# limit superior and inferior 2

Let $$x_n$$ be a sequence. If $$M =\limsup x_n$$ then there is some subsequence $$x_k$$ with $$M = \lim x_k$$. Then $$\left|x_k - M \right|< \epsilon \iff M - \epsilon < x_k < M+\epsilon \quad \forall k>K,$$

Then how are we saying that $$x_k>M-\epsilon$$ for infinitely many $$k$$ but $$x_k < M+\epsilon$$ for $$k>K$$?

You can have that $$x_k > M-\epsilon$$ and $$x_k < M+\epsilon$$ at the same time. For example, let $$M=2$$, $$\epsilon=0.1$$ and $$x_k = 1.99$$. These values satisfy the inequality $$2-0.1 < 1.99 < 2+0.1$$. When $$\epsilon$$ starts getting smaller and smaller, you just need $$x_k$$ to get closer and closer to $$M$$ (this may require that $$k$$ increases too).
Not sure what your question is exactly. In your case, there is an interval $$(M-\epsilon,M+\epsilon)$$, where $$\epsilon$$ depends on the specific $$K$$, such that if you take any $$k>K$$, then all members of the subsequence $$\left\{x_{n_k}\right\}$$ will fall inside this interval.
It is expected that as $$K \to \infty$$ you will have $$\epsilon \to 0$$.