What is the difference between the limit of a sequence and a limit point of a set? I always thought they were the same thing. The limit of a sequence is a point such that every neighborhood around it contains infinitely many terms of the sequence. The limit point of a set is a point such that every neighborhood around it contains infinitely many points of the set. So is the limit of a sequence also a limit point of a set that contains it? what's the difference?
 A: Let's work in the real line, for concreteness. Consider a sequence $(x_n)_{n \geq 1}$. A limit point of the sequence is a limit point of the set $\{x_n \mid n\geq 1\}$. When you say the limit point, it means that the set $\{x_n \mid n\geq 1\}$ has only one limit point, say, $L$. And this $L$ is the element who satisfies the definition we all know and love: $$\forall \ \epsilon > 0, \ \exists \ n_0 \geq 1, {\rm s.t.} \ n > n_0 \implies |x_n - L| < \epsilon.$$
If the set $\{x_n \mid n\geq 1\}$ has more than one limit point, then the sequence $(x_n)_{n \geq 1}$ does not converge.
A: In a general metric space X, it depends on the range of the sequence whether both the notions are same or not. Let $x_n$ be a convergent sequence i.e. its limit exists and is unique, say $x$. Then its range is a set, say E $\subseteq$ X, which contains all the distinct elements in the sequence $x_n$.
If the set E is finite, then it is obvious that one of the terms in E, suppose $y$, occurs infinitely many times in the sequence $x_n$. Then the limit of the sequence is, of course, $y$. Also, the $y$ is unique since the sequence is convergent so every subsequence has same subsequential limit. So, $y = x$.  But note that E has no limit point since it is finite (a point is a limit point of a subset of a metric space if all of its neighbourhoods contain infinitely many points of the subset) and in particular, all of its points are isolated points.
Now if the range set E is countably infinite (it can't be uncountable since the sequence itself is countable), then E has a unique limit point equal to $x$. Hence, every neighbourhood has infinitely many points of E and hence of $x_n$. Then for a given $\varepsilon \gt 0,$ $$\exists N \in \Bbb N \ni d(x_n,x)\lt \varepsilon\quad \forall\;n \ge N$$ Limit point is unique because if it was not the case, then there would be another point say $z\, (\neq x)$ in $\overline{E}$ which has the above property. Hence the limit of the sequence would not be unique. Thus, if the sequence $x_n$ has infinitely distinct elements, then both the notions are same.
P.S: If you consider a subset A such that A $\supseteq$ E, then as commented by littleO, A can have many limit points, different from the limit of the sequence, no matter whether E is finite or infinite. But for this case, the question of differentiating between the two notions also becomes irrelevant.
A: Let's take elements of a sequence as a set say S. S has many subsets, which form many subsequences of the primary sequence.
S has a limit that would be the limit of sequence & some of it's subsets too have limits those would be the limits of their respective subsequences & the limit point of whole sequence
Can say that the set of all limits of subsets would be limit points of the sequence. 
