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I have this limit but Mathematica can't solve it??!!

Limit[(f/n (-2  ArcTanh[(f/n  x)/Sqrt[4  a (1 - 2/(3 n)) + b x^2]]-Log[a (1 - 2/(3 n)) + k x^2]) + 2 Sqrt[b] Log[ Sqrt[b] x + Sqrt[4 a (1 - 2/(3 n)) + b x^2]]),  a -> 0]
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  • $\begingroup$ Mathematica can solve it in 12.3.1: the result is: ConditionalExpression[-(( f (2 ArcTanh[(f x)/(n Sqrt[b x^2])] + Log[k x^2]))/n) + 2 Sqrt[b] Log[Sqrt[b] x + Sqrt[b x^2]], And[ Re[b x^2] > 0, Re[k x^2] > 0, Inequality[-1, Less, Re[f n^(-1) x (b x^2)^Rational[-1, 2]], Less, 1], Re[b^Rational[1, 2] x + (b x^2)^Rational[1, 2]] > 0]] $\endgroup$
    – flinty
    Oct 16, 2021 at 13:05
  • $\begingroup$ Add this line:Assumptions -> {n > 0, x > 0, k > 0, n > 0, f > 0, b > 0} to Limit ? $\endgroup$ Oct 16, 2021 at 13:08
  • $\begingroup$ @MariuszIwaniuk: FullSimplify[ Limit[(f/n (-2 ArcTanh[(f/n x)/Sqrt[4 a (1 - 2/(3 n)) + b x^2]] - Log[a (1 - 2/(3 n)) + k x^2]) + 2 Sqrt[b] Log[Sqrt[b] x + Sqrt[4 a (1 - 2/(3 n)) + b x^2]]), a -> 0, Assumptions -> {n > 0, x > 0, k > 0, n > 0, f > 0, b > 0}], Assumptions -> {n > 0, x > 0, k > 0, n > 0, f > 0, b > 0}] // InputForm $\endgroup$
    – user64494
    Oct 16, 2021 at 13:16
  • $\begingroup$ results in the end of math {((2*I)*(f*ArcTan[f/n] + n*Log[(2*I)*x]) - f*Log[k*x^2])/n, ConditionalExpression[Log[4] + 2*Log[x] - (f*(2*ArcTanh[f/n] + Log[k*x^2]))/n, f < n], ConditionalExpression[Sqrt[2]*Log[8*x^2] - (f*(2*ArcCoth[(Sqrt[2]*n)/f] + Log[k*x^2]))/n, Sqrt[2]*f < 2*n] in 12.3.1 on Windows 10. $\endgroup$
    – user64494
    Oct 16, 2021 at 13:17
  • $\begingroup$ @user64494 - with 12.3.1 on a Mac Assuming[{n > 0, x > 0, k > 0, n > 0, f > 0, b > 0}, Limit[(f/n (-2 ArcTanh[(f/n x)/Sqrt[4 a (1 - 2/(3 n)) + b x^2]] - Log[a (1 - 2/(3 n)) + k x^2]) + 2 Sqrt[b] Log[Sqrt[b] x + Sqrt[4 a (1 - 2/(3 n)) + b x^2]]), a -> 0] // FullSimplify] evaluates to ConditionalExpression[Sqrt[b]*Log[4*b*x^2] - (f*(2*ArcCoth[(Sqrt[b]*n)/f] + Log[k*x^2]))/n, f < Sqrt[b]*n] $\endgroup$
    – Bob Hanlon
    Oct 16, 2021 at 16:56

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