lim ln(1 1 x)
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x趋于负无穷e^x趋于01+e^x趋于1则分子趋于ln1=0分母是无穷大所以极限是0
这是两道题吗?1、两边同时乘(x-1)(x-2)(x-3)答案是x1=-1,x2=22、分母因式分解分别为(x-7)(2x+3)和(x-5)(x-7)然后同时乘(x-5)(x-7)(2x+3)答案是x
1x+2x+3x+4x+5x+6x+7x+8x+9x+10x+11x+12x+13x+14x+15x=550120x=550x=55/12=4.583
lim[ln(1+x)-lnx]/x=limln[(1+x)/x]/x=limln(1+1/x)/x=0.
x²+2x=4x²+2x+1=4+1(x+1)²=5x+1=±√5x=-1-√5,x=-1+√5
【解法一】【解法二】5x(x+7)(x+4)=5x(x^2+11X)5x(x²+11x+28)-5x(x²+11x)=05x^3+55x^2+140x=5x^3+55x^25x(x
3x*x+11x+10=(3x+5)(x+2)
6x^3+7x^2-13x+11=6x^3+3x^2-15x+4x^2+2x-10+21=3x(2x*x+x-5)+2(2x*x+x-5)+21=3x*0+2*0+21=21
5x(x+7)(x+4)=5x(x^2+11X)5x^3+55x^2+140x=5x^3+55x^2140x=0x=0
x(2x-3)+3x+3x(x-2)=5x(x-4)+112x²-3x+3x+3x²-6x=5x²-20x+115x²-6x=5x²-20x+115x
[1/(x-1)(x+2)]+[1/(x+2)(x+5)]+[1/(x+5)(x+8)]+[1/(x+8)(x+11)]=(1/3x-3)-1/24(1/3)[1/(x-1)-1/(x+11)]=1/
(2013+2012)X
当x>=13时|x+11|+|x-12|+|x+13|+|x-13|=x+11+x-12+x+13+x-13=4x-1>=51(代入x=13)当13=>x>=12时|x+11|+|x-12|+|x+1
原式=[x(x-3)/(x-3)²][(x-5)(x-6)/x(x-5)]-1=[x/(x-3)][(x-6)/x]-1=(x-6)/(x-3)-1=(x-6-x+3)/(x-3)=-3/(
lim(x→0+)ln(sin3x)/ln(sinx)=lim(x→0+)[3cos3x/(sin3x)/[cosx/sinx]=lim(x→0+)(3sinx/sin3x=1再问:[3cos3x/(
用换元法设x*x+11x-8=A然后再做
看三个分母的因式可以知道:首先,每个提出一个常数,把分子化成简单的常数,即化为:1+1/(x^2+6x+9)+1-2/(x^2-9)+2-1/(x^2+4x+3)接着,每个因式分母提出(x+3),即为
9x-12x+5x=-3x+5x=2x3x-5x-2+5x+8=-2x+5x-2+8=3x+63x+5+2x+3-7-11x=3x+2x-11x+5+3-7=5x-11x+8-7=-6x+1
首先我们可以确定一下常数项常数项的结果肯定是-1*n=-6那么n=6再看2次方项-1+m=-6那么m=-5所以m+n=1
通过泰勒公式可以在0点展开ln(x+√(1+x^2):ln(x+√(1+x^2)=x+o(x)o(x)表示余项是x的高阶无穷小所以代入原式=limln(x+√(1+x^2))/x=lim[x+o(x)