Suppose $D$ is a set of real numbers, $f : D \rightarrow R$ with $a \in D$.
Definition: We say that $f$ is continuous at $a$ if and only if $f(a)$ is the limit of $f$ as $x$
approaches $a$, that is, $$ \lim_{x \rightarrow a} f(x) = f(a) $$ or that the
following condition is satisfied:
For any $ \epsilon \gt 0$, there exists a positive number $\delta
(\ \gt 0)$ where if $x \in D$ and $ |x - a| < \delta$ then $|f(x) - f(a)|
< \epsilon$.
We can break the analytic definition for continuity down into three steps needed to verify it being correctly applied.
- "For any $ \epsilon \gt 0$": First consider a positive real
number, $\epsilon$, that is arbitrarily chosen and usually considered
to be small, that is "close to $0$".
The number $\epsilon$ will be a control on the values of the function $f$, ensuring that these values are "close to $f(a)$".
In fact, we want $|f(x) - f(a)| < \epsilon$; that is, we want the values
of $f$, $f(x)$, to be within the interval $(f(a)- \epsilon, f(a) +
\epsilon)$.
- "there exists a positive number $\delta (\ \gt 0)$": Now
based on the chosen $\epsilon$, the task is to find a second
positive number, $\delta$. The number $\delta$ provides a control on the
number $x$ in the domain of $f$ to keep $x$ "close to $a$".
In fact the constraint $|x - a| < \delta$ amounts to having $x$ in the interval $(a-\delta,a+\delta)$.
- "if $x \in D$ and $ |x - a| < \delta$ then $|f(x) - f(a)| < \epsilon$": The number $\delta$ works in the definition for the given $\epsilon$ only if when $x$ is a number with $x \in D$ and $ |x - a| < \delta$ then it can be demonstrated that
$|f(x) - f(a)| < \epsilon$. The demonstration will connect the
constraint on $x$ with the actual function's definition and an analysis
that eventually results in a conclusion connecting to the inequality $|f(x) - f(a)| < \epsilon$.