# Limit Points of a Sequence

A number $$l$$ is said to be a **limit point** of a sequence $$u$$ if every neighborhood $${N_l}$$ of $$l$$ is such that $${u_n} \in {N_l}$$, for infinitely many values of $$n \in \mathbb{N}$$, i.e. for any $$\varepsilon > 0$$, $${u_n} \in \left( {l – \varepsilon ,l + \varepsilon } \right)$$, for finitely many values of $$n \in \mathbb{N}$$. Evidently, if $$l = {u_n}$$ for infinitely many values of $$n$$ then $$l$$ is a limit point of the sequence $$u$$.

As in the case of sets of real numbers, limit points of a sequence may also be called accumulation, cluster or condensation points. The limit points of a sequence may be classified into two types: **(i)** those for which $$l = {u_n}$$ for infinitely many values of $$n \in \mathbb{N}$$, **(ii)** those for which $$l = {u_n}$$ for only a finite number of values of $$n \in \mathbb{N}$$. But this distinction is not necessary. As such, we do not distinguish the above mentioned two types of limit points of sequences by different titles. It should be noted that every limit point $$l$$ of the range set $$R\left\{ u \right\}$$ of a sequence $$u$$ is also a limit point of the sequence $$u$$ , because every neighborhood of $$l$$ contains infinitely many points of $$R\left\{ u \right\}$$, and so of the sequence $$u$$. On the other hand, a limit point of $$u$$ may or may nor be a limit point of $$R\left\{ u \right\}$$.

If the values of only a finite number of terms of $$u$$ are not distinct, then evidently the limit points of $$u$$ are the same as those of the set $$R\left\{ u \right\}$$. Conclusively, it follows that the limit points of a sequence $$u$$ are either the points or the limit points of the set $$R\left\{ u \right\}$$.

** Example 1:** If a sequence $$u$$ is defined by $${n_n} = 1$$, then $$1$$ is the only limit point of

sequence.

** Solution:** For any $$\varepsilon > 0$$, $${u_n} = 1 \in \left( {1 – \varepsilon ,1 + \varepsilon } \right)$$ $$\forall n \in \mathbb{N}$$. Therefore, $$1$$ is a limit pint of the sequence. Let $$\alpha \in \mathbb{R}$$ and $$\alpha \ne 1$$. Then for all $$n$$, $$\left| {{u_n} – \alpha } \right| = \left| {1 – \alpha } \right|\not < \varepsilon $$. When $$\left| {1 – \alpha } \right| < \varepsilon < 0$$. Thus no point $$\alpha $$ other than $$1$$ is the limit point of the sequence. Note that the limit point of the sequence $$u$$ is not a limit point of the range $$R\left\{ u \right\} = \left\{ 1 \right\}$$.

** Example 2:** If $${u_n} = \frac{1}{n}$$, then $$0$$ is the only limit point of the sequence $$u$$.

**Sufficient conditions for number $$l$$ to be or not to be a limit point of a sequence $$u$$:**

- If for every $$\varepsilon > 0{\text{ }}\exists m \in \mathbb{N}$$ such that $${u_n} \in \left( {l – \varepsilon ,l + \varepsilon } \right)$$, $$\forall n \geqslant m$$ or equivalently $$\left| {{u_n} – l} \right| < \varepsilon $$ $$\forall n \geqslant m$$, then $$l$$ is a limit point of the sequence $$u$$ .In such a case it can be easily seen that $$l$$ is the only limit point of the sequence. The above condition is not necessary as it can be seen that for the sequence $$\left\langle {1,\frac{1}{2},1,\frac{1}{3},1,\frac{1}{4} \ldots } \right\rangle $$,$$1$$ is a limit point of this sequence but the above condition is not satisfied.
- If $$\varepsilon > 0$$, $${u_n} = 1 \in \left( {1 – \varepsilon ,1 + \varepsilon } \right)$$ for only a finite number of values of $$n$$ then $$l$$ is not a limit point of $$u$$. This condition is also necessary for a number $$l$$ not to be a limit point of the sequence $$u$$.

__Remarks__

- Whenever we simply write $$\varepsilon > 0$$ it is implied that $$\varepsilon $$ may be howsoever small positive number.
- A positive number $$\eta $$ is said to be arbitrarily small if given any $$\varepsilon > 0$$, $$\eta $$ may be chosen such that $$0 < \eta < \varepsilon $$.
- If $$\eta $$ is an arbitrary small positive number and given any $$k > 0$$ then $$k\eta $$ is also an arbitrary small positive number, this follows immediately if we take $$0 < \eta < \varepsilon /2k$$ for an $$\varepsilon > 0$$.
- If $${\varepsilon _1},{\varepsilon _2}$$ are two arbitrary small positive numbers then it readily follows that $$l$$ is a limit point of a sequence $$u$$ if and only if $${u_n} \in \left( {l – \varepsilon ,l + \varepsilon } \right)$$ for infinitely many values of $$\eta $$.

** Example 3:** Every bounded sequence $$u$$ has at least one limit point.

** Example 4:** The set of limit points of a bounded sequence $$u$$ is bounded.

** Theorem:** The set of limit points $$E$$ of every sequence $$u$$ is a closed set.

** Corollary: **Every bounded set $$E$$, of limit points of a sequence $$u$$, contains the smallest and greatest members.