Given, `underset(xto(pi//6))"lim"(sqrt(3)sinx-cos)/(x-pi/6)=underset(xto(pi//6))"lim"(2(sqrt(3)/2sinx-1/2cosx))/(x-pi/6)`
`=underset(xto(pi//6))"lim"(2(sinxcospi/6-cosxsinpi/6))/(x-pi/6)=2 underset(xto(pi//6))"lim"(sin(x-pi/6))/(x-pi/6)`
=2 `[therefore sinAcosB-cosAsinB=sin(A-B)]`
`[therefore underset(xto0)"lim"(sinx)/(x)=1]`
`[therefore xto pi/6 rArr (x-pi/6) to 0]`