When the right side approaches 1, the result is 1-e,

So it is discontinuous.

It should be pointed out that:

When x tends to infinity, (1+ 1/x) x = e, and e is a natural logarithm.

Solution process:

1, molecular Kaiping variance: (1-x N) (1+x N),

2. When x approaches 1 (both sides can approach), lim(n approaches infinity) (1+x n)/(1+x 2n) =1,

3. When x approaches 1 to the left,: lim(n approaches infinity) (1-x n) =1-/e;

When x approaches 1 to the right,: lim(n approaches infinity) (1-x n) =1-e;

This topic has its limitations and needs special care.

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Leave some character for LP. I hope her interview goes well!

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