第八题怎么写
1个回答
展开全部
x->0
分母 :
[(1-x)(1+x)]^(1/3) =(1-x^2)^(1/3) ~ 1- (1/3)x^2
[(1-x)(1+x)]^(1/3) -1 ~ -(1/3)x^2
分子 :
e^(3x) - e^(2x) - e^x +1
~ [ 1+ 3x + (1/2) (3x)^2] - [ 1+ 2x + (1/2) (2x)^2] -[ 1+ x + (1/2) (x)^2] +1
=2x^2
lim(x->0) [e^(3x) - e^(2x) - e^x +1 ]/ {[(1-x)(1+x)]^(1/3) -1}
=lim(x->0) 2x^2/ [-(1/3)x^2]
=-6
分母 :
[(1-x)(1+x)]^(1/3) =(1-x^2)^(1/3) ~ 1- (1/3)x^2
[(1-x)(1+x)]^(1/3) -1 ~ -(1/3)x^2
分子 :
e^(3x) - e^(2x) - e^x +1
~ [ 1+ 3x + (1/2) (3x)^2] - [ 1+ 2x + (1/2) (2x)^2] -[ 1+ x + (1/2) (x)^2] +1
=2x^2
lim(x->0) [e^(3x) - e^(2x) - e^x +1 ]/ {[(1-x)(1+x)]^(1/3) -1}
=lim(x->0) 2x^2/ [-(1/3)x^2]
=-6
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