Show that limit of sequence exists with the help of subsequence. Let $(x_n)^\infty_{n=1}$ be a sequence of real numbers. Suppose that there is a real number $L$ such that
$$L=\lim_{n\to\infty}x_{3n-1}=\lim_{n\to\infty}x_{3n+1}=\lim_{n\to\infty}x_{3n}$$
Show that $\lim_{n\to\infty}x_n$ exists and equals to $L$.

I dont even have a clue how to start.
I know every bounded sequence of real numbers has a convergent subsequence.
so is it if I can show $\lim_{n\to\infty}x_{3n-1}$ , $\lim_{n\to\infty}x_{3n+1}$ , $\lim_{n\to\infty}x_{3n}$ are convergent $\rightarrow$ I can show $x_n$ is bounded $\rightarrow$ $\lim_{n\to\infty}x_n$ exists?????
As I already know $x_{3n-1}$, $x_{3n+1}$ and $x_{3n}$ converge, is that enough to show $x_n$ converges too????
Thank you!!
 A: Your main problem is that you’ve not yet correctly absorbed the definition of limit. The statement that $\lim_{x\to\infty}x_n=L$ does not mean that $|x_n-L|<\epsilon$. This actually isn’t even a meaningful statement. What is $\epsilon$? For what values of $n$?
The statement that $\lim_{x\to\infty}x_n=L$ means that for each $\epsilon>0$ there is some natural number $m_\epsilon$ such that $|x_n-L|<\epsilon$ whenever $n\ge m_\epsilon$. To show that $\lim_{x\to\infty}x_n=L$, you must show that no matter how small a positive number $\epsilon$ I choose, you can find some $m_\epsilon\in\Bbb N$ such that all of the terms $x_{m_\epsilon},x_{m_\epsilon+1},x_{m_\epsilon+2},\dots$ are less between $L-\epsilon$ and $L+\epsilon$. We sometimes say that by taking a tail of the sequence starting far enough out, we can ensure that the whose tail is within $\epsilon$ units of $L$.
In this problem you’re given that the sequence $\langle x_{3n}:n\in\Bbb N\rangle$ converges to $L$, i.e., that $\lim_{x\to\infty}x_{3n}=L$. According to the definition above, that means that for each $\epsilon>0$ there is an $m_\epsilon^{(0)}\in\Bbb N$ such that $|x_{3n}-L|<\epsilon$ whenever $n\ge m_\epsilon^{(0)}$. (The superscript is not an exponent; it’s just a label.)
You’re also given that $\lim_{x\to\infty}x_{3n+1}=L$, which means that for each $\epsilon>0$ there is some $m_\epsilon^{(1)}\in\Bbb N$ such that $|x_{3n+1}-L|<\epsilon$ whenever $n\ge m_\epsilon^{(1)}$.
What happens if you take $n\ge\max\{m_\epsilon^{(0)},m_\epsilon^{(1)}\}$? Then $|x_{3n}-L|<\epsilon$ and $|x_{3n+1}-L|<\epsilon$. Every integer is of one of the forms $3n$, $3n+1$, and $3n+2$. If you could also get $|x_{3n+2}-L|<\epsilon$ for all $n\ge\text{ something}$, you could get $|x_n-L|<\epsilon$ for a whole tail of the sequence $\langle x_n:n\in\Bbb N\rangle$. Look again at the last two paragraphs, and see if you can see how to ensure that $|x_{3n}-L|<\epsilon$, $|x_{3n+1}-L|<\epsilon$, and $|x_{3n_2}-L|<\epsilon$ for all $n$ from some point on.
