lim_(x→0) (tan2x⋅cosx−tan2x)/(sinx⋅tan²x)=....

Jawabannya adalah -1 Konsep : 1-cos x = 2 sin² 1/2 x lim x→0 sin ax/tan bx = a/b lim x→0 tan ax/sin bx = a/b Jawab : lim_(x→0) (tan2x⋅cosx−tan2x)/(sinx⋅tan²x) = lim_(x→0) -tan 2x(1-cos x)/(sinx⋅tan²x) = lim_(x→0) -tan 2x(2 sin² 1/2 x)/(sinx⋅tan²x) = lim_(x→0) -tan 2x(2 sin 1/2 x. sin 1/2 x)/(sinx⋅tanx tan x) = lim_(x→0) 2. -tan 2x/sin x. sin 1/2 x/tan x. sin 1/2 x/ tan x = 2. -2. 1/2. 1/2 = -4/4 = -1 Jadi lim_(x→0) (tan2x⋅cosx−tan2x)/(sinx⋅tan²x)= -1