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 $|f(x) - f(a)| < \epsilon$.