Question

设f(x)=∫(0--sinx) ln(1+t^2)dt,g(x)=x^3+tan^4 x,则当x--0时,f（x）是g

Answer 1

f'(x)=cosx *ln(1+t^2)=cosx*ln[1+(sinx)^2 ] (sinx)^2~x^2 等价无穷小

g'(x)=3x^2+4(tanx)^3*1/cosx^2~3x^2

所以lim(x->0) f(x)/g(x)= lim(x->0)f'(x)/g'(x)=lim(x->0)x^2/3x^2=1/3

即x-->0时,f(x)是g(x)的同阶无穷小