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Question Number 84492 by john santu last updated on 13/Mar/20

lim_(x→(π/3))  ((sin (x−(π/3)))/(1−2cos (x))) =

limxπ3sin(xπ3)12cos(x)=

Commented by john santu last updated on 13/Mar/20

lim_(x→0)  ((sin (x))/(1−2cos (x+(π/3)))) =  lim_(x→0)  ((cos (x))/(2sin (x+(π/3)))) = (1/(2×((√3)/2))) = (1/(√3))

limx0sin(x)12cos(x+π3)=limx0cos(x)2sin(x+π3)=12×32=13

Commented by john santu last updated on 13/Mar/20

other way  sin (x−(π/3)) ≈ (x−(π/3))  cos (x) ≈ (1/2)−((√3)/2)(x−(π/3))  lim_(x→(π/3))  ((x−(π/3))/(1−2[(1/2)−((√3)/2)(x−(π/3))])) =  lim_(x→(π/3))  (((x−(π/3)))/((√3)(x−(π/3)))) = (1/(√3)) .

otherwaysin(xπ3)(xπ3)cos(x)1232(xπ3)limxπ3xπ312[1232(xπ3)]=limxπ3(xπ3)3(xπ3)=13.

Commented by mathmax by abdo last updated on 13/Mar/20

let l(x)=((sin(x−(π/3)))/(1−2cosx))  changement x−(π/3)=t give  l(x)=A(t)=((sint)/(1−2cos(t+(π/3)))) =((sint)/(1−2(costcos(π/3)−sint sin((π/3)))))  =((sint)/(1−2((1/2)cost−((√3)/2)sint))) =((sint)/(1−cost+(√3)sint))  x→(π/3) ⇔t→0  and   A(t)∼ (t/((t^2 /2) +(√3)t)) =(1/((t/2)+(√3))) ⇒lim_(t→0)  A(t)=(1/(√3))  ⇒lim_(x→0)   l(x)=(1/(√3))

letl(x)=sin(xπ3)12cosxchangementxπ3=tgivel(x)=A(t)=sint12cos(t+π3)=sint12(costcosπ3sintsin(π3))=sint12(12cost32sint)=sint1cost+3sintxπ3t0andA(t)tt22+3t=1t2+3limt0A(t)=13limx0l(x)=13

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