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Question Number 131769 by liberty last updated on 08/Feb/21
   lim_(x→π/2) ((2^(−cos x)  −1)/(x(x−(π/2)))) =?
limxπ/22cosx1x(xπ2)=?
Answered by bemath last updated on 08/Feb/21
 lim_(x→π/2)  (1/x) . lim_(x→π/2)  ((2^(sin (x−(π/2))) −1)/(x−(π/2))) =   (2/π) . lim_(h→0)  ((2^(sin h) −1)/h) = (2/π) . lim_(h→0)  ((ln 2. 2^(sin h) .cos h)/1)   = ((2ln 2)/π) = π^(−1) .ln 4
limxπ/21x.limxπ/22sin(xπ2)1xπ2=2π.limh02sinh1h=2π.limh0ln2.2sinh.cosh1=2ln2π=π1.ln4
Answered by Dwaipayan Shikari last updated on 08/Feb/21
lim_(x→(π/2)) ((2^(−cos(x)) −1)/(x(x−(π/2))))=((2^(−sin((π/2)−x)) −1)/(x(x−(π/2))))=((1−((π/2)−x)log(2)−1)/((π/2)(x−(π/2))))  =((2log(2))/π)=((log(4))/π)              lim_(x→0) a^x =1+xlog(a)
limxπ22cos(x)1x(xπ2)=2sin(π2x)1x(xπ2)=1(π2x)log(2)1π2(xπ2)=2log(2)π=log(4)πlimx0ax=1+xlog(a)
Answered by mathmax by abdo last updated on 08/Feb/21
f(x)=((2^(−cosx) −1)/(x(x−(π/2)))) we do the changement x−(π/2)=t(so t→o) and  f(x)=f(t+(π/2)) =((2^(sint) −1)/(t(t+(π/2)))) we have 2^(sint)  =e^(sint ln(2))   sint ∼t−(t^3 /6) ⇒e^(ln(2)sint)  ∼e^(ln(2)(t−(t^3 /6)))  ∼1+ln(2)(t−(t^3 /6)) ⇒  f(t+(π/2))∼((ln(2)(t−(t^3 /6)))/(t(t+(π/2)))) =((ln(2)(1−(t^2 /6)))/(t+(π/2)))→((2ln(2))/π) (t→0) ⇒  lim_(x→(π/2))  f(x)=((2ln(2))/π)
f(x)=2cosx1x(xπ2)wedothechangementxπ2=t(soto)andf(x)=f(t+π2)=2sint1t(t+π2)wehave2sint=esintln(2)sinttt36eln(2)sinteln(2)(tt36)1+ln(2)(tt36)f(t+π2)ln(2)(tt36)t(t+π2)=ln(2)(1t26)t+π22ln(2)π(t0)limxπ2f(x)=2ln(2)π

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