Subsequence and sequences 
Suppose $(x_n)_{n=1}^\infty$ is a sequence in $\mathbb R$, and that $L_k$ are real numbers with $\lim_{k\to\infty}L_k=L$. If for each $k\geq 1$, there is a subsequence of $(x_n)_{n=1}^\infty$ converging to $L_k$, show that some subsequence converges to $L$. HINT: Find an increasing sequences $n_k$ such that $|x_{n_k}-L|<1/k$.

Can someone tell me what $L_k$ actually are? Is that a sequence or is it something else? I thought it was a sequence first, but the following sentence suggests it isn't (you can't converges to a sequence)
 A: If you are uncomfortable with the multiple indexing, maybe try an indirect approach: Assume there is no such subsequence, hence there is an $\epsilon>0$ such that only finitely many terms of the seuence are between $L-\epsilon$ and $L+\epsilon$. Now make use of $L_k\to L$ to find an $L_k$ closer than $\frac\epsilon2$ to $L$. Then make use of the subsequence of $(x_k)$ that converges to $L_k$ ...
A: $L_k$ seem to be terms of a sequence $\left(L_k\right)_{k=1}^\infty$ with limit $L$. The question is basically asking you to prove that if the sequence $\left(x_n\right)_{n=1}^\infty$ has convergent subsequences to each term $L_k$ then it also has a subsequence which converges to $L$. 
A: 
"...$L_k$ are real numbers with $\lim_{k\to\infty}L_k=L$."

So for each $k$, $L_k\in\mathbb R$. $L_k$ is the $k$th term of the sequence $(L_k)_{k=1}^\infty$.
A: $(L_k)_{k=1}^\infty$ is a sequence that converges to the limit $L$.   So the question supposes that for each $k$ there is a corresponding subsequence of $(x_n)_{n=1}^\infty$ which has limit each $L_k$.
A: The set of all subsequential limits of a sequence is closed.Since ${L_k}$ is a sequence in that closed set and it converges to $L$,so $L$ must be in that set.
