即化简得lim(sin6x-6x)/x³
然后多次利用洛必达法则即可得极限为-36
再根据极限的四则运算可得
所求极限为36追问
lim(sin6x-6x)/x³怎么来的？
追答lim(sin6x+xf(x))/x³-(6+f(x))/x²
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.
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.
若limx趋近于0,(sin6x+xf(x))\/x^3=0,求limx趋近于0,(6+f(x))\/x^2...
lim x→0,[sin6x + xf(x)]\/x³=0+α，其中lim x→0,α=0 即f(x)\/x² = -sin6x\/x³ + α 从而lim x→0,[6+f(x)]\/x²=lim x→0,( 6\/x² - sin6x\/x³ + α )=lim x→0,(6x-sin6x)\/x³，用洛必达法则 =lim x→0,[...
已知{limx趋近0 [(sin6x)+xf(x)]\/x^3}=0 求limx趋近0 [6+f(x)]\/x^...
f有二阶导数吧？条件要写全了，否则很难做题的。Taylor展式最简单，sin6x=6x-(6x)^3\/6+小o(x^3)，xf(x)=x(f(0)+f'(0)x+f''(0)\/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...
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
x趋于0,lim x-o ( sin6x+xf(x))\/x3=0 ,lim x-o (6+f(x))\/x2=?
实际上，所要求的极限就是原极限的变形！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)]\/...
已知,lim x→0 [sin6x+xf(x)]\/x^3=0 求:lim x→0 [6+f(x)]\/x^2=...
lim(x-->0) [sin(6x) + xf(x)]\/x³ = 0，要弄做不定式，分子和分母都要趋向0 sin(6x) + xf(x) = 0 xf(x) = - sin(6x)f(x) = - (sin6x)\/x lim(x-->0) [6 + f(x)]\/x²= lim(x-->0) [6 - (sin6x)\/x]\/x²= lim(x-->0) [6x - ...
...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)...
...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
