高数问题,坐等大神解答!问：确定常数a和b的值,使f(x)=x-(a+be^(x^2))sinx当x→0时是x的五阶无穷小

Question

高数问题,坐等大神解答! 问：确定常数a和b的值,使f(x)=x-(a+be^(x^2))sinx当x→0时是x的五阶无穷小

Answer 1

limf(x)/x^5=lim(x-(a+b*e^(x^2))*sin x)/x^5
=lim(x-(a+b*e^(x^2))*x)/x^5 (sin x与x是等价无穷小)
=lim(1-(a+b*e^(x^2)))/x^4 ①
=lim(2*x*b*e^(x^2))/(4*x^3) (洛必塔法则)
=lim(2*b*e^(x^2))/(4*x^2) (约掉x)
=lim(4*x*b*e^(x^2))/(8*x) (洛必塔法则)
=lim(b*e^(x^2))/2=b/2=1（代入x=0）
∴ b=2
根据①式有：
∵ (1-(a+b*e^(x^2)))与x^4是等价无穷小
所以1-(a+b*e^(0^2))=0 ∴1=a+b
a=-1