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lim-x-pi-3-1-2cos-x-pi-3x-




Question Number 90210 by manuel__ last updated on 22/Apr/20
lim_(x→(π/3))  (((1−2cos (x))/(π−3x)))=?
limxπ3(12cos(x)π3x)=?
Commented by mathmax by abdo last updated on 22/Apr/20
f(x)=((2cosx−1)/(3x−π)) =((2cosx−1)/(3(x−(π/3)))) we do the changement x−(π/3)=t ⇒  (x→(π/3)⇔ t→0) and f(x)=((2cos(t+(π/3))−1)/(3t))=g(t)  =((2(cost cos((π/3))−sint sin((π/3)))−1)/(3t))  =((cost−(√3)sint −1)/(3t)) =−((1−cost)/(3t)) −((√3)/3) ((sint)/t)  1−cost ∼(t^2 /2)   and ((sint)/t) ∼ 1  ⇒g(t) ∼−(1/(3t))×(t^2 /2)=−(t/6)→0  ⇒lim_(t→0)   g(t) =−((√3)/3) =lim_(x→(π/3))   f(x)
f(x)=2cosx13xπ=2cosx13(xπ3)wedothechangementxπ3=t(xπ3t0)andf(x)=2cos(t+π3)13t=g(t)=2(costcos(π3)sintsin(π3))13t=cost3sint13t=1cost3t33sintt1costt22andsintt1g(t)13t×t22=t60limt0g(t)=33=limxπ3f(x)
Answered by john santu last updated on 22/Apr/20
L′hopital   lim_(x→(π/3))  ((2sin x)/(−3)) = −(2/3)×sin ((π/3))  = −(2/3)×((√3)/2) = −((√3)/3)
Lhopitallimxπ32sinx3=23×sin(π3)=23×32=33

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