有：lim[sinx^n/(sinx)^m](x→0)=lim(sinx^n)'/[(sinx)^m]'(x→0)
=nx^(n-1)cosx/[m(sinx)^m-1]cosx=nx^(n-1)/m(sinx)^(m-1)连续用洛必达法则
=n(n-1)x^(n-2)/m(m-1)(sinx)^(m-2)
......
......
......
当n＜m时，lim[sinx^n/(sinx)^m](x→0)=0；
当n=m时，lim[sinx^n/(sinx)^m](x→0)=n!/m!=1;
当n＞m时，limsinx^n/(sinx)^m x不存在.本回答被提问者和网友采纳
1，n＝m
0，n＜m
∞，n＞m
求极限sinx^n\/(sinx)^m x趋近0
当x→0时sinx^n→0,cosx→1,(sinx)^m→0,故sinx^n\/(sinx)^m为0\/0型,用洛必达法则 有：lim[sinx^n\/(sinx)^m](x→0)=lim(sinx^n)'\/[(sinx)^m]'(x→0) =nx^(n-1)cosx\/[m(sinx)^m-1]cosx=nx^(n-1)\/m(sinx)^(m-1)连续用洛...
求极限sinx^n\/(sinx)^m x趋近0
当x→0时sinx^n→0,cosx→1,(sinx)^m→0，故sinx^n\/(sinx)^m为0\/0型，用 洛必达法则 有:lim[sinx^n\/(sinx)^m](x→0)=lim(sinx^n)'\/[(sinx)^m]'(x→0)=nx^(n-1)cosx\/[m(sinx)^m-1]cosx=nx^(n-1)\/m(sinx)^(m-1)连续用洛必达法则 =n(n-1)x^(n-2)\/m(m...
x→0时,sin(x^n)\/(sinx^m)为的极限值是多少?
1、本题是无穷小\/无穷小型的不定式问题；2、解答本题的最快捷方法是运用等价无穷小代换；3、在代换的过程中，要分成三种情况讨论。具体解答如下：
limsinx(n次方)\/(sinx)的m次方在x趋近于0时的极限为什么不能直接等于...
lim(x->0) sin(x^n)\/(sinx)^m =lim(x->0) x^n\/x^m =lim(x->0) x^(n-m)case 1: m<n , lim(x->0) sin(x^n)\/(sinx)^m=lim(x->0) x^(n-m) =0 case 2: m=n , lim(x->0) sin(x^n)\/(sinx)^m=lim(x->0) x^(n-m) =1 case 3: m>n ,...
limsinx^n\/sinx^m(x趋向于0)(m,n为正整数)
lim(x->0) sin(x^n)\/sin(x^m) (0\/0)=lim(x->0) nx^(n-1) .cos(x^n)\/mx^(m-1) cos(x^m)if n>m lim(x->0) sin(x^n)\/sin(x^m)=lim(x->0) nx^(n-1) .cos(x^n)\/mx^(m-1) cos(x^m)=0 if n=m lim(x->0) sin(x^n)\/sin(x^m)=li...
极限高数, lim[sin(x^n)\/(sinx)^m] x→0 (n,m为正整数)
∵原式=lim(x->0){[sin(x^n)\/(x^n)]*[(x\/sinx)^m]*[x^(n-m)]} =lim(x->0)[sin(x^n)\/(x^n)]*lim(x->0)[(x\/sinx)^m]*lim(x->0)[x^(n-m)]=1*(1^m)*lim(x->0)[x^(n-m)] (应用重要极限lim(z->0)(sinz\/z)=1)=lim(x->0)[x^(n-m)]∴当m0...
极限高数, lim[sin(x^n)\/(sinx)^m] x→0 (n,m为正整数)
∵原式=lim(x->0){[sin(x^n)\/(x^n)]*[(x\/sinx)^m]*[x^(n-m)]} =lim(x->0)[sin(x^n)\/(x^n)]*lim(x->0)[(x\/sinx)^m]*lim(x->0)[x^(n-m)]=1*(1^m)*lim(x->0)[x^(n-m)] (应用重要极限lim(z->0)(sinz\/z)=1)=lim(x->0)[x^(n-m)]∴当m0...
limx趋近于0sin(xn)\/(sinx)m
lim<x→0> sin(x^n)\/(sinx)^m = lim<x→0> x^n\/x^m 当 n>m ， 该极限为 0；当 n=m ， 该极限为 1；当 n<m ， 该极限为 ∞.
lim((sin(x^n))\/((sinx)^m)),x→0,求极限
原式=limx^n\/x^m（分子，分母同时用等价无穷小代换）=limx^(n-m)= 0 n>m 1 n=m 无穷大 n<m
