解：原式=lim(x->0)[(sinx-xcosx)/(sinx)^3]
=lim(x->0)[(cosx-cosx+xsinx)/(3sin²x)]
(0/0型极限，应用罗比达法则)
=lim(x->0)[x/(3sinx)]
(化简)
=(1/3)lim(x->0)(x/sinx)
=(1/3)*1
(应用重要极限lim(x->0)(sinx/x)=1)
=1/3.
limx->0 (sinx-xcosx)\/x^3 极限 0.0谢谢
解：原式=lim(x->0)[(sinx-xcosx)\/(sinx)^3]=lim(x->0)[(cosx-cosx+xsinx)\/(3sin²x)] (0\/0型极限，应用罗比达法则)=lim(x->0)[x\/(3sinx)] (化简)=(1\/3)lim(x->0)(x\/sinx)=(1\/3)*1 (应用重要极限lim(x->0)(sinx\/x)=1)=1\/3.
limx->0 (sinx-xcosx)\/x^3 极限 0.0谢谢
解：原式=lim(x->0)[(sinx-xcosx)\/(sinx)^3]=lim(x->0)[(cosx-cosx+xsinx)\/(3sin²x)](0\/0型极限，应用罗比达法则)=lim(x->0)[x\/(3sinx)](化简)=(1\/3)lim(x->0)(x\/sinx)=(1\/3)*1 (应用重要极限lim(x->0)(sinx\/x)=1)=1\/3.
lim(x趋于0) (sinx-xcosx)\/x^3=lim(x趋于0)sinx\/3x 是怎么来的,求详解...
lim(x->0) (sinx - x cosx ) \/ x³ o\/o 利用洛必达法则 = lim(x->0) ( x sinx ) \/ (3 x²)= lim(x->0) sinx \/ (3 x)= 1\/3
lim下面x→0,(sinx-xcosx)\/sin³x?
先把分母等价替换成x^3, 这样解起来方便，洛必达的话，分子求导得cosx-cosx+xsinx=xsinx,分母求导得3x^2, 约分再利用第一个重要极限可以得到最后的结果是1\/3.
当X→0时,求lim(sinx-xcosx)\/x^3
2013-03-23 求lim x趋向0 (sinx-xcosx)\/x^3可不可以... 1 2011-01-31 速求极限问题 limx→0 (sinx-xcosx)\/sin... 9 2019-04-07 limx→0 (x-sinxcosx)\/x^3 1 2017-01-13 当x -> 0+时,求极限lim (lnx-xcosx)\/x... 2019-09-20 求极限:x→0时,(x-sinxcosx)\/x³ 3 ...
lim(sinx-x*cosx)\/(sinx)^3,x趋于0,求极限
lim(x→0) (sinx-x*cosx)\/(sinx)^3 =lim(x→0) (sinx-x*cosx)\/x^3 =lim(x→0) (cosx-cosx+xsinx)\/(3x^2)=lim(x→0) (xsinx)\/(3x^2)=1\/3
lim→0(sinx-xcos)\/x²sinx
简单计算一下即可，答案如图所示
求下列极限 lim(x→0) (sinx-xcosx)\/(x-sinx)
0\/0型,用洛必达法则 分子求导=cosx-cosx+xsinx=xsinx 分母求导=1-cosx 还是0\/0型,继续用洛必达法则 分子求导=sinx+xcosx 分母求导=sinx 所以=(sinx+xcosx)\/sinx =1+xcosx\/sinx =1+cosx\/(sinx\/x)x→0,sinx\/x极限是1 所以原来极限=1+cos0\/1=2 ...
limsinx-xcosx\/sin^3x x趋于0
x->0 sinx ~ x -(1\/6)x^3 xcosx ~ x -(1\/2)x^3 sinx -xcosx ~ (1\/3)x^3 --- lim(x->0) (sinx-xcosx)\/(sinx)^3 =lim(x->0) (1\/3)x^3\/x^3 =1\/3
为什么(sinx-xcosx)\/x^3不可以拆开进行无穷小替换计算呢?
lim(x→0) (sinx-xcosx)\/x^3 =lim(x→0) cosx(tanx-x)\/x^3 =lim(x→0) (tanx-x)\/x^3 =lim(x→0) (sec^2x-1)\/(3x^2)=lim(x→0) tan^2x\/(3x^2)=1\/3
