=((2*(1\2)-1)/sinπ= 0/0
Замена переменной
х-(π/6)=t
x=t+(π/6)
если x→ π/6 , то t→0
lim_(x→ π/6)(2sinx-1)/sin6x=
=lim_(t→ 0)(2sin(t+(π/6))-1)/sin6*(t+(π/6))=
=lim_(t→ 0)(2sin(t+(π/6))-1)/(sin(6t+π))=
=lim_(t→ 0)(sqrt(3)sint+cost-1)/(-sin6t)=-sqrt(3)/6
или по правилу Лопиталя:
lim_(x→ π/6)(2sinx-1)/sin6x=(0/0)=
=lim_(x→ π/6)(2sinx-1)`/(sin6x)`=
=lim_(x→ π/6)(2cosx)/(6cos6x)=(2*sqrt(3)/2)/-6=
=-sqrt(3)/6