Solution: According to the meaning of the question, there is | f (x) | < =| e x-1|.

Then | f (x)/e x- 1 |

When x tends to 0, both sides take the limit at the same time: | lim f (x)/e x-1| < =1

Namely: lim | a1ln (1+x)+a2ln (1+2x)+...+anln (1+NX)/ex-1|.

That is, there are two ways to deal with the above formula: 1. If you have studied Taylor formula, let ln( 1+x)=x+O(x), ln( 1+2x)=2x+O(2x), ... = 1

2. If you haven't studied Taylor's formula, you can take apart the molecules of the above formula one by one to find the limit, and then you can get the answer by L'H?pital.