Question

lim根号下（x^2-x+1） -ax-b =0 x趋向于正无穷求a，b

Answer 1

解：显然，a≠-1
∵lim(x->∞)[√(x²-x+1)-ax-b]=0 ==>lim(x->∞){[x²-x+1-(ax+b)²]/[√(x²-x+1)+ax+b]}=0
==>lim(x->∞){[(1-a²)x²-(2ab+1)x+(1-b²)]/[√(x²-x+1)+ax+b]}=0
==>1-a²=0..........(1)
==>lim(x->∞){[(1-b²)-(2ab+1)x]/[√(x²-x+1)+ax+b]}=0
==>lim(x->∞){[(1-b²)/x-(2ab+1)]/[√(1-1/x+1/x²)+a+b/x]}=0
==>-(2ab+1)/(1+a)=0..........(2)
∴解方程组(1)与(2)，得a=1，b=-1/2。