设f(x)当x->0时满足limf(x)/x^2=-1,当x->0时,lncosx^2是比x^nf(x)高阶的无穷小量,而x^nf(x)是比e^(sinx)^2-1的高阶无穷小量,则正整数n等于A.1B.2C.3D.4

Question

设f(x)当x->0时满足limf(x)/x^2=-1,当x->0时,lncosx^2是比x^nf(x)高阶的无穷小量,而x^nf(x)是比
e^(sinx)^2-1的高阶无穷小量,则正整数n等于
A.1 B.2 C.3 D.4

Answer 1

x->0时：
f(x)~-x^2
lncosx^2=ln(1+cosx^2-1) cosx^2-1 (-x^4)/2
e^(sinx)^2-1 (sinx)^2 x^2
所以x^nf(x) x^n*(-x^2)
所以n=1