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I'm stuck with this proof I just can't get my head around and I would really appreciate any sort of help. The problem is as follows:

Problem: Let $f$ be a function defined in the neighborhood of the point $x=x_{0}$. It is given that for all $\epsilon>0$ there exists a number $\delta>0$ such that for all $x, y$, if $0<|x-x_{0}|<\delta$ and $0<|y-x_{0}|<\delta$, then $|f(x)-f(y)|<\epsilon$. Show that the limit of $f$ exists at $x_{0}$.


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Are you familiar with Cauchy sequences? – Harald Hanche-Olsen Nov 10 '12 at 9:51

$\def\abs#1{\left|#1\right|}$Let $x_n \to x_0$, $x_n \ne x_0$ a sequence. Given $\epsilon > 0$, choose a $\delta$ by assumption. Let $N \in \mathbb N$ be such, that for $n \ge M$ we have $\abs{x_n - x_0}<\delta$ for $n \ge N$. Then by definition of $\delta$, we have $\abs{f(x_n) - f(x_m)} < \epsilon$ for $n,m \ge N$. So $\bigl(f(x_n)\bigr)$ is a Cauchy-sequence. As $\mathbb R$ is complete, there is an $\alpha \in \mathbb R$ with $f(x_n) \to \alpha$. So for all sequences $x_n \to x_0$, $x_n \ne x_0$ we have that $\lim_{n \to\infty} f(x_n)$ exists, therefore it has to be the same $\alpha$ for all sequences, we can see this by 'mixinig'. Suppose there are $x_n, y_n \to x_0$ with $f(x_n) \to \alpha$, $f(y_n) \to \beta$. Then $(z_n) := (x_1, y_1, x_2, y_2, \ldots)$ has $z_n \to x_0$, but $f(z_n)$ doesn't converge, contradiction.

That is for all $x_n \to x_0$, $x_n \ne x_0$ we have that $\alpha = \lim_{n\to \infty} f(x_n)$ exists. Hence $\alpha = \lim_{x\to x_0, x\ne x_0} f(x)$ exists.

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Thnx, but I must prove this without Cauchy sequences. – Gal Nov 11 '12 at 21:01

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