Lt_(x rarr pi/6)((3sin x-sqrt(3)cos x)/(6x-pi))=

lim_(x rarr pi/6) ((3sinx - sqrt3 cos x)/(6x-pi)) =

underset(x to pi//6)lim ((3sin x-sqrt3cosx)/(6x-pi))=

lim_ (x rarr (pi) / (6)) (3sin x-sqrt (3) cos x) / (6x-pi)

lim_(x rarr pi/6)(sqrt(3)sin x-cos x)/(x-(pi)/(6))

The value of lim _(x to (pi)/(6)) (3 sin x- sqrt3 cos x)/(6x -pi) is equal to-

lim_(x to (pi)/(6)) [(3sinx - sqrt(3)cos x)/(6x -pi)] is equal to

lim_(x rarr pi/4)(sqrt(cos x)-sqrt(sin x))/(x-(pi)/(4))

If f(x) ={((3sinx-sqrt(3)cosx)/(6x-pi),",",x != (pi)/6),(a, ",",x = pi/6):} is continuous at x = (pi)/6 , then a =