Confusion on the definition of accumulation points I've been trying to learn a bit about limits of sequences and accumulation points to get a better intuition behind the workings for calculus and I got confused on the definitions of limits,limit points and accumulation points of sequences and sets.
My first question is a limit of a sequence the same as the accumulation point and is that the same as the limit point i looked online and its all very vague.
My second confusion is that is the limit of a sequence the same as the limit of a set if not is there some proof or intuitive explanation as to why not?.
I know this is probably a very simple and probably trivial concept for all of you on here but its confused me a lot.
Thanks in advance
 A: A limit point is the same thing as an accumulation point, and its definition is that:

A point $x$ is a limit point of a set $A$ if for every neighbourhood $S$ of $x$ there exists $y \in S$ such that $y \in A$, $y \neq x$.

I strongly prefer the name "accumulation point", because you are not actually taking limits here... it's the other way around! In order to be able to do limits you normally require accumulation points, since the topological definition of a limit requires taking neighbourhoods and computing the function there.
About your second question:

A point $x$ is an accumulation point for a sequence $\{x_n\}$ if any neighbourhood $S$ of $x$ is such that there are infinitely many indices $n$ such that $x_n \in S$.

It is essentially the same definition as above, but you take $A=\{x_n\}_{n \in \mathbb{N}}$. However, a point is a limit point for a sequence if all the indices after a certain $n$ are in any neighbourhood. Formally:

A point $x$ is the limit of a sequence $\{x_n\}$ if any neighbourhood $S$ of $x$ is such that there exists $N \in \mathbb{N}$ such that $x_n \in S$ for all $n>N$.

And this is stronger than simply being an accumulation point: you can see the difference by considering the sequence $x_n = \frac{(-1)^n n}{n+1}$. Any neighbourhood of $1$ contains infinitely many points of this sequence, namely all the $x_{2n}$ after a certain $n$. Similarly, any neighbourhood of $-1$ will contain all the $x_{2n+1}$ after a certain $n$, so both $1$ and $-1$ are cluster points for $x_n$. However, there is no limit (in fact limits are unique, if they exist).
A: There is a difference between limit and limit point.
The concept is defined for sequences and functions but limit point is defined for sets, as mentioned in above answer. A sequence may have limit point but no limit. For example let $\{a_n\}$ is defined as $$1+\frac{1}{n} , (-1)+ \frac{1}{n},... $$
That $a_n=1+\frac{1}{n} $ for odd n's and $a_n=-1+\frac{1}{n} $ for evens.
In this sequence both $1$ and $-1$ are limit point but the sequence is not convergent and there is no limit.
