A basic doubt on a problem on sequence Suppose a sequence of "distinct" elements $\{x_n\}$ converges to $x$ and $x$ is not in the sequence. Then given an $N$ can we find an $\epsilon >0$ s.t. there exists $M >N$ with the property that $\forall n \geq M$ $|x_n-x| <\epsilon$ and the set $\{x_n:n\geq M\} = \{x_n:|x_n-x| < \epsilon\}$ ? This looks intuitively possible, but I am not able to formally write it. 
Actually, I want to find an $\epsilon >0$ such that all the points in the sequence which are $\epsilon$ distance apart from $x$ are some $x_n$ where $n > N$ and $N$ is the given integer. Because I can get rid of the points before $x_N$ by choosing $\epsilon<min(x_i,x); i <N$ it looks like it is possible to find such an $\epsilon$ 
In other words, given an $N$ I want to find an $\epsilon$ such that all the sequence elements starts becoming less than $\epsilon$ distance apart after $N$. Am I making sense now ? 

Ok so firstly the definition of a sequence $\{x_n\}$ converging to $x$ is that for every $\epsilon>0$ there exists $N$ such that for all $n>N$, $|x_n-x|<\epsilon$. I write this as the title mentions confusion regarding the definition.
What you write is not equivalent to the definition, for instance take the sequence $0,1,1/2,1/3,1/4,\dots$ and take $N=2$. Then no matter how small (or big) you pick $\epsilon>0, \{x_n\,:\,|x_n-x|<\epsilon\}$ always contains $0=x_1$, however for each $M>N$, $\{x_n\,:\, n>M\}$ does not contain $0$. So these sets aren't equal.
