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Am I allowed to set an upper bound to delta in the epsilon-delta definition of limit? Why would it still be equivalent to the original definition ? For example, if the function is defined for all $x$ in $\Bbb R$ :

$f : D(f) = \Bbb R \rightarrow \Bbb R$.

Let $\epsilon > 0, \exists \delta > 0 \text{ s.t. } \forall x : 0 < |x-x_o| < \delta < 3 \Rightarrow |f(x) - l| < \epsilon$

$\Leftarrow\text{?}\Rightarrow$

Let $\epsilon > 0, \exists \delta > 0 \text{ s.t. } \forall x : 0 < |x-x_o| < \delta \Rightarrow |f(x) - l| < \epsilon$.

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    $\begingroup$ Lets suppose you found an $\; \delta \; $ then the definition still holds for any $\delta^* \le \delta$. Hence, you can choose your $\; \delta^* \le 3$ $\endgroup$ – XPenguen Dec 26 '15 at 20:16
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    $\begingroup$ If you find an epsilon which beats a small delta, you've found an epsilon which beats a big one. $\endgroup$ – Mark Bennet Dec 26 '15 at 20:16
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$$ \forall\varepsilon > 0\ \exists \delta > 0\ \forall x \Big( 0 < |x-x_o| < \delta < 3 \Rightarrow |f(x) - \ell| < \varepsilon\Big) $$

I would phrase this differently:

$$ \forall\varepsilon > 0\ \exists \delta \in(0,3)\ \forall x \Big( 0 < |x-x_o| < \delta \Rightarrow |f(x) - \ell| < \varepsilon\Big) $$

Now the question is whether that is equivalent to this:

$$ \forall\varepsilon > 0\ \exists \delta>0\ \forall x \Big( 0 < |x-x_o| < \delta \Rightarrow |f(x) - \ell| < \varepsilon\Big) $$

If it is true that $\forall\varepsilon>0\ \exists\delta\in(0,3)\ \cdots\cdots$ then it is true that $\forall\varepsilon>0\ \exists\delta>0\ \cdots\cdots$, simply because every number in $(0,3)$ is $>0$.

Now the question is whether the converse holds. If it is true that $\forall\varepsilon>0\ \exists\delta>0\ \cdots\cdots$, does it necessarily follow that $\forall\varepsilon>0\ \exists\delta\in(0,3)\ \cdots\cdots\,{}$? Here the answer in general is “no”. I.e. there are some things you could put in place of $\text{“}\cdots\cdots\text{''}$ for which the first statement would be true and the second false. However, in the definition of "limit" the thing in place of $\text{“}\cdots\cdots\text{''}$ is “if a certain thing is $<\delta$, then a certain thing follows.” Given a certain $\varepsilon>0$ suppose we know there exists $\delta>0$ such that if a certain thing is less than $\delta$ then a certain conclusion follows. If the given value of $\delta$ is small enough, then the minimum of that value of $\delta$ and $2.9$ is small enough, because everything less than $\min\{\delta,2.9\}$ is less than $\delta$. So when the statement in place of $\text{“}\cdots\cdots\text{''}$ has the form “if a certain thing is $<\delta$, then a certain thing follows.”, then it is true that if $\forall\varepsilon>0\ \exists\delta>0\ \cdots\cdots$ then $\forall\varepsilon>0\ \exists\delta\in(0,3)\ \cdots\cdots$.

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Yes, of course. If the first condition is satisfied, then clearly the second is too. Conversely, if the second condition is satisfied, then one can replace $\delta$ by $\min(\delta, 3)$ and so the first condition is satisfied.

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Yes, as you have to exhibit an $\delta$ that works for every $\epsilon$, then you can be as artificial as you want if that works for you. Always in existence proofs the same rule applies, construct all you need and show that it does work.

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Yes you can. The idea is that some proposition $P(x,\epsilon)$ is true for all $x$ that are sufficiently close to $x_0$. So if it holds for all $x\in (-d+x_0,d+x_0)$ then it holds for all $x\in (-d'+x_0,d'+x_0)$ whenever $0<d'<d$. Sometimes it is useful to know how large $d$ can be, for a given $\epsilon.$

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