\int_0^{\pi/2} x^r \cos x \, dx = \frac{1}{r + 1} \int_0^{\pi/2} x^{r+1} \sin x \, dx = \frac{f(r + 1)}{r + 1}.
\]  
Thus setting \(c = -1\) in the given limit yields  
\[
\lim_{r \to \infty} \frac{(r + 1)f(r)}{r f(r + 1)} = \frac{2}{\pi},
\]  
as desired.