求极限这一步是怎么来的?3是怎么得来的?
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分式通分:1/(x+1)-3/(x³+1)=[1/(x+1)]-[3/(x+1)(x²-x+1)]
=[(x²-x+1)-3]/[(x+1)(x²-x+1)]=(x²-x-2)/(x+1)(x²-x+1)=(x-2)(x+1)/(x+1)(x²-x+1)
=(x-2)/(x²-x+1);
∴x→-1lim[1/(x+1)-3/(x³+1)]=x→-1lim[(x-2)/(x²-x+1)]=-3/3=-1
=[(x²-x+1)-3]/[(x+1)(x²-x+1)]=(x²-x-2)/(x+1)(x²-x+1)=(x-2)(x+1)/(x+1)(x²-x+1)
=(x-2)/(x²-x+1);
∴x→-1lim[1/(x+1)-3/(x³+1)]=x→-1lim[(x-2)/(x²-x+1)]=-3/3=-1
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