Diseuss the continuity of the function :
`{:(f(x)=(sin 2x)/(sqrt(1-cos2x)),",for"0ltx le(pi)/(2)),(=(cosx)/(pi-2x),",for" (pi)/(2)ltx ltpi):}}at x=(pi)/(2)*`

Explanation

`f(x)-(sin 2x)/(sqrt(1-cos2x)),"for"0ltx lepi/2`
`thereforef((pi)/(2))=(sin pi)/(sqrt(1-cospi))=(0)/(sqrt(1-(-1)))=0" "...(1)`
`underset(xto((pi)/(2))^(+))limf(x)=lim (cosx)/(pi-2x)`
Put `x=pi/2+theta.Then as x to (pi)/(2), thetato0`
`=(sinpi)/(sqrt(1-cospi))=(0)/(sqrt(-1(-1)))=0" "...(2)`
Put `x =pi/2+theta.` Then as ` x to pi/2,theta to 0`
Also `pi -2x =pi-2((pi)/(2)+theta)=-2theta`
`thetaunderset(x to((pi)/(2)))limf(x)=underset(thetato0)lim(cos ((pi)/(2)+theta))/(-2theta)`
`=underset(thetato0)lim (-sin theta)/(-2theta)=1/2underset(thetato0)lim(sin theta)/(theta)`
`=1/2xx1=1/2" "...(3)`
From (2) and (3),
`underset(xto ((pi)/(2))^(-))limf(x)ne underset(xto ((pi)/(2))^(+))limf(x)`
`thereforef` is discontinous at `x=pi/2*`