极限limx->0[(sinx)^2-xsinxcosx]/x^4,我用等价无穷小量算是(x^2-x*xcosx]/x^4=[x^2(1-cosx)[/x^4=1/2x^4/x^4=1/2用罗比达算出来是[cosx-cosx+xsinx]/(3x^2)=1/3

Question

极限limx->0[(sinx)^2-xsinxcosx]/x^4,
我用等价无穷小量算是(x^2-x*xcosx]/x^4=[x^2(1-cosx)[/x^4=1/2x^4/x^4=1/2
用罗比达算出来是[cosx-cosx+xsinx]/(3x^2)=1/3

Answer 1

用等价无穷小量算
limx->0[(sinx)^2-xsinxcosx]/x^4
＝limsinxcosx﹙tanx－x﹚]/x⁴
＝lim﹙tanx－x﹚]/x³
＝lim﹙x³/3﹚/x³
＝1/3