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Whenever we are supposed to prove that a limit for a given function at a given point doesn't exist, using the $\epsilon-\delta$ criterion, we assume that the limit exists, show that for ONE value of $\epsilon$, we arrive at a contradiction, and thus prove that the limit does not exist.

I am having trouble choosing the value of $\epsilon$. Supposing we know that the graph is discontinuous at the given point, what should we look for to assume the value of $\epsilon$? Or is the value completely arbitrary? Or do we arrive at a contradiction for more than one values of $\epsilon$?

Just to make myself clear, here's an example:

$$f(x)=x\;\;\;\;\;\;\;\;\;\;\;if\;x<1$$ $$\;\;\;\;\;\;\;\;=3-x\;\;\;\;\;\; if\;x\ge1$$

To prove that the limit doesn't exist at $x=1$, the value of $\epsilon$ taken was $\frac12$. I won't go into the details of the solution, but we arrive at a contradiction as $1<1$.

How did we know that $\epsilon=\frac12$ would work?

Tejas
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    Near $x=1$, but to the left of it, $f(x)$ is awfully close to $1$. On the right, it is awfully close to $2$. The limit, if it existed, would have to be simultaneously awfully close to $1$ and to $2$. But the gap between $1$ and $2$ is $1$. We cannot be simultaneously within $(1)(1/2)$ of $1$ and of $2$. It would be even more obvious if we take $\epsilon=1/10$. – André Nicolas Sep 06 '15 at 15:37
  • @Chilango: So for any such problem, any $\epsilon$ between $0$ and the difference between left and right hand limits will work? – Tejas Sep 06 '15 at 15:38
  • There was a little typo, it should have said $0\lt \epsilon\le 1/2$, half the difference. (The $=1/2$ possibility is sensitive to the exact definition of limit.) – André Nicolas Sep 06 '15 at 15:41
  • @AndréNicolas: Ok. So half the difference, always? Without any exception? – Tejas Sep 06 '15 at 15:43
  • Yes there was a typo. I meant: because the difference between the left and right hand limit is 1, so in fact any 0<ϵ<1/2 will do. so any ϵ< half the distance will work. If you draw the picture you wll see why this is so. – Matematleta Sep 06 '15 at 15:49
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    @Tejas: Not quite. In the particular case where the left and right limits are $a$ and $b$, then any number less than $\frac{|a-b|}{2}$ is fine, no exceptions. Exactly $1/2|(a-b)|$ works in this case because we are approaching $1$ from below, and $2$ from above. – André Nicolas Sep 06 '15 at 15:55

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