For any $ \epsilon \gt 0$, there exists a positive number $\delta (\ \gt 0)$ where if $x \in D $ and $0 < |x - a| < \delta$ then $|f(x) - L| < \epsilon$.

We can break down the analytic definition for a number $L$ to be a limit for $f$ as $x$ approaches $a$ into

**"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 $L$".

In fact, we want $|f(x) - L| < \epsilon$; that is, we want the values of $f$, $f(x)$, to be within the interval $(L- \epsilon, L + \epsilon)$.

**"there exists a positive number $\delta (\ \gt 0)$":**Now based on the chosen $\epsilon$, the task is to find a second positive number, $\delta \gt 0 $. 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 $0 < |x - a| < \delta$ amounts to having $x \ne a$ and $x$ in the interval $(a-\delta,a+\delta)$.**"if****$x \in D $ and $ |x - a| < \delta$ then $|f(x) - L| < \epsilon$****":**The number $\delta$ works in the definition for the given $\epsilon$**only if**when $x$ is a number with $x \in D $ and $ 0 < |x - a| < \delta$ then it can be demonstrated that $|f(x) - L| < \epsilon$. 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 $|f(x) - L| < \epsilon$.