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I've asked the question below before with no answer, but I would like to stress that this time it is not a homework question (and also that I've spent hours trying to come up with a solution).

This is the question:

Let f be a function defined around $x_o$. For every $\epsilon>0$ there's some $\delta>0$ such that if $0<|x-x_0|<\delta$ and $0<|y-x_0|<\delta$ then $|f(x)-f(y)|<\epsilon$.

And what's needed to be proven is that $\lim_{x\to x_0}f(x)$ exists.

I've been told that there are two ways to do so: One is quite easy and requires Cauchy sequences (I haven't learned sequences yet, but I think I'll look it up sometime soon and try to solve it this way).

The second way is a direct way, which I've been told is cumbersome and unrecommended, but since this is the way I tried solving it so far, I am really curious as to how the proof goes and this is the way I'm asking about. I tried applying all kinds of inequalities but with no success.

Even a little hint/direction would be swell. Thank you in advance.

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    @ChristianBlatter I was thinking of using as much of the delta-epsilon limit definition as possible, and also inequalities that involve delta & epsilon; Nevertheless I wouldn't mind listening to any sort of proof. I guess this problem really does require things beyond what I've mentioned above. edit: To put it in another way, I was curious to see the correlation between what was given and what was to be proved.2012-11-11

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The following proof is essentially taken from Zorich, Mathematical analysis I.

Let $\mathcal{B}$ denote the family of all deleted neighborhoods of $x_0$. For each $B \in \mathcal{B}$, define $ m_B = \inf_{x \in B} f(x), \quad M_B=\sup_{x \in B}f(x). $ Since $ m_{B_1} \leq m_{B_1 \cap B_2} \leq M_{B_1 \cap B_2} \leq M_{B_2} $ for any elements $B_1$ and $B_2$ of $\mathcal{B}$, the axiom of completeness of $\mathbb{R}$ implies the existence of some $A \in \mathbb{R}$ that separates the numerical sets $\{m_B\}_B$ and $\{M_B\}_B$ as $B$ ranges over $\mathcal{B}$. By assumption, given $\varepsilon>0$ there exists $B \in \mathcal{B}$ such that $M_B-m_b < \varepsilon$. Then $|f(x)-A| < \varepsilon$ whenever $x \in B$.

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    Since |f(y)-f(x)|<\varepsilon for all $x$ and $y$ close to $x_0$, you deduce that M_B-m_B<\varepsilon for some $B \in \mathcal{B}$.2012-11-12
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O.k.; here is a proof not using sequences:

For each $\epsilon>0$ choose a point $x_\epsilon$ with $0<|x_\epsilon-x_0|<\delta$, where $\delta=\delta(\epsilon)$ is described in the statement. Put $a_\epsilon:=f(x_\epsilon)-\epsilon\ ,\quad b_\epsilon:=f(x_\epsilon)+\epsilon\ .$ Given $\epsilon$, $\epsilon'$ we have $a_\epsilon\leq f(x)\leq b_{\epsilon'}$ for any $x$ with $0<|x-x_0|<\min\{\delta,\delta'\}$. Keeping $\epsilon'$ fixed we see that $a:=\sup_{\epsilon>0}a_\epsilon\leq b_{\epsilon'}\ ,$ and since this is true for all $\epsilon'>0$ we conclude that $b:=\inf_{\epsilon'>0} b_{\epsilon'}\geq a\ .$

Now for each $\epsilon>0$ we have $f(x_\epsilon)-\epsilon=a_\epsilon\leq a\leq b\leq b_\epsilon=f(x_\epsilon)+\epsilon\ .\qquad(*)$ Therefore $0\leq b-a\leq 2\epsilon$, and since this is true for all $\epsilon>0$ we conclude that $a=b$.

Consider now any $x$ with $0<|x-x_0|<\delta$. Then $f(x)$ lies also within the two bounds $f(x_\epsilon)\pm\epsilon$ given in $(*)$. Therefore $|f(x)-a|\leq 2\epsilon$. Since $\epsilon>0$ was arbitrary we have proven that $\lim_{x\to x_0} f(x)=a$.