Question

已知(x*2+1/x+1 ) –ax–b的极限=0求a,b.亲,

Answer 1

是(x^2+1)/(x+1 )–ax–b在x→∞的极限么?

原式=x-1+2/(x+1)-ax-b=(1-a)x-1-b+2/(x+1),当x→∞时2/(x+1)→0.而等式极限为0所以1-a=0且-1-b=0,有a=1,b=-1