已知 lim{[sin6x-xf(x)]/x³}=0，说明 lim{sin6x-xf(x)}=k*x^(3+m)=0（m>0），即当x→0 时xf(x) 与sin6x 十分接近，所以lim{xf(x)}=0，那么lim{6-xf(x)}=6，于是题目要求的极限式 lim{[6-xf(x)]/x²}=lim{6/x²}=+∞；来自：求助得到的回答
若极限lim(x-0)[sin6x+xf(x)]\/x^3=0,则lim(x-0)[6+f(x)]\/x^2=?
简单计算一下即可，答案如图所示
lim(x趋近于0)[sin6x+xf(x)]\/x^3=0,则lim(x趋近于0)[6+f(x)]\/x^2=?
,lim(x趋近于0)[sin6x+xf(x)]\/x^3=0,则知道f(x) 是低天X^2的。所以（6+F(X)）是低于X^2的。从而 lim(x趋近于0)[6+f(x)]\/x^2=0.
x趋于0,lim x-o ( sin6x+xf(x))\/x3=0 ,lim x-o (6+f(x))\/x2=?
原式＝lim(x→0)[sin6x＋xf(x)]\/x³＝lim(x→0)［x(sin6x)\/x＋f(x)］\/x³＝lim(x→0)［x(6sin6x)\/6x＋f(x)］\/x³＝lim(x→0)[6＋f(x)]\/x²＝0 导数第二步就是所要求的。即 lim(x→0)[6＋f(x)]\/x²＝0 ...
lim(x趋近于0)[sin6x+xf(x)]\/x^3=0,则lim(x趋近于0)[6+f(x)]\/x^2=...
lim(x趋近于0)[6+f(x)]\/x^2=lim(x趋近于0)6\/x^2+lim(x趋近于0)f(x)\/x^2=lim(x趋近于0)sin6x\/x^3+lim(x趋近于0)xf(x)\/x^3=lim(x趋近于0)[sin6x+xf(x)]\/x^3=0.
lim x→0 [sin6x+xf(x)]\/x^3=0,求 lim x→0 [6+f(x)]\/x^2
因为lim x→0 [sin6x\/(6x)]=1 所以,lim x→0 [sin6x+xf(x)]\/x^3 =lim x→0 [6x+xf(x)]\/x^3 =lim x→0 [6+f(x)]\/x^2 =0
已知lim x→0 [sin6x+xf(x)]\/x^3=0, 求 lim x→0 [6+f(x)]\/x^2...
lim x→0 [sinx-x]\/x^3，如果按照你的那种做法，显然结果是0。实际上答案是-1\/6.此处应用的是一个很重要的公式——泰勒公式（只展开有限项目，后边的高阶项可视为高阶无穷小）sinx=x-1\/6*x^3.回到你的这道题，lim [sin6x+xf(x)]\/x^3=0 也就是 lim[6x-1\/6*(6x)^3+xf(x)]...
已知{limx趋近0 [(sin6x)+xf(x)]\/x^3}=0 求limx趋近0 [6+f(x)]\/x^...
\/2x^2+小o(x^2))=xf(0)+x^2f'(0)+0.5x^3f''(0)+小o(x^3)，于是由条件知 f(0)+6=0,f'(0)=0,.-36+0.5f''(0)=0，f(0)=-6,f'(0)=0,f''(0)=72，(6+f(x))\/x^2=【6+f(0)+xf'(0)+0.5x^2f''(0)+小o(x^2)】\/x^2=36+小o(1)，极限是36 ...
设x->0时,lim((sin6x+xf(x))\/x^3)=0,求x->0时,lim((6+f(x))\/x^2...
lim((sin6x+xf(x))\/x^3)=lim((sin6x-6x+6x+xf(x))\/x^3)=lim((sin6x-6x)\/x^3+(6x+xf(x))\/x^3)=-108 + lim((6+f(x))\/x^2) =0 所以lim((6+f(x))\/x^2) =108
已知lim x→0 [sin6x+xf(x)]\/x^3=0, 求 lim x→0 [6+f(x)]\/x^2...
简单计算一下即可，答案如图所示
lim((sin6x)+xf(x))\/(x^3)=0求lim((6+f(x))\/(x^2)),x趋0
当x趋近0时,sin6x是趋近6x,上下约去一个x,就是等于0 O(∩_∩)O 高等数学!
