Evaluate: lim_x→π/6 [(2sin² x + sinx - 1) / (2sin² x - 3sin x + 1)]

2sin² x + sinx - 1 = (2sin x - 1)(sinx + 1)

2sin² x - 3sin x + 1 = (2sin x - 1)(sinx - 1)

∴ lim_x→π/6 [(2sin² x + sinx - 1) / (2sin² x - 3 sinx + 1)] = lim_x→π/6 [((2sinx - 1)(sinx + 1) ) / ( (2sinx - 1)(sinx - 1) )]

= lim_x→π/6 [(sinx + 1) / (sinx - 1) ]   (since 2sinx - 1 ≠ 0)

= (1 + sin(π/6)) / (sin(π/6) - 1) = -3