Question

高数求解……

怎么做？谢谢了

Answer 1

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Answer 2

x->0
e^x = 1+ x +(1/2)x^2 +(1/6)x^3 +o(x^3)
(1+ax)/(1+bx)
=(1+ax) . [ 1-bx + b^2.x^2 -b^3.x^3+o(x^3) ]
=(1-bx + b^2.x^2 -b^3.x^3) + ax.(1-bx + b^2.x^2) +o(x^3)
=1+ (a-b)x +(b^2- ab)x^2 +( -b^3+ ab^2)x^3 +o(x^3)
e^x - (1+ax)/(1+bx)
=[1-(a-b)]x +[ 1/2 -(b^2- ab)] x^2 + [ 1/6 -( -b^3+ ab^2) ] x^3 +o(x^3)
e^x - (1+ax)/(1+bx) 为3价无穷小
=>
coef.of x =0 and coef. of x^2 =0
1-(a-b) = 0 (1)
1/2 -(b^2- ab) =0 (2)
sub (1) into (2)
1/2 -(b^2- ab) =0
1/2 -[b^2- (1+b)b ] =0
1/2 +b = 0
b=-1/2
from (1)
1-(a-b) = 0
1- ( a+1/2) = 0
a= 1/2
(a,b)= (1/2, -1/2) , e^x - (1+ax)/(1+bx) 为3价无穷小

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