Q:

Help explain existence of a limit point of a sequence implies infinitely many $m$ where $d(x,x_m)<\epsilon$

I don't understand the phrase "...all but finitely many elements...". What does this mean exactly and how does the conclusion "Infinitely many elementsof the sequence $\{x_k\}$ must also be within $\epsilon$ of $x$"?
I do understand that since the subsequence $\{x_{m(k)}\}$, converges to $x$, there will be some $K$ such that $k\ge K$ implies $|x_{m(k)}-x|<\epsilon$. Is the paragraph saying that there is only a few elements in the subsequence $\{x_{m(k)}\}$ that satisfies this inequality? I don't think this can be true since $\{x_{m(k)}\}$ is an infinite subset of $\{x_k\}$ 

A:

The paragraph is saying that there are only finitely many elements that do not satisfy the inequality $|x_{m(k)} - x| < \varepsilon$, namely, all elements with $k < K$. This then implies that the inequality is satisfied for infinitely many $k$ (for $k \ge K$).

