求lim [（1/x+1)-(3/x^3+1)]= x→-1
简单分析一下，详情如图所示
求lim [（1/x+1)-(3/x^3+1)]= x→-1
函式应该这样吧：1/(x+1)-3/(x^3+1)
lim [1/(x+1)-3/(x^3+1)]
= lim [ (x^3-3x-2)/(x+1)(x^3+1)]
=lim ( 3x^2-3)/[(x^3+1)+(x+1)3x^2]（罗比达）
=lim [( 3x^2-3)/(4x^3+3x^2+1)]（罗比达）
=lim[6x/(12x^2+6x)]
=-6/6=-1
lim(x→-1) [3/(x^3+1)-1/(x+1)］
x3+1＝(x+1)(x2-x+1)
求:当x→-1时,lim {（1/x+1)-[3/（x^3+1)]}=
分式化简：（）=1/(x+1) -3/[(x+1)(x^2-x+1)]=[(x+1)(x-2)]/[(x+1)(x^2-x+1)]=(x-2)/(x^2-x+1)
=(x-2)/[(x+1)(x-2)+3]=1/[(x+1)+3/(x-2)]
当x→-1时,结果为-1
求极限lim下面是x-->-1 [1/(x+1)-3/(x^3+1)]
先通分，然后用洛必塔法则
lim(x->-1) [1/(x+1)-3/(x^3+1)]
=lim(x->-1) [(x^2-x+1)/(x^3+1)-3/(x^3+1)]
=lim(x->-1) [(x^2-x-2)/(x^3+1)]
=lim(x->-1) [(2x-1)/(3x^2)]
=-1
Lim[1/(x+1)-3/(x3+1)] x→-1
原式=lim[(x^2-x+1-3)/(x^3+1)]
=lim[(x+1)(x-2)/(x+1)(x^2-x+1)]
=lim[(x-2)/(x^2-x+1)]=-1
lim[1/(x+1)-3/(x3+1)]
要把问题说得清楚点x趋于什么？是不是-1?
如果是这样的话，就应该是这个样子的。
lim[1/(x+1)-3/(x3+1)]=lim[(x^3+1-3x-3)/(x^3+1)]=lim[(x+1)(x^2-x-2)/(x+1)(x^2-x+1)]=lim[(x^2-x-2)/(x^2-x+1)]=-2
lim（x→-1）［3/（x³+1）-1/（x+1）］
lim(x→-1) [3/(x^3+1)-1/(x+1)］
=lim(x→-1) [3- (x^2-x+1) ]/[(x+1)(x^2-x+1) ]
=lim(x→-1) -(x^2-x-2)/[(x+1)(x^2-x+1) ]
=lim(x→-1) -(x+1)(x-2)/[(x+1)(x^2-x+1) ]
=lim(x→-1) -(x-2)/(x^2-x+1)
=-(-1-2)/(1+1+1)
=1
正数x规定f x=x/x+1如f3=3/3+1=3/4f1/3=1/3/1/3+1=1/4试算f1
2012+1/2过程原式=1/2014+1/2013+1/2012+·····+1/4+1/3+1/2+2/3+3/4+4/5+····+2012/2013+2013/2014首尾依次合并=1；故2012+1/2f1/2013=1/2014,f2013=2013/2014相加为一f1/2012一样f1/1/2同样f1=1/2追问：看不太懂诶补充：真的假的，将第一项算出为1/2014，最后一项算出为2013/2014,第二项与倒数第二项和也为一，这样的项一直到第2012项即f1/2=1/3,中间一项f1=1/2不能合并，故共2012个1加一个1/2
化简：（X-1）/(X^2/3+X^1/3+1)+（X+1）/(X^1/3+1)-(X-X^1/3)/(X^1/3-1)
原式=(x^1/3-1)(x^2/3+x^1/3+1)/(x^2/3+x^1/3+1)+(x^1/3+1)(x^2/3-x^1/3+1)/(x^1/3+1)-x^1/3(x^2/3-1)/(x^1/3-1)
=(x^1/3-1)+(x^2/3-x^1/3+1)-x^1/3(x^1/3+1)(x^1/3-1)/(x^1/3-1)
=x^1/3-1+x^2/3-x^1/3+1-x^1/3(x^1/3+1)
=x^2/3-x^2/3-x^1/3
=-x^1/3
