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For every $\epsilon > 0$ there exists a $\delta > 0$ such that $|x - c| < \delta$ implies $|f(x) - f(c)| < \epsilon$.

Start with $|f(x) - f(c)| < \epsilon$ which gives $|x - c| < \epsilon$. We also know $|x - c| < \delta$ but how can we connect $\epsilon$ and $\delta$?

user2850514
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    Choose $;\delta=\epsilon;$ and you're done...:) – DonAntonio Feb 05 '14 at 14:40
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    To add on to DonAntonio, the key to these sort of proofs is that given a positive $\epsilon$, we need to tell everyone else what is $\delta$ (which can depend on $\epsilon$) to ensure that $|f(x)-f(c)|<\epsilon$ whenever $|x-c|<\delta$. – Kelvin Soh Feb 05 '14 at 14:42
  • Right, and we choose $\epsilon = \delta$ as it is the only possible case where $|x - c| < \epsilon$ and $|x - c| < \delta$? – user2850514 Feb 05 '14 at 14:48
  • Just aside to everybody: it would be less confusing to write (as DonAntonio did) $\delta = \epsilon$ as $\delta$ depends on $\epsilon$ ;-) – Thomas Produit Feb 05 '14 at 14:50
  • @user2850514 not necessarily. Sure, $\delta=\varepsilon$ is the most natural choice for $\delta$, but letting $\delta$ be any positive number less than $\varepsilon$ will still allow you to deduce that $\left|f(x)-f(c)\right|<\varepsilon$ whenever $x$ satisfies $0<|x-c|<\delta$. – Alann Rosas Aug 27 '21 at 20:10

2 Answers2

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choose $\epsilon=\delta$.

That will do it.

tattwamasi amrutam
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    $\delta=\varepsilon$ is a better way of saying it because $\delta=\delta(\varepsilon)$ – Alex Feb 05 '14 at 14:53
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(For future visitors) You can prove it as follows.

For any $x \in \mathbb R$ we have:

$x_n \to x_0 \implies f\left(x_n\right) = x_n \to x_0 = f\left(x_0\right)$

Meaning, if sequence $x_n$ somehow approaches $x_0$, then corresponding sequence $f\left(x_n\right)$ always approaches $f\left(x_0\right)$, which is the definition of continuity. Taking in account that we chose any $x$ from $\mathbb R$ (in the beginning), we've proven that $\lim_{x \to x_0} f\left(x\right) = f\left(x_0\right)$ for any $x \in \mathbb R$. $\blacksquare$

mathgeek
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