求极限(sinx-xcosx)/(xIn(1+x^2))(Sinx/x-coax)/x^2=(1-cosx)/x^2=(2sin(x/2)^2/(x/2)^2)*1/4=1/2哪里错了

Question

求极限(sinx-xcosx)/(xIn(1+x^2))
(Sinx/x-coax)/x^2=(1-cosx)/x^2=(2sin(x/2)^2/(x/2)^2)*1/4=1/2哪里错了

Answer 1

lim(sinx-xcosx)/(xln(1+x^2))
=lim(cosx-cosx+xsinx)/[ln(1+x^2)+2x^2/(1+x^2)]
=limxsinx/[ln(1+x^2)+2x^2)=lim(sinx+xcosx)/[2x/(1+x^2)+4x)
=lim(sinx/x+cosx)/[2/(1+x^2)+4]=(1+1)/(2+4)=1/3
lim(sinx/x-cosx)/x^2=lim(sinx-xcosx)/x^3
=lim(cosx-cosx+xsinx)/3x^2=1/3limsinx/x=1/3
之所以算错,是因应为2阶的无穷小在过程中被忽略了,但分母中恰好是2阶的无穷小