The value of $\underset{m\to \mathrm{\infty }}{lim} \frac{{\int }_0^{\pi /2} {\sin }^{2m}x\mathrm{d}x}{{\int }_0^{\pi /2} {\sin }^{2m+1}x\mathrm{d}x}$

 

We know that $I_{2n}={\int }_0^{\pi /2} {\sin }^{2n}x\mathrm{d}x$

$\begin{array}{l}=\frac{2n-1}{2n}\times \frac{2n-3}{2n-2}\times \dots \times \frac{1}{2}\times \frac{\pi }{2},\\ I_{2n+1}={\int }_0^{\pi /2} {\sin }^{2n+1}x\mathrm{d}x\end{array}$

$=\frac{2n}{2n+1}\times \frac{2n-2}{2n-1}\times \dots \times \frac{2}{3}$ and

Also, $I_{2m+1}=\frac{2m}{2m+1}{I}_{2m-1}$

For all $x\in (0,\pi /2),{\sin }^{2m-1}x>{\sin }^{2m}x>{\sin }^{2m+1}x$

Integrating from 0 to $\pi /2$, we get $I_{2m-1}\ge I_{2m}\ge I_{2m+1}$

hence $\frac{I_{2m-1}}{I_{2m+1}}\ge \frac{I_{2m}}{I_{2m+1}}\ge 1$                (i)

Also   $\frac{I_{2m-1}}{I_{2m+1}}=\frac{2m+1}{2m}$

Hence $\underset{m\to \mathrm{\infty }}{lim} \frac{I_{2m-1}}{I_{2m+1}}=\underset{m\to \mathrm{\infty }}{lim} \frac{2m+1}{2m}=1$

From (i) and using sandwitch theorem we have

$\underset{m\to \mathrm{\infty }}{lim} \frac{I_{2m}}{I_{2m+1}}=1$