# m $\in \Bbb N$ so that $\lim_{m+n \to \infty} x_n = a$. Prove that $\lim_{n \to \infty} x_{n}=a$

Let $$(x_n)_{n\in \Bbb N}$$ be a sequence in real numbers and a$$\in \Bbb R$$. If exists m $$\in \Bbb N$$ so that $$lim_{m+n \to \infty} x_n = a$$. Prove that $$lim_{n \to \infty} x_{n}=a$$

It is a problem that, I think it involves the definition of a limit as $$\forall \epsilon \gt 0, \exists N \in \Bbb N / |x_n-a|< \epsilon, \forall n\ge N$$ is the same as $$lim_{n \to \infty} x_n=a$$ An answer using this definition would be appreciated, as I have tried to find a moment N so that it becomes true.

• $\lim_{m+n\to\infty}x_n = \lim_{k\to\infty}x_{k-m}$. – amsmath Oct 4 '19 at 18:02

Given $$\lim_{m+n \to \infty}x_{n}=a$$ $$\implies \forall \epsilon >0, \exists N \in \mathbb{N} ~ \text{s.t}~ |x_{n}-a|< \epsilon ~ \text{for} ~ m+n \ge N.$$ $$\text{Choose}~ M=N-m, ~ \text{then} ~ \forall \epsilon >0, \exists M \in \mathbb{N} ~ \text{s.t} ~ |x_{n}-a|< \epsilon ~ \text{for} ~ n \ge M.$$ $$\text{Therefore,} ~ \lim_{n \to \infty}x_{n}=a$$
If you want to use the definition it enough to replace $$N$$ by $$N+m$$ but the idea is that you only care when $$n$$ is large, so it doesn't matter if you "drop" the first $$m$$ terms of the sequence, (the limit won't change).