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求lim [(1/x+1)-(3/x^3+1)]= x→-1

lim(x->-1)[(x-x-2)/((x+1)(x-x+1))].=lim(x->-1)[((x-2)(x+1))/((x+1)(x-x+1))].=(-1-2)/(1-(-1)+1).=-1.与常数a的接近程度.ε越小,表示接近得越近;而正数ε可以任意地变小,说明xn与常数a可以接近到任何不断地靠近的程度.但是,尽管ε有

:原式=lim(x->-1)[1/(x+1)-3/((x+1)(x-x+1))]=lim(x->-1)[(x-x-2)/((x+1)(x-x+1))]=lim(x->-1)[((x-2)(x+1))/((x+1)(x-x+1))]=lim(x->-1)[(x-2)/(x-x+1)]=(-1-2)/(1-(-1)+1)=-1.

函数应该这样吧: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

函数应该这样吧: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

这个是大一的题目吧,等于零.因为在x→0时分母很大,分子很小

根据立方差公式得:1-x^3=(1-x)(1+x+x^2),所以lim【1/(1-x)-3/(1-x^3)】=lim【(x^2+x-2)/(1-x^3)】,当x趋于1时,分子分母都分别趋于0,此时采用罗必塔法则:lim【1/(1-x)-3/(1-x^3)】=lim【(x^2+x-2)/(1-x^3)】=lim【-(2x+1)/(3x^2)]=-1.

-31/(x+1)-3/(x^3+1)=1/(x+1)-3/[(x+1)(x-x+1)]=[1/(x+1)]*[1-3/(x-x+1)]

1/(x+1)-3/(x^3+1)=(x^2-x-2)/(x^3+1)=(x-2)/(x^2-x+1)-3/3=-1

答案应该是-1lim[(1/1-x)-3/(1-x)(1+x+x^2)]=lim [(1+x+x^2)-3]/(1-x)(1+x+x^2)=lim (x-1)(x+2)/(1-x)(1+x+x^2)=-lim(x+2)/(1+x+x^2)∵x→1∴极限为-1

x-1和x^3-1都是分线下的吧 lim(x→1)(1/x-1)-(3/x^3-1)=lim(x→1) 1/(x-1)-3/(x-1)(x^2+x+1)=lim(x→1) (x^2+x+1-3)/(x-1)(x^2+x+1)=lim(x→1) (x+2)(x-1)/ (x-1)(x^2+x+1)=lim(x→1) (x+2)/(x^2+x+1)=3/3=1

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