Question

设当x-》0时，(xf(x)+ln(1-2x))/x^2的极限 =4,则当x趋近于0时，(f(x)-2)/x的极限是多少。

这里我把题目中的ln用等价无穷小代替了，结果发现两者相等，得出答案为4，为什么不对

Answer 1

简单计算一下即可，答案如图所示

Answer 2

lim(x->0) (xf(x)+ln(1-2x))/x^2 (0/0)
=lim(x->0) [ xf'(x)+f(x) -2/(1-2x) ]/(2x) (0/0)

=> 0.f'(0)+f(0) -2/(1-2(0)) =0
=>f(0)- 2=0
=>f(0)=2

lim(x->0) [ xf'(x)+f(x) -2/(1-2x) ]/(2x) (0/0)
lim(x->0) [ xf''(x) + 2f'(x) - 2/(1-2x)^2 ]/2 =4
=>
0.f''(0) + 2f'(0) - 2/(1-2(0))^2 =8
2f'(0) - 2 =8
f'(0) = 5

lim(x->0)( f(x) -2)/x (0/0)
=lim(x->0)f'(x)
=f'(0)
=5

追问

答案是6，而且f(x)不一定可导

追答

f(x)不一定可导，那就没法计算！

lim(x->0) (xf(x)+ln(1-2x))/x^2 (0/0)
=lim(x->0) [ xf'(x)+f(x) -2/(1-2x) ]/(2x) (0/0)

=> 0.f'(0)+f(0) -2/(1-2(0)) =0
=>f(0)- 2=0
=>f(0)=2

lim(x->0) [ xf'(x)+f(x) -2/(1-2x) ]/(2x) (0/0)
lim(x->0) [ xf''(x) + 2f'(x) - 4/(1-2x)^2 ]/2 =4
=>
0.f''(0) + 2f'(0) - 4/(1-2(0))^2 =8
2f'(0) - 4 =8
f'(0) = 6

lim(x->0)( f(x) -2)/x (0/0)
=lim(x->0)f'(x)
=f'(0)
=6