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Continuity is defined in my book Basic Analysis (by Lebl pg 86) like this:

Let $S \subset \mathbb R$, $f : S\rightarrow \mathbb R$ be a function, and let $c \in S$ be a number. We say that $f$ is continuous at c if for every $\epsilon > 0$ there is a $\delta > 0$ such that whenever $x \in S$ and $|x − c| < \delta$ , then $| f (x) − f (c)| < \epsilon$.

When $f : S \rightarrow \mathbb R$ is continuous at all $c \in S$, then we simply say that f is a continuous function.

But how does this definition ensure continuity?

If, for $c=0,\epsilon = 10$ there exists $\delta = 5$ for which $|x-0| = |x| < 5, |f(x)-f(c)| < 10$ , and for another $\epsilon=30$, there exists $\delta = 1$ for which $|x| < 1, |f(x)-f(c)| < 30$, wouldnt this increase chances of a function that wildly jumping between values, and perhaps indicating a noncontinous function? I just picked these values out of a hat of course, but the definition does not say anything about how values are structured, if they become smaller while approaching $c$ for example.

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    My b$a$d.. :) The function I mentioned should $b$e graph #2, how's that look? And I was just going to ask if graph #1 (even if it belonged to another function) would be continuous, Didier answered that above. Thanks guys.2012-01-17

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Think of $\epsilon$ as a margin of error. If we are close enough to $c$, within $\delta$, the function value $f(x)$ is within the margin of error $\epsilon$. As Didier says it is important to let $\epsilon$ be arbitrarily small. For each margin of error $\epsilon$ there is a closeness $\delta$, such that all points within a distance $\delta$ of $c$, all the values of these points are within the margin of error $\epsilon$. Thus $f(x)$ is close to $f(c)$ if $x$ is close to $c$. This is the idea of continuity. A nice picture is at the wikipedia page.

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    @Didier: Thanks2012-01-17
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The definition says that no matter how small a ‘target’ you set around $f(c)$, you can guarantee that $f(x)$ is inside that target whenever $x$ is sufficiently close to $c$. Do you want to guarantee that $f(x)$ is closer than $0.001$ to $f(c)$? Take $\epsilon=0.001$, and the definition of continuity tells you that there’s some positive $\delta$, possibly very small, such that $f(x)$ is less than $0.001$ away from $f(c)$ provided that $x$ is less than $\delta$ away from $c$. Do you want to be sure that $f(x)$ is less than $10^{-10}$ away from $f(c)$? Apply the definition with $\epsilon=10^{-10}$: there is some $\delta$ such that as long as you choose $x$ between $c-\delta$ and $c+\delta$, $f(x)$ will be less than $10^{-10}$ away from $f(c)$. In short, it’s really the small values of $\epsilon$ that you should be interested in.

This doesn’t prevent $f$ from oscillating more and more frequently as $x$ approaches $c$, but it does force the magnitude of the oscillations to approach $0$. Take a look, for instance, at the graph of the function

$f(x)=\begin{cases} x\sin\frac1x,&\text{if }x\ne 0\\ 0,&\text{if }x=0\;, \end{cases}$

which is shown here. As you can see, they oscillate infinitely often as $x$ approaches $0$, but because the amplitude of the oscillations approaches $0$, both functions are continuous at $0$. In fact, no matter what $\epsilon>0$ you pick as a ‘target’, the same number will work for $\delta$: it’s pretty straightforward to check that if $|x|<\epsilon$, then $|f(x)<\epsilon$ as well, since $|\sin(1/x)|\le 1$ for all $x$.