lim[x^2+ax]^1/2-(bx^2-1)^1/2=1,求a和b,x趋近正无穷

Question

如题所述

Answer 1

分子有理化
原式=lim (x^2+ax-bx^2+1)/{[x^2+ax]^1/2+(bx^2-1)^1/2}
=lim [(1-b)x^2+ax+1]/{[x^2+ax]^1/2+(bx^2-1)^1/2}
=lim [(1-b)x+a+1/x]/{[1+a/x]^1/2+[b-1/x^2]^1/2}
=lim [(1-b)x+a]/(1+b^1/2)
=1
所以b=1 a=1+b^1/2=2