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

Explanation

`underset(x to pi//6)lim f(x)=underset(x to pi//6)lim(1-sqrt3tanx)/(pi-6x)`
Put ` x=pi/6+h.` Then as `xtopi/6,h to 0.`
Also `pi-6x=pi-6((pi)/(6)+h)=-6h`
`and tan x=tan ((pi)/(6)+h )=(tan""(pi)/(6)+tan h)/(1-tan""(pi)/(6)*tanh)`
`=((1)/(sqrt3)+tanh)/(1-(1)/(sqrt3)*tanh)=(1+sqrt3tanh)/(sqrt3-tanh)`
`thereforeunderset(x to pi//6)limf(x)=underset(h to 0)lim (1-sqrt3((1+sqrt3tanh))/(sqrt3-tan h))/(-6h)`
`=underset(h to 0) lim (sqrt3-tan h-sqrt3-3tan h)/(-6h (sqrt3-tanh))`
`underset(hto0)lim(-4tanh)/(-6h(sqrt3-tanh))`
`=2/3underset(h to 0)lim (tan h)/(h)*(1)/(sqrt3-tan h)`
`=2/3 underset(x to 0)lim(tan h)/(h)xx(1)/(underset(h to 0)lim(sqrt3-tanh))`
`=2/3 (1)xx(1)/(sqrt3-0)=(2)/(3sqrt3)`
Now, f is continous at `x=pi/6`
`therefore f((pi)/(6))=underset(x to pi//6)limf(x)`
`therefore f((pi)/(6))=(2)/(3sqrt3)*`