One may solve the problem go the OP by an application of the following result
Proposition: Suppose $(\Omega,\mathscr{F},\mu)$ is a finite measure space and $f\in L_\infty$. Define $\alpha_p =\int_X |f|^p\,d\mu$. Then
$$\frac{\alpha_{p+1}}{\alpha_p}\xrightarrow{p\rightarrow\infty}\|f\|_\infty$$
to the case $([0,\pi/2],\mathscr{B}([0,\pi/2]),\mu)$, where $\mu(dx)=\mathbb{1}_{[0,\pi/2]}(x)\,\sin(x)\,dx$, and $f(x)=x$. In this case, after an application of integration by parts, we obtain
$$ \lim_{r\to \infty}\frac{\int_{0}^{\pi/2}x^{r+1}\sin x\,dx}{\int_{0}^{\pi/2}x^r\sin x\,dx} = \lim_{r\to\infty} \frac{(r+1)\int^{\pi/2}_0 x^r\cos x\,dx}{\int^{\pi/2}_0 x^r\sin x\,dx}=\|f\|_\infty=\frac{\pi}{2} $$
This gives possible values $L=\frac{2}{\pi}$ and $C=-1$.
Note:
- A proof of the Proposition above can be found here. The proof is based on Holder's inequality along with the well known limit $\lim_{p\rightarrow\infty}\|f\|_p=\|f\|_\infty$.