Question

Let `L=lim_(xtoa) (|2sinx-1|)/(2sinx-1)`. Then
A. `1//2`
B. `-1//3`
C. `-1//6`
D. 3

Answer 1

Correct Answer - A::B::C
`L=underset(xtoa)lim(|2sinx-1|)/(2sinx-1)`
For `a=pi//6`.
`L.H.L.=underset(xto(pi^(-))/(6))lim(1-2sinx)/(2sinx-1)=-1`
`R.H.L.=underset(xto(pi^(+))/(6))lim(2sinx-1)/(2sinx-1)=1`
Hence, the limit does not exist.
For `a=pi,underset(xto pi)lim(1-2sinx)/(2sinx-1)=-1" "`(as in neighborhood of `pi, sinx` is less than `1/2`).
For `a=(pi)/(2),underset(xto pi//2)lim(2sinx-1)/(2sinx-1)=1" "`(as in neighborhood of `pi/2, sin x` approaches 1).