If `(x)=(1-sqrt3tanx)/(pi-6x),"for"x nepi/6` is continous at `x=pi/6, "find " f ((pi)/(6)).`

Explanation

Given f(x) is continuous at `x=(pi)/(6)`……….
`:.f((pi)/(6))=underset(xto""pi//6)limf(x)=underset(xto""pi//6)lim(1-sqrt3tanx)/(pi-6x)`
Let `x=(pi)/(6)+h," as "xto(pi)/(6),hto0`
`:.underset(xto""pi//6)limf(x)=underset(hto0)lim(1-sqrt3tan((pi)/(6)+h))/(pi-6((pi)/(6)+h))`
`=underset(hto0)lim(1-sqrt3[(tan""(pi)/(6)+tanh)/(1-tan""(pi)/(6)+h)])/(pi-pi-6h)`
`underset(hto0)lim((-(1)/(sqrt3)-sqrt3)tanh)/(-6h(1-(1)/(sqrt3)tanh))`
`=underset(hto0)lim(-(4)/(sqrt3)tanh)/(-6h.(1-(1)/(sqrt3)tanh))`
`=underset(hto0)lim(2)/(3sqrt3)(tanh)/(h)(1)/((1-(1)/(sqrt3)tanh))`
`=(2)/(3sqrt3)underset(hto0)lim(tanh)/(h)underset(htoo)lim(1)/(1-(1)/(sqrt3)tanh)`
`=(2)/(3sqrt3)xx1xx(1)/(1-(1)/(sqrt3)tan0)`
`=(2)/(3sqrt3)`
`:.f((pi)/(6))=(2)/(3sqrt3)`