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本回答被提问者采纳
lim((6+f(0))x+f'(0)x^2+(0.5f''(0)-36)x^3+小o(x^3))/x^3=0，因此必须有f(0)=-6,f'(0)=0,f''(0)=72。代入可得结果。
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(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.
已知,lim x→0 [sin6x+xf(x)]\/x^3=0 求:lim x→0 [6+f(x)]\/x^2=...
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 - sin(6x)]\/x³= lim(x-->0) [6 - 6cos(6x)]\/(3x²) <== 洛必达法则 =...
若limx趋近于0,(sin6x+xf(x))\/x^3=0,求limx趋近于0,(6+f(x))\/x^2...
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,[6(1-cos6x)]\/3x²，用等价无穷小lim x→0,(1-cosx)等价于lim x...
...sin6x)+xf(x)]\/x^3}=0 求limx趋近0 [6+f(x)]\/x^2=? 答案是36.希望...
\/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 ...
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lim [sin6x+xf(x)]\/x^3=0,则lim [6+f(x)]\/x^2是多少? (x是趋近0)
属于0-0型，可以应用洛必答法则：(x→0)lim[6cos6x+f(x)+xf'(x)]\/(3x^2)=0 (x→0)lim[-36sin6x+f'(x)+f'(x)+xf''(x)]\/(6x)=0 (x→0)lim[-216cos6x+2f''(x)+f''(x)+xf'''(x)]\/6=0 所以，x→0时：3f''(x)+xf'''(x)=216 3f''(x)=216 f''(x)...
求极限当x→0若lim[sin6x+x f(x)]\/x3=0,求lim[6+ f(x)]\/x2
代入得到 lim[sin6x+xf(x)]\/x^3=6x-(6x)^3\/3!+o(x^3)+f(0)x+f'(0)x^2+1\/2f''(0)x^3+o(x^3)\/x^3=0 x→0 整理得lim[6x+f(0)x+f'(0)x^2]\/x^3+1\/2f''(0)-36=0 从而f(0)=-6 f'(0)=0 1\/2f''(0)-36=0 f''(0)=72 lim[6+ f(x)]\/x^2=...
