Show me your Limits

Note that

$\begin{aligned} I_s(r)&=\int_0^{\frac \pi 2}x^r\sin x\ dx \\&=-\int_0^{\frac \pi 2}x^rd\cos x \\&=-\left(x^r\cos x\bigg|_0^{\frac \pi 2}-\int_0^{\frac \pi 2}rx^{r-1}\cos x\ dx\right) \\&=r\int_0^{\frac \pi 2}x^{r-1}\cos x\ dx \\&=r\int_0^{\frac \pi 2}x^{r-1}d\sin x \\&=r\left(x^{r-1}\sin x\bigg|_0^{\frac \pi 2}-\int_0^{\frac \pi 2}(r-1)x^{r-2}\sin x\ dx\right) \\&=r\left(\frac \pi 2\right)^{r-1}-r(r-1)\int_0^{\frac \pi 2}x^{r-2}\sin x\ dx \\&=r\left(\frac \pi 2\right)^{r-1}-r(r-1)I_s(r-2) \end{aligned}$

Similarly,

$\displaystyle I_c(r)=\int_0^{\frac \pi 2}x^r\cos x\ dx=\left(\frac \pi 2\right)^r-r(r-1)I_c(r-2)$

Therefore,

$\begin{aligned} L&=\lim_{r\to\infty}\frac{r^c\displaystyle\int_0^{\frac\pi 2}\sin x\ dx}{\displaystyle\int_0^{\frac\pi 2}\cos x\ dx} \\&=\lim_{r\to\infty}\frac{r^cI_s(r)}{I_c(r)} \\&=\lim_{r\to\infty}\frac{\left(\frac \pi 2\right)^{r-1}-(r-1)I_s(r-2)}{\left(\frac \pi 2\right)^r-r(r-1)I_c(r-2)}r^{c+1} \\&=\lim_{r\to\infty}\frac{\left(\frac \pi 2\right)^{r-1}-(r-1)\left((r-2)\left(\frac\pi 2\right)^{r-3}-(r-2)(r-3)I_s(r-4)\right)}{\left(\frac \pi 2\right)^r-r(r-1)\left(\left(\frac\pi 2\right)^{r-2}-(r-2)(r-3)I_c(r-4)\right)}r^{c+1} \\&=\cdots \\&=\lim_{r\to\infty}\frac 2\pi r^{c+1} \end{aligned}$

Since the limit exists, we have $c+1=0$ , that is, $c=-1$ . Thus, $L=\frac 2\pi$ , and the answer is $\pi L-c=\boxed 3$ .

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Remark: Try a similar problem here .