The value of limm→∞ ∫0π/2 sin2m⁡xdx∫0π/2 sin2m+1⁡xdx

We know that I2n=∫0π/2 sin2n⁡xdx=2n−12n×2n−32n−2×…×12×π2,I2n+1=∫0π/2 sin2n+1⁡xdx=2n2n+1×2n−22n−1×…×23 andAlso, I2m+1=2m2m+1I2m−1For all x∈(0,π/2),sin2m−1⁡x>sin2m⁡x>sin2m+1⁡xIntegrating from 0 to π/2, we get I2m−1≥I2m≥I2m+1hence I2m−1I2m+1≥I2mI2m+1≥1                (i)Also   I2m−1I2m+1=2m+12mHence limm→∞ I2m−1I2m+1=limm→∞ 2m+12m=1From (i) and using sandwitch theorem we havelimm→∞ I2mI2m+1=1