To find f(x) is to find the limit. Is it obvious that x should be discussed?

| x |< in 1, lim

X 2n = 0, so f (x) =-1;

| x | >； 1, the numerator denominator is divided by x^2n and then the limit is found, and f (x) =1;

F(x)=0 when | x | = 1.

For example:

[ 1/(n^2- 1]-0]= 1/(n^2- 1)，

For any δ > 0, the limit | n | > 1,

If | 1/(n 2- 1) | < δ,

Only n > 1/δ+ 1 is needed to solve it.

For any δ > 0, there is n = [1/δ+1]+1,for any n≥N, | xn-a | < δ,

Certificate of completion.

Note: [X] means rounding X,

For example, take 0.3 1. 56.6 take 57.

Extended data:

Generally speaking, n increases with the decrease of ε, so n is often written as N(ε) to emphasize the dependence of n on ε change. But this does not mean that n > is exclusively determined by ε: (for example, if n >; N makes | xn-a | < ε hold, then obviously n & gtN+ 1, n >2N and so on also make | xn-a |.

"when n >; When n, there is inequality | xn-a |

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