You met the ε–δ definition of a limit in Calculus 1. It is the same idea as the ε–N definition, with "far enough out in the sequence" replaced by "close enough to $a$." This time we treat it as a working tool: given $\varepsilon$, find a $\delta$ and check it.
Key idea
$\lim_{x \to a} f(x) = L$ means: for every $\varepsilon > 0$ there is a $\delta > 0$ such that $|f(x) - L| < \varepsilon$ whenever $0 < |x - a| < \delta$ (and $x$ is in the domain). If a $\delta$ works, so does every smaller positive $\delta$. The condition $0 < |x - a|$ means the value $f(a)$, if any, plays no role.
Why it works
The $\varepsilon$ is a horizontal band around $L$; $\delta$ is a vertical window around $a$. The definition asks that the graph over the window (with the point $x = a$ removed) stays inside the band. Narrowing the window can only remove points, so it never breaks the condition, which is why any smaller $\delta$ works. The largest $\delta$ that works is set by the nearest place where the graph leaves the band.
Worked examples
Worked example: A line
For $f(x) = 3x - 1$, $\lim_{x \to 2} f(x) = 5$. Find the largest $\delta$ for $\varepsilon = 0.06$.
- $|f(x) - 5| = |3x - 6| = 3|x - 2|$.
- $3|x - 2| < 0.06$ exactly when $|x - 2| < 0.02$.
Answer: $\delta = 0.02$; any $\delta$ with $0 < \delta \le 0.02$ works.
Worked example: A removable hole
$f(x) = \frac{x^{2} - 9}{x - 3}$ for $x \ne 3$. Find a $\delta$ for $\varepsilon = 0.01$ at $a = 3$, where the limit is $6$.
- For $x \ne 3$, $f(x) = x + 3$, so $|f(x) - 6| = |x - 3|$.
- This is less than $0.01$ exactly when $|x - 3| < 0.01$.
Answer: $\delta = 0.01$. The point $x = 3$ itself is excluded by $0 < |x - 3|$, which is lucky: $f(3)$ is undefined.
Worked example: The square root, where the sides differ
For $f(x) = \sqrt{x}$ at $a = 9$ (limit $3$) and $\varepsilon = 0.1$, find the largest $\delta$.
- $|\sqrt{x} - 3| < 0.1$ means $2.9 < \sqrt{x} < 3.1$, that is, $8.41 < x < 9.61$.
- The distances from $9$ to the ends are $0.59$ (left) and $0.61$ (right).
- A symmetric window must fit inside $(8.41, 9.61)$, so it can reach only $0.59$.
Answer: $\delta = 0.59$, set by the steeper side.
Common mistake
"The δ." There is no single correct $\delta$: the definition asks for some $\delta$, and if one works, all smaller positive ones do too. A second slip is a $\delta$ that depends on $x$, like "$\delta = \frac{\varepsilon}{|x + 2|}$": the window must be chosen before you look at individual points. Check a proposed $\delta$ by testing points near both ends of the window, on both sides of $a$.