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The following thought came to my mind: Given we have a function $f$, and for arbitrary $\varepsilon>0$, $f(a+\varepsilon)= 100\,000$ while $f(a) = 1$. Why is or isn't this function continuous? I thought that with the epsilon-delta definition, where we would chose delta just bigger then $100 000$, the function could be shown to be continuous. Am I wrong?

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    $\begingroup$ In addition to my answer below, I'd like to point out that your use of $\varepsilon$ and $\delta$ are the opposite of the normal use. Usually, $\varepsilon$ refers to the distance between $y$ values, and $\delta$ refers to the distance between $x$ values. The way you used them in your question is the other way around. $\endgroup$ – Yoni Rozenshein Mar 14 '13 at 10:19
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The function is discontinuous because there exists $\epsilon = 27$ such that for any $\delta > 0$, given $x \in (a, a + \delta), x \ne a$, we have that $|f(x) - f(a)| = 99999 \ge 27$.

Edit: Since your question is about understanding the basic concept of the epsilon-delta definition, I've added a short explanation:

To prove a function is continuous you must show that for any $\epsilon > 0$ (not of your "choosing"), you can choose a $\delta > 0$ (which you may choose after you learn of $\epsilon$, so it may depend on $\epsilon$), such that if $x$ is within $\delta$ of $a$, then $f(x)$ is within $\epsilon$ of $f(a)$.

To prove a function is discontinuous you must show that the above cannot be solved for every $\epsilon$, which means you need to point out just one value of $\epsilon$ for which, no matter how $\delta$ is chosen, there is still some $x$ within $\delta$ of $a$ where $f(x)$ is not within $\epsilon$ of $f(a)$.

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If $f$ is continuous, by the intermediate value theorem, there is some $y>0$ with $f(a+y)=2$. But you've said $f(a+\epsilon)=100~000$ for every $\epsilon>0$, so this is impossible.

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Hint:

Choose $0<r<10000-1$

Then $\forall~\epsilon>0,~|f(a+\epsilon)-f(a)|=10000-1>r.$

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