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certificate-lim-n-0-1-n-1-n-2-x-2-e-x-3-dx-pi-2-




Question Number 212600 by MrGaster last updated on 18/Oct/24
                    certificate:             lim_(n→∞) ∫_0 ^1 (n/(1+n^2 x^2 ))e^x^3  dx=(π/2).
certificate:limn01n1+n2x2ex3dx=π2.
Answered by Berbere last updated on 21/Oct/24
=lim_(n→∞) ∫^1 _0 ((d(nx))/(1+(nx)^2 ))e^x^3  A=lim_(n→∞) {[^1 _0 tan^(−1) (nx)e^x^3  ]−∫_0 ^1 3x^2 e^x^3  tan^(−1) (nx)dx}  =(π/2)e−lim_(n→∞) ∫_0 ^1 3x^2 e^x^3  tan^(−1) (nx)dx  (π/2)−(1/x)≤tan^(−1) (x)=(π/2)−tan^(−1) ((1/x))≤(π/2)  ((π/2)−(1/(nx)))3x^2 e^x^3  ≤3x^2 e^x^3  tan^(−1) (nx)≤(π/2).3x^2 e^x^3    lim_(n→∞) (π/2)e−(π/2)−(1/n)∫_0 ^1 3xe^x^3  ≤u=lim_(n→∞) ∫_0 ^1 3x^2 e^x^3  tan^(−1) (nx)≤(π/2)e−(π/2)  (π/2)e−(π/2)−(1/n)∫_0 ^1 3xe^x^2  dx≤(π/2)e−(π/2)−(1/n)∫_0 ^1 3xe^x^3  ≤u≤(π/2)e−(π/2)  lim_(n→∞) (π/2)e−(π/2)−(3/(2n))(e−1)≤u≤(π/2)e−(π/2)  u=(π/2)e−(π/2);A=(π/2)e−u=(π/2)
Extra \left or missing \right=π2elimn013x2ex3tan1(nx)dxπ21xtan1(x)=π2tan1(1x)π2(π21nx)3x2ex33x2ex3tan1(nx)π2.3x2ex3limnπ2eπ21n013xex3u=limn013x2ex3tan1(nx)π2eπ2π2eπ21n013xex2dxπ2eπ21n013xex3uπ2eπ2limnπ2eπ232n(e1)uπ2eπ2u=π2eπ2;A=π2eu=π2

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