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Question Number 161583 by cortano last updated on 19/Dec/21
  lim_(x→2)  ((1−sin (π/x))/(x^2 −4x+4)) =?   lim_(x→2)  ((x^3 −8+sin πx)/(2−x))=?   lim_(x→1)  ((x^2 −1)/(cos ((π/(x+1)))))=?
limx21sinπxx24x+4=?limx2x38+sinπx2x=?limx1x21cos(πx+1)=?
Commented by blackmamba last updated on 20/Dec/21
(1) lim_(x→2)  ((sin ((π/2))−sin ((π/x)))/((x−2)^2 ))        = lim_(x→2)  ((2cos (((πx+2π)/(4x)))sin (((πx−2π)/(4x))))/((x−2)^2 ))        = lim_(x→2)  ((2sin ((π/2)−(((πx+2π)/(4x))))sin (π/(4x))(x−2))/((x−2)^2 ))       = lim_(x→2)  ((2sin (π/(4x))(x−2) sin (π/(4x))(x−2))/((x−2)^2 ))     = ((2π^2 )/(64)) = (π^2 /(32))
(1)limx2sin(π2)sin(πx)(x2)2=limx22cos(πx+2π4x)sin(πx2π4x)(x2)2=limx22sin(π2(πx+2π4x))sinπ4x(x2)(x2)2=limx22sinπ4x(x2)sinπ4x(x2)(x2)2=2π264=π232
Answered by Ar Brandon last updated on 20/Dec/21
lim_(x→1) ((x^2 −1)/(cos((π/(x+1)))))=lim_(x→1) (((x+1)(x−1))/(cos((π/(x+1)))))=lim_(x→1) (((x+1)(x−1))/(sin((π/2)−(π/(x+1)))))  lim_(x→1) (((x+1)(x−1))/(sin(((π(x−1))/(2(x+1))))))=lim_(x→1) ((2(x+1)^2 )/π)∙(((π/2)(((x−1)/(x+1))))/(sin((π/2)(((x−1)/(x+1))))))  =lim_(x→1) ((2(x+1)^2 )/π)∙lim_(x→1) ((((π/2)(((x−1)/(x+1))))/(sin((π/2)(((x−1)/(x+1)))))))=(8/π)×1= determinant (((8/π)))
limx1x21cos(πx+1)=limx1(x+1)(x1)cos(πx+1)=limx1(x+1)(x1)sin(π2πx+1)limx1(x+1)(x1)sin(π(x1)2(x+1))=limx12(x+1)2ππ2(x1x+1)sin(π2(x1x+1))=limx12(x+1)2πlimx1(π2(x1x+1)sin(π2(x1x+1)))=8π×1=8π
Answered by Ar Brandon last updated on 20/Dec/21
lim_(x→2) ((x^3 −8+sinπx)/(2−x))=lim_(x→2) (((x−2)(x^2 +2x+4))/(2−x))+lim_(u→0) ((sin(π(2−u)))/u)  =−12+lim_(u→0) ((sin(−πu))/u)=−12−lim_(u→0) π∙((sin(πu))/(πu))=−(π+12)
limx2x38+sinπx2x=limx2(x2)(x2+2x+4)2x+limu0sin(π(2u))u=12+limu0sin(πu)u=12limu0πsin(πu)πu=(π+12)
Answered by mathmax by abdo last updated on 20/Dec/21
f(x)=((1−sin((π/x)))/(x^2 −4x+4))  changement (π/x)=t give x=(π/t)  f(x)=((1−sint)/((π^2 /t^2 )−((4π)/t)+4))=((t^2 (1−sint))/(π^2 −4πt +4t^2 ))   (t→(π/2))  =_(t−(π/2)=z)      (((z+(π/2))^2 (1−cosz))/(4(z+(π/2))^2 −4π(z+(π/2))+π^2 ))=Ψ(z) (z→0)  Ψ(z)∼(z^2 /2)×(((z+(π/2))^2 )/(4(z^2 +πz+(π^2 /4))−4πz−π^2 )) ⇒  Ψ(z)∼((z^2 (z+(π/2))^2 )/(2{4z^2 +4πz+π^2 −4πz−π^2 }))∼(1/8)(z+(π/2))^2   ⇒lim_(z→0) Ψ(z)=(1/8)(π^2 /4)=(π^2 /(32))=limf(x)
f(x)=1sin(πx)x24x+4changementπx=tgivex=πtf(x)=1sintπ2t24πt+4=t2(1sint)π24πt+4t2(tπ2)=tπ2=z(z+π2)2(1cosz)4(z+π2)24π(z+π2)+π2=Ψ(z)(z0)Ψ(z)z22×(z+π2)24(z2+πz+π24)4πzπ2Ψ(z)z2(z+π2)22{4z2+4πz+π24πzπ2}18(z+π2)2limz0Ψ(z)=18π24=π232=limf(x)

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