How many limits can a sequence have

Let us calculate the limit of. Notice that:. Since , , the numerator converges to , and since and , the limit of denominator is we apply the theorem on arithmetic of limits. Therefore again we apply the theorem. A sequence is bounded, if there exists , such that for any. We have the following theorem: if converges to zero and is bounded, then also converges to zero.

Obviously , where is bounded, and converges to zero. If is such that for all there exists , such that for all we have , so if for arbitrarily large number from a given place on the elements of sequence are greater then this number we say that the sequence tends to infinity and write. If a sequence is such that for any there exists , such that for all we have , we say that the sequence tends to minus infinity and write.

This sequence tends to infinity, because if , , let. Then for all , we get. Let and be sequences of real numbers. Notice that this theorem does not tell anything about some types of limit operation, in which we can tell nothing about the limits.

In this cases usually some further calculations are needed to be able to use the above theorem. Since ,. Because , we get that. Imagine that you have a bank account with interest rate of a year! But you have only in this account. If interest rates are calculated every year, after one year you will have dollars. But if the interest rates were calculated twice a year you will get a half of your money twice a year, so at the end of the year you will have.

If it would be calculated four times a year, you will get , monthly. It is clear that the final amount of money increases if the frequency of calculating interest rate increases. How big this amount can be? Obviously we ask about the limit of sequence ,. Active 6 years, 2 months ago. Viewed 9k times. Alec Teal 5, 3 3 gold badges 28 28 silver badges 52 52 bronze badges.

Pburg Pburg 1, 8 8 silver badges 14 14 bronze badges. As you're using extended reals there are bounds for both. I'm just happy I was on the right lines! Wikipedia lists this definition: "Formally, a sequence can be defined as a function whose domain is a countable totally ordered set, such as the natural numbers.

Show 1 more comment. Active Oldest Votes. Introduce infinitely many limit points, by giving names to each one: 1,2,3, Construct the simplest sequence converging to each one : the constant sequence.

Pack the sequences into one sequence using the greedy algorithm. It may be worth noticing that this counterexample works only for limit sets that are countable or that admit a countable dense subset. Take the reals with the discrete metric.

This sequence still has an infinite number of accumulation points. It does not have an accumulation point at every real, however. Otherwise uncountable would be countable. But you can have a single sequence whose subsequences have limit points that are uncountable. They just can't be all eventually constant. The others are only "approached".. Add a comment. I am assuming a subsequence is the original sequence with some elements deleted and the original order maintained otherwise.

I guess I should have included this in the original answer, let me add this. Show 9 more comments. But a bit of explanation might be welcome. Evpok Evpok 4 4 silver badges 17 17 bronze badges. Community Bot 1. Zhanxiong Zhanxiong And we won't even erase numbers that have already appeared! Benjamin Dickman Benjamin Dickman The deviation from minimalism in this answer is that the constant subsequences can be loosened slightly to convergent subsequences with the same limits.

If I figure out how to articulate it, I will post an edit. If it isn't, the set can be countable, but not uncountable. I'm not sure about using diagonalization as an argument. Sign up or log in Sign up using Google. Sign up using Facebook. Sign up using Email and Password.

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