Exercise 1-8, advanced mathematics 6th edition, question 4, X=- 1 is discontinuous. I know ~ but I don't know how to get the piecewise function ~ - Training Enrollment Network

Exercise 1-8, advanced mathematics 6th edition, question 4, X=- 1 is discontinuous. I know ~ but I don't know how to get the piecewise function ~

Now let's judge the limit of x^2n.

When the limit of | x | > x 2n is positive infinity at 1

lim[( 1-x^2n)/( 1+x^2n)]*x=[( 1/x^2n- 1/( 1/x^2n+ 1)]*x=-x

You can get it in the same way.

|x|= 1 Limit is 0.

The limit of | x | <1is X.

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