Questions
The value of
detailed solution
Correct option is D
We know that I2n=∫0π/2 sin2nxdx=2n−12n×2n−32n−2×…×12×π2,I2n+1=∫0π/2 sin2n+1xdx=2n2n+1×2n−22n−1×…×23 andAlso, I2m+1=2m2m+1I2m−1For all x∈(0,π/2),sin2m−1x>sin2mx>sin2m+1xIntegrating 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=1Similar Questions
The value of is
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