Question

当x→0时,(x^2+x-2)^2lnx与a(x-1)^n是等价无穷小,则常数a=，n =？

Answer 1

x->1
(x^2+x-2)^2. lnx
=[(x-1)(x+2)]^2. ln[1+(x-1)]
=(x+2)^2. (x-1)^3 +o[(x-1)^3]
//
lim(x->1) (x^2+x-2)^2. lnx / [a(x-1)^n] = 1
lim(x->1) (x+2)^2. (x-1)^3/ [a(x-1)^n] = 1
lim(x->1) 9(x-1)^3 / [a(x-1)^n] = 1
=>
a=9 and n=3

追问

(x+2)^2. (x-1)^3这怎么来的？

追答

(x^2+x-2)^2 = [(x+2)(x-1)]^2 =(x+2)^2.(x-1)^2
x->1
lnx = ln( 1+ (x-1) ) 等价于 （x-1）
=>
(x^2+x-2)^2 .lnx 等价于 (x+2)^2. (x-1)^3

Answer 2

lnx = ln(1+x-1) ~ x-1
(x^2+x-2)^2 = （x+2)^2(x-1)^2 ~ 9(x-1)^2
(x^2+x-2)^2lnx ~ 9 (x-1)^3
a=9, n=3