I came across a problem that I couldn't solve in a mathematical analysis textbook：
Let $\alpha, L \in \mathbb{R}$ and $L \neq 0$:$$\mbox{If}\quad\lim_{n \to \infty}\dfrac{\displaystyle n^{\alpha}\int_{0}^{\pi/2}x^{n}\sin\left(x\right)\,\mathrm{d}x}{\displaystyle\int_{0}^{\pi/2}x^{n}\cos\left(x\right)\,\mathrm{d}x} = L,\quad\mbox{then find}\quad\alpha, L$$
My attempt：Let $a_n=\int_0^{\frac{\pi}{2}} x^n \sin x \mathrm{~d} x$，$b_n=\int_0^{\frac{\pi}{2}} x^n \cos x \mathrm{~d} x$,then we get\begin{aligned} a_n&=-\int_0^{\frac{\pi}{2}} x^n \mathrm{~d} (\cos x)=-x^n\cos x\big |_0^{\frac{\pi}{2}}+n\int_0^{\frac{\pi}{2}} x^{n-1} \cos x \mathrm{~d} x=nb_{n-1}\\ b_n&=\int_0^{\frac{\pi}{2}} x^n \mathrm{~d} (\sin x)=x^n\sin x\big |_0^{\frac{\pi}{2}}-n\int_0^{\frac{\pi}{2}} x^{n-1} \sin x \mathrm{~d} x=\left(\frac{\pi}{2}\right)^{n}-na_{n-1}\\&=\left(\frac{\pi}{2}\right)^{n}-n(n-1)b_{n-2}.\end{aligned}
But the subsequent iterations were so complex that I couldn't figure out the answer. Are there any other ideas to solve this problem?