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lim(x->0)sin^2(pi/(2-p x))^(sec^2((pi/(2...

`lim_(x->0)sin^2(pi/(2-p x))^(sec^2((pi/(2-p x)))`

A

-1

B

2

C

`sqrt(5)`

D

`e^(-p^(2)//q^(2))`

Text Solution

Verified by Experts

The correct Answer is:
D

Given limit taks `1^(oo)` form. Therefore.
`L=underset(xto0)lim{sin^(2)((pi)/(2-px))}^(sec^(2)((pi)/(2-qx)))`
`=exp{underset(xto0)lim[sin^(2)((pi)/(2-px))-1]sec^(2)((pi)/(2-qx))}`
`=exp{-underset(xto0)lim("cos"^(2)((pi)/(2-px)))/("cos^(2)((pi)/(2-qx)))}`
`=exp{-underset(xto0)lim("sin"((pi)/(2)-(pi)/(2-px)))/("sin"^(2)((pi)/(2)-(pi)/(2-qx)))}`
Now,`=underset(xto0)lim("sin"^(2)((pi)/(2)-(pi)/(2-px)))/("sin"^(2)((pi)/(2)-(pi)/(2-qx)))`
`=underset(xto0)lim(("sin"^(2)((pi)/(2)-(pi)/(2-px)))/(((pi)/(2)-(pi)/(2-px))^(2)))/(("sin"^(2)((pi)/(2)-(pi)/(2-qx)))/(((pi)/(2)-(pi)/(2-qx))^(2)))xx(((pi)/(2)-(pi)/(2-px))^(2))/(((pi)/(2)-(pi)/(2-qx))^(2))`
`=underset(xto0)lim(((pi)/(2)-(pi)/(2-px))^(2))/(((pi)/(2)-(pi)/(2-px))^(2))`
`=underset(xto0)lim(((-pipx)/(2(2-px)))^(2))/(((-piqx)/(2(2-qx)))^(2))`
`=p^(2)//q^(2)`
`:.L=e^(-p^(2)//q^(2))`
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