已知{limx趋近0 [(sin6x)+xf(x)]/x^3}=0 求limx趋近0 [6+f(x)]/x^2=? 答案是36.希望大家给点步骤,谢了

如题所述

第1个回答  2011-12-13
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)=-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

已知{limx趋近0 [(sin6x)+xf(x)]\/x^3}=0 求limx趋近0 [6+f(x)]\/x^...
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

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: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 - s...

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.

...6x+xf(x)]\/x^3=0,则lim(x趋近于0)[6+f(x)]\/x^2=? rt
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.

...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→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

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

...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 ...

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