You can limit the upper bound of ε, but not the lower bound of ε, such as ε < 1, ε < 1/2, etc. , does not affect its "arbitrarily small" characteristics. It can also be understood that n can be found for smaller ε, so when ε is larger, the original n can still be taken, and | xn-a | < ε can still be guaranteed.
For n, when |Xn-a| is simple, n > n can be directly obtained from | xn-a | < ε; Otherwise, you can first enlarge |Xn-a| to a simple formula related to n, such as a multiple of 1/n, and you can get | xn-a | < 1/n in this question, and use this formula to determine that n is less than ε.
For this topic, if | xn-a | < 1/n is selected, then ε does not need to be limited to less than 1, and the process is as follows:
Because | xn-a | < 1/n, for any small positive integer ε, make | xn-a | < ε exactly 1/n < ε, that is, n > 1/ε, and choose a positive integer n = [/kloc-0]. So the limit of the sequence {Xn} is 0.