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Consider the information below.

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My main confusion stems from the link between stating that the limit of $X_n$ is $L$ and that the limit of $X_{n+1}$ is also $L.$

Could someone explain explicitly why this so?

Michael Hardy
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Charlz97
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    Have you tried to work this out using the $\epsilon,\delta$ definition of limits? If so, where did you get stuck? – Lee Mosher Apr 14 '18 at 13:59
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    There is even a more general statement: limit of a subsequence is the same as the limit of the sequence if both limits exist. It can also be easily proven using $\varepsilon,\delta$ – Brian Cheung Apr 14 '18 at 14:04
  • @LeeMosher : I'm surprised that people are suggesting looking at this via any sort of rigorous definition of limits, and that a comment to that effect gets several up-votes. If someone finds it unexpected that these limits would be the same, it seems obvious that they missed something OTHER than a rigorous definition. – Michael Hardy Apr 14 '18 at 14:20
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    @MichaelHardy you're right. Your answer was the "common sense" answer I was looking for - the link I was missing. Though rigorous definitions might help too if I wanted to read on! – Charlz97 Apr 14 '18 at 14:22
  • @MichaelHardy: I'm unsure of the nature of your surprise, although I suspect this is simply a pedagogical debate. There are multiple aspects of this question: intuitive points of view; rigorous points of view. I'm happy to find that the intuitive point of view is what the OP was looking for. When teaching this material I find that students often arrive at their own intuition by working through the rigor, but they are often quite blocked when trying to start the rigor. So my first question to them is usually: have you tried working through the rigor? – Lee Mosher Apr 14 '18 at 15:24
  • @LeeMosher : From the way the question is phrased, it seemed as if the poster found it unexpected that those two limits would be the same. Is that not how it appeared to you? – Michael Hardy Apr 14 '18 at 15:30
  • @LeeMosher : Mathematicians toss around the word "intuitive" in as wrecklessly promiscuous a way as non-mathematicians toss around the word "equation", and with as much thought about what they're saying, or maybe less. – Michael Hardy Apr 14 '18 at 15:31
  • @MichaelHardy I wouldn't say "unexpected" per se...more of "unsure". – Charlz97 Apr 14 '18 at 15:32
  • @MichaelHardy but i do see your point – Charlz97 Apr 14 '18 at 15:33

3 Answers3

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Let's define $Y_n = X_{n+1}$, so we need to show that $\lim_{n \to \infty} Y_n = L$.

Given $\epsilon > 0$, we need to find an $N$ such that if $n \ge N$ then $|Y_n - L| < \epsilon$, equivalently $|X_{n+1} - L| < \epsilon$.

We already know that $\lim_{m \to \infty} X_m = L$ (notice the change of variables, to avoid any clash of variables later), so we may conclude that there is an $M$ such that if $m \ge M$ then $|X_m - L| < \epsilon$.

But now we've found $N$, namely $N = M-1$: if $n \ge N$ then $n+1 \ge M$ and so $|X_{n+1}-L| < \epsilon$ and therefore $|Y_n-L| < \epsilon$.

Lee Mosher
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Fix $\varepsilon > 0$. By definition, if $x_n \to L$, then there exists $N$ such that $$m > N \implies |x_m - L| < \varepsilon.$$ So, using the same $N$, we have, $$n > N \implies n + 1 > N \implies |x_{n+1} - L| < \varepsilon.$$

Theo Bendit
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$$ \begin{array}{c|ccccccl} n & 1 & 2 & 3 & 4 & 5 & \ldots & \\ x_n & x_1 & x_2 & x_3 & x_4 & x_5 & \ldots & \longleftarrow \text{Why does this sequence have} \\ x_{n+1} & x_2 & x_3 & x_ 4 & x_5 & x_6 & \ldots & \longleftarrow \text{the same limit as this sequence?} \end{array} $$ When $n=4$ then $x_n=x_4$ and $x_{n+1}=x_5,$ etc.

These two sequences have the same limit because they are the same sequence except one of them has one more term at the beginning. Changing one term at the beginning does not alter the limit.

Michael Hardy
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