Given the constants A and B, LIM (x - > 0) (2arctanx ln ((1 + x) / (1-x)) / SiNx ^ a = b The answer given in the book is a = 3, B = - 4 / 3

Given the constants A and B, LIM (x - > 0) (2arctanx ln ((1 + x) / (1-x)) / SiNx ^ a = b The answer given in the book is a = 3, B = - 4 / 3

If B = 0, then a = 0
If B ≠ 0, then a ≠ 0
=LIM (x - > 0) (2arctanx ln ((1 + x) / (1-x)) / x ^ a [Equivalent Infinitesimal Substitution]
=lim(x->0)(2arctanx -ln(1+x)+ ln(1-x))/x^A
=LIM (x - > 0) (2 / (1 + X & # 178;) - 1 / (1 + x) + 1 / (x-1)) / [a · x ^ (A-1)] [lobita rule]
=lim(x->0)(2(x²-1) +( (x+1)-(x-1) )·(1+x²) )/[A·(x²-1)·(x²+1)x^(A-1)]
=lim(x->0)(4x²)/[A·(x²-1)·(x²+1)x^(A-1)]
=4·lim(x->0) 1/[A·(x²-1)·(x²+1)x^(A-3)]
If this limit exists, then there must be A-3 = 0
A=3.
Then B = 4 · LIM (x - > 0) 1 / [3 · (X & # 178; - 1) · (X & # 178; + 1) · 1] = - 4 / 3