# Epsilon-delta definition

## Main Question or Discussion Point

please i am new to math. I donot know exact meanings of epsilon-delta definition. i dont comprehend it. Would Anybody help me. thanks in advance

## Answers and Replies

vanhees71
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What particularly are you not understanding? Do you mean the $\epsilon$-$\delta$ definition of continuity of a function at a point?

In this case it reads:

A function $f:D \rightarrow \mathbb{R}$ is continuous at $x_0 \in D$ iff for any $\epsilon>0$ there exists a $\delta>0$ such that for all $x \in D$ with $|x-x_0|<\delta$
$$|f(x)-f(x_0)|<\epsilon.$$
This just says in a formal way that the graph of the function doesn't jump at $(x_0,f(x_0))$.

pwsnafu
$0<|x-x_0|<\delta$ actually.

WannabeNewton
jbunniii
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$0<|x-x_0|<\delta$ actually.
There's no reason to exclude the point $x = x_0$. We trivially have $|f(x_0) - f(x_0)| < \epsilon$ for any $\epsilon$.

arildno
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There's no reason to exclude the point $x = x_0$. We trivially have $|f(x_0) - f(x_0)| < \epsilon$ for any $\epsilon$.
It sure is a reason.
Otherwise, discontinuous functions would be deprived of limit values at the point of discontinuity. Thus, the limit concept would be conflated with the continuity concept.

jbunniii
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It sure is a reason.
Otherwise, discontinuous functions would be deprived of limit values at the point of discontinuity. Thus, the limit concept would be conflated with the continuity concept.
But vanhees71 was giving the definition of continuity at $x_0$, not the definition of the existence of a limit at $x_0$. If the function is to be continuous at $x_0$, then it must be defined at $x_0$ and have the correct value!

arildno
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But vanhees71 was giving the definition of continuity at $x_0$, not the definition of the existence of a limit at $x_0$. If the function is to be continuous at $x_0$, then it must be defined at $x_0$ and have the correct value!
Hmm..no read it again.
What he posted was the definition in terms of the LIMIT concept in which L=f(x_0).
In particular, he writes d>0

jbunniii
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Hmm..no read it again.
What he posted was the definition in terms of the LIMIT concept in which L=f(x_0).
In particular, he writes d>0
One more remark and then I'll shut up. :tongue: Here's a discussion from Spivak's Calculus (chapter 6, before Theorem 2) which may clarify:

Spivak said:
If we translate the equation $\lim_{x \rightarrow a} f(x) = f(a)$ according to the definition of limits, we obtain:

For every $\epsilon > 0$ there is a $\delta > 0$ such that, for all $x$, if $0 < |x - a| < \delta$, then $|f(x) - f(a)| < \epsilon$.

But in this case, where the limit is $f(a)$, the phrase $0 < |x-a| < \delta$ may be changed to the simpler condition $|x-a| < \delta$, since if $x = a$ it is certainly true that $|f(x) - f(a)| < \epsilon$.

arildno