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Question Number 55592 by gunawan last updated on 27/Feb/19
lim_(x→π/3)  ((cos x−sin (π/6))/((π/6)−(x/2)))=..
limxπ/3cosxsinπ6π6x2=..
Commented by maxmathsup by imad last updated on 27/Feb/19
let A(x)=((cosx−sin((π/6)))/((π/6)−(x/2))) ⇒ A(x)=6((cosx−(1/2))/(π−3x)) =3  ((2cosx−1)/(3((π/3)−x)))  =((1−2cosx)/(x−(π/3)))   changement x−(π/3) =t give lim_(x→(π/3)) A(x)  =lim_(t→0)      ((1−2cos(t+(π/3)))/t) =lim_(t→0)   ((1−2{(1/2)cost −((√3)/2)sint)})/t)  =lim_(t→0)   ((1−cost +(√3)sint)/t) =lim_(t→0)   ((1−cost)/t) +lim_(t→0)   (√3)((sint)/t)  but cost ∼1−(t^2 /2) ⇒((1−cost)/t) ∼ (t/2) →0(t→0)  also sint ∼t ⇒lim_(t→0)  ((sint)/t) =1 ⇒  lim_(x→(π/3))    A(x)=(√3).
letA(x)=cosxsin(π6)π6x2A(x)=6cosx12π3x=32cosx13(π3x)=12cosxxπ3changementxπ3=tgivelimxπ3A(x)=limt012cos(t+π3)t=limt012{12cost32sint)}t=limt01cost+3sintt=limt01costt+limt03sinttbutcost1t221costtt20(t0)alsosinttlimt0sintt=1limxπ3A(x)=3.
Answered by kaivan.ahmadi last updated on 27/Feb/19
hopital  lim_(x→(π/3)) ((−sinx)/(−(1/2)))=lim_(x→(π/3)) 2sinx=2sin(π/3)=2×((√3)/2)=(√3)
hopitallimxπ3sinx12=limxπ32sinx=2sinπ3=2×32=3
Answered by math1967 last updated on 27/Feb/19
lim_(x→(π/3)) ((cosx−cos((π/2)−(π/6)))/((π/6)−(x/2)))   lim_(x→(π/3))  ((2sin((π/6)−(x/2))sin((x/2)+(π/6)))/((π/6)−(x/2))) ★  2×1×sin((π/6)+(π/6))=2×1×((√3)/2)=(√3)   ★lim_(((π/6)−(x/2))→0)  ((sin((π/6)−(x/2)))/(((π/6)−(x/2))))=1  ∵x→(π/3)  ∴(x/2)→(π/6) ∴((π/6)−(x/2))→0
limxπ3cosxcos(π2π6)π6x2limxπ32sin(π6x2)sin(x2+π6)π6x22×1×sin(π6+π6)=2×1×32=3lim(π6x2)0sin(π6x2)(π6x2)=1xπ3x2π6(π6x2)0

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