lim(x→0) Incos2x/Incos3x
作者&投稿:邵疤 (若有异议请与网页底部的电邮联系)
0/0型可以用洛比达法则,分子分母分别求导得2sin2xcos3x/3sin3xcos2x,cos3x和cos2x在x趋于0时为1,对2sin2x/3sin3x分子分母再求导得4cos2x/9cos3x,所以答案是4/9。
lim(x->0) ln(cos2x)/ln(cos3x) (0/0)
=lim(x->0) [-2sin2x/(cos2x)]/[-3sin3x/(cos3x)]
=lim(x->0) 2tan2x/(3tan3x)
=4/9
lim(x→0) Incos2x/Incos3x= 2/3(tan2x/tan3x)=2/3(2x/3x)=4/9
最后一步使用的是无穷小代换tan2x ~2x,tan3x~3x
lim(x->0) In(cos2x)/In(cos3x) (∞/∞)
=lim(x->0) [-2(sin2x)/(cos2x)]/[-3(sin3x)/(cos3x)]
=(2/3)lim(x->0) tan2x/tan3x
=(2/3)(2/3)
=4/9
用等阶无穷小啊!
x趋向于0时,lncos2x∽2x;lncos3x∽3x!
所以,极限为2/3
徵选绿萼:[答案] 是0吧,不是正无穷 这是0/0型,用洛必达法则 =lim(x→0)[1/cos2x*(-sin2x)*2]/[1/cos3x*(-sin3x)*3] =lim(x→0)(2/3)*tan2x/tan3x 用等价无穷小替换 =lim(x→0)(2/3)*2x/3x =4/9
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