求大神讲一道高数题
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你还记得立方差公式吗？！
把(1-x+x²)看做是[(1-x+x²)^(1/3)]³，则：
1-(1-x+x²)=[1-(1-x+x²)^(1/3)]·[1+(1-x+x²)^(1/3)+(1-x+x²)^(2/3)]
即：x-x²=[1-(1-x+x²)^(1/3)]·[1+(1-x+x²)^(1/3)+(1-x+x²)^(2/3)]
所以：分子1-(1-x+x²)^(1/3)=(x-x²)/[1+(1-x+x²)^(1/3)+(1-x+x²)^(2/3)]
故，原式=lim<x→0>(1-x)/[1+(1-x+x²)^(1/3)+(1-x+x²)^(2/3)]
=(1-0)/(1+1+1)
=1/3
把(1-x+x²)看做是[(1-x+x²)^(1/3)]³，则：
1-(1-x+x²)=[1-(1-x+x²)^(1/3)]·[1+(1-x+x²)^(1/3)+(1-x+x²)^(2/3)]
即：x-x²=[1-(1-x+x²)^(1/3)]·[1+(1-x+x²)^(1/3)+(1-x+x²)^(2/3)]
所以：分子1-(1-x+x²)^(1/3)=(x-x²)/[1+(1-x+x²)^(1/3)+(1-x+x²)^(2/3)]
故，原式=lim<x→0>(1-x)/[1+(1-x+x²)^(1/3)+(1-x+x²)^(2/3)]
=(1-0)/(1+1+1)
=1/3
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