Question

lim根号下（x^2-x+1） -ax-b =0 x趋向于正无穷求a,b

Answer 1

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.