分子是0,如果B=0則必有A=0.
若B≠0,那么A≠0.此時原式
=lim(x->0)(2arctanx-ln((1+x)/(1-x))/x^A 【等價無窮小代換】
=lim(x->0)(2arctanx -ln(1+x)+ ln(1-x))/x^A
=lim(x->0)(2/(1+x²) -1/(1+x)+ 1/(x-1))/[A·x^(A-1)] 【洛筆答法則】
=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)]
如果這個極限存在,則必有A-3=0.
A=3.
則B=4·lim(x->0) 1/[3·(x²-1)·(x²+1)·1]=-4/3.