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I have trouble understanding why, when finding the limit of a recurrently defined sequence, we can assume that $x_n=x_{n+1}$ to find the actual limit.

I figured it's got something to do with the fact that $\lim\limits_{n \to \infty} x_n=\lim\limits_{n \to \infty} x_{n+1}$ and intuitively that makes perfect sense, but it's the theory behind it that evades my understanding (we can't just say that "because $|x_n-x_{n+1}|$ for high enough $n$ is very close to zero, we might aswell make them equal", right?)

Example: In $x_1=0$ , $ x_{n+1}=\frac{1}{1+x_n}$ I'd use the equality $ L=\frac{1}{1+L}$

This question probably isn't very well formulated, but that comes from my lack of understanding of the problem, sorry! (and thanks for the answer :)

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up vote 3 down vote accepted

We don't really assume that $x_n=x_{n+1}$. Instead, suppose that the limit exists, call it $x$. Then $\lim_{n\to\infty}x_n=x$. This, by definition of limit, means that for every $\epsilon>0$ there is a $N\in\mathbb N$ such that for all $n>N$ we have $|x_n-x|<\epsilon$. Intuitively, for every $\epsilon>0$, the sequence is $\epsilon$-close to $x$ from some place on.

Now, we will show that it follows that $\lim_{n\to\infty}x_{n+1}=x$. Let $\epsilon>0$. Then for $n>N$ we have $n+1>N$ and therefore by the previous paragraph, $|x_{n+1}-x|<\epsilon$. So, for all $n>N$, the inequality $|x_{n+1}-x|<\epsilon$ holds. This means that for every $\epsilon>0$, the number $x_{n+1}$ is also $\epsilon$-close to $x$ for all $n>N$, i.e. the limit exists and is equal to $x$.

Similarly, we can show that if $\lim_{n\to\infty}x_{n+1}=x$, then $\lim_{n\to\infty}x_n=x$. To see this, just write out what $\lim_{n\to\infty}x_{n+1}=x$ means in terms of $\epsilon$, $N$ and $n$. Then you will be able to show that $x_n$ is $\epsilon$-close to $x$ for all $n>N+1$.

What all this means is that $\lim_{n\to\infty}x_n=x$ holds precisely if $\lim_{n\to\infty}x_{n+1}=x$ holds, i.e. the conditions are equivalent.

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Perfect, thanks a lot! – Dahn Jahn Nov 11 '12 at 13:09

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