Question

已知lim x→0(（（根号下1+x+x^2) - （1+ax））/x^2)=b,求常数a、b的值

Answer 1

∵lim(x->0){[√(1+x+x²)-(1+ax)]/x²}=b

==>lim(x->0){[x+(1-a)]/[x√(1+x+x²)+(1+ax)]}=b (分子有理化化简)

∴1-a=0 ==>a=1

==>lim(x->0){1/[√(1+x+x²)+(1+x)]}=b

==>1/2=b

故a=1,b=1/2.