MHB Can you find the limit of a complex function containing integrals and exponents?

[Ackbach]

Here is this week's POTW:

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

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[Ackbach]

No one answered this week's POTW, which was Problem A-3 in the 2011 Putnam archive. The solution, attributed to Kiran Kedlaya and associates, follows:

[sp]
We claim that $(c,L) = (-1,2/\pi)$ works.
Write $\displaystyle f(r) = \int_0^{\pi/2} x^r\sin(x)\,dx$. 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.
[/sp]