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I understand the epsilon-delta definition of a limit but I do not understand how that relates to continuity

Definition:

Let $f:X \rightarrow \Bbb R$ be a real function, and suppose that $a \in X$ then f is continuous at a if, given any $\epsilon$ in $\Bbb R^+$, there exists $\delta \in \Bbb R^+$ such that $$\lvert f(x)-f(a)\rvert \lt \epsilon\ \text{whenever}\ x \in X \ \text{and}\ \lvert x-a\rvert \lt \epsilon$$

Would you be able to explain the intuition behind this definition? How does this show that a function is continuous?

Thanks!

  • "How does this show that a function is continuous?": it doesn't. It defines what it means for a function to be continuous (Weierstrass-continuous that is). –  Nov 11 '15 at 18:15
  • What is the intuition behind this definition? – PeakingFriendly Nov 11 '15 at 18:16
  • You can find a lot of articles on the Internet, videos on YouTube and questions & answers on this site tackling this issue. –  Nov 11 '15 at 18:16
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    @PeakingFriendly. Perhaps you may want to try out this applet: http://www.geogebra.org/student/m1129949 – imranfat Nov 11 '15 at 18:17
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    @imranfat Perfect! Exactly the type of thing I was looking for! Thank you! – PeakingFriendly Nov 11 '15 at 18:18
  • http://math.stackexchange.com/questions/1398928/how-is-the-epsilon-delta-definition-of-continuity-equivalent-to-the-following-st this might help to understand – Babai Nov 11 '15 at 18:24
  • Hi PeakingFriendly, welcome to the site. You might have noticed we have a search feature. It can be very helpful. For example, searching epsilon delta intuition brings up a bunch of existing questions that may help you. – Antonio Vargas Nov 11 '15 at 18:30
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    Hi guys, I understand the definition of epsilon-delta to a limit. However the part that I don't understand is how this all relates to continuity – PeakingFriendly Nov 11 '15 at 18:36
  • If you can, get your hands on Spivak's Calculus. He has a long, helpful conversation about this definition and its motivation. – Simon S Nov 11 '15 at 19:01

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The idea is that if you get very close to the point $a$, the values of $f$ will be very close to the values of $f(a)$.

So, if you prescribe a distance from $f(a)$ (the $\varepsilon$) then you can always find a distance (the $\delta)$ such that if $x$ is at distance less than $\delta$ from $a$, then $f(x)$ is at distance less than $\varepsilon$ from $f(a)$.

Martin Argerami
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Well....

Continuous intuitively means as you go from $x$ to $w$ the values of $f(x)$ to $f(w)$ can't have a huge discontinuous jump. That means if you look at all the points really close to $f(w)$, call them $\alpha's$, they all correspond to points, call them x'es, that each x is really close to w and $f(x) = \alpha$ for some $\alpha$.

In other words. $\alpha$ really close to $f(w)$ => there is an x such that $f(x) = \alpha$ and x really close to w.

This means by making x and w really close we can "force" f(x) and f(w) to be really close. So if we want f(x) and f(w) to be "that close" we can do that by making sure x and w are "this close". (Imagine my holding my index finger and thumb together and squinting.)

Well, we've got to put this in real math term....

f(x) and f(w) being "that close" means that we can want them to be such $|f(x) - f(w)| < \epsilon$ where $\epsilon$ is an arbitrarily small number that we want to force them to be within. What we want to show is that we can force this relation by finding another small number, call it $\delta_{\epsilon}$, so that if we force x and w to be $\delta_{\epsilon}$ close to each other than it has to follow that $f(x)$ and $f(w)$ are within $\epsilon$ of each other.

(If we can not force the function to do this, than the function is "jumping" a big distance from f(x) to f(w) while x and w are a really small distance apart. This means the function isn't continuous.)

So we define a function as continuous at w if, for any small $\epsilon > 0$ we can find a $\delta > 0$ such that whenever $|x - w| < \delta$ it has to follow that $|f(x) - f(w)| < \epsilon.$

fleablood
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An intuitive approch is as a garantuee. If FOR ALL $\epsilon >0$ THERE IS SOME $\delta >0$ such that $\mid x-y \mid < \delta $ implies (or "its garantueed") that $\mid f(x)-f(y) \mid < \epsilon$

Note the importance of the "for all" and "for some", these are very important in logical statments

user123124
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I note you may be confusing the definition of a limit and the definition of continuity.

Given a function $f:[a,b]→ℝ$ and $x∈[a,b]$, we say:

  1. The limit of $f(y)$ as $y$ approaches $x$ exists and equals $L$, $\lim_{y→ x} f(y) = L$ if for any $ε>0$ there is a $δ>0$ such that if $0<|y-x|<δ$ then $|f(y) - L|<δ$.
  2. $f$ is continuous at $x$ if for any $ε>0$ there is a $δ>0$ such that if $0<|y-x|<δ$ then $|f(y) - f(x)|<δ$.

In particular a function can have limits at $x$ without being defined at $x$. A function is continuous at $x$ iff $\lim_{y→ x}f(y)$ exists and equals $f(x)$.

Calvin Khor
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I think, if I remember correctly, that this definition was an attempt by 19th century Germans to make precise the notion of a graph that can be drawn without lifting one's pencil.

Turns out that this definition doesn't quite capture that intuition, but it's a good first attempt. I'll leave it to you to ponder why this definition is a good attempt.

Quinn Culver
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