The Epsilon-Delta Definition of a Limit

The epsilon-delta definition is not a historical curiosity; it is the operating logic beneath continuity, differentiability, and integration, and beneath the definitions of limits of sequences and series via the analogous $\varepsilon$-$N$ formulation. In real analysis it underlies the rigorous proofs of the Intermediate Value Theorem, the Extreme Value Theorem, and the Mean Value Theorem. In topology, the same idea generalizes to open sets and neighborhoods, giving the definition of continuity between arbitrary topological spaces without reference to distance at all. In numerical analysis and scientific computing, error-tolerance reasoning ("if the input error is below this threshold, the output error is below that threshold") is a direct engineering incarnation of the $\varepsilon$-$\delta$ pattern, showing up in stability and convergence analyses of algorithms. In cryptography and theoretical computer science, closely related "for every adversary strategy there is a defense" quantifier structures appear in security definitions and in the theory of approximation algorithms. In physics and engineering, control theory's notion of stability — small perturbations to a system's input or initial conditions produce boundedly small changes in output — is essentially an $\varepsilon$-$\delta$ statement about a dynamical system.

Ask the tutor: Can you walk me through choosing $\delta$ for a limit involving a square root, like $\lim_{x\to 4}\sqrt{x} = 2$?

Read the definition as a game between an adversary and you. The adversary names a tolerance $\varepsilon > 0$ — perhaps a very small one, like $\varepsilon = 0.0001$ — and challenges you to keep $f(x)$ within $\varepsilon$ of $L$. You respond by naming a tolerance $\delta > 0$ on the $x$-side, and you win the round if every $x \neq c$ within $\delta$ of $c$ (and in the domain) produces an output within $\varepsilon$ of $L$. The statement $\lim_{x\to c} f(x) = L$ is true precisely when you have a winning strategy no matter what $\varepsilon$ the adversary picks — that is, you can always find a good $\delta$, however small $\varepsilon$ gets. Two features of the definition deserve emphasis. First, $|x - c|$ and $|f(x)-L|$ are literally distances on the number line — this is exactly the Absolute value as distance idea, so "close" has a numerical, checkable meaning. Second, the condition $0 < |x-c|$ excludes $x = c$ itself: the limit describes the behavior of $f$ near $c$, and is entirely independent of (and does not require) the value $f(c)$, or even that $f$ be defined at $c$ at all. This is what allows limits to detect and repair removable discontinuities, and it is exactly the gap that the definition of Continuity of Functions later closes by additionally demanding $f(c) = L$. The picture below shows the standard "epsilon-band, delta-band" diagram for $f(x) = 2x - 1$ near $c = 3$, where $L = 5$. A horizontal strip of height $2\varepsilon$ around $L$ on the $y$-axis corresponds, through the graph, to a vertical strip of width $2\delta$ around $c$ on the $x$-axis; the definition demands that the graph, restricted to that vertical strip (punctured at $c$), stay inside the horizontal strip.

Notice how narrowing the yellow $\varepsilon$-band forces the blue $\delta$-band to narrow too (here $\delta = \varepsilon/2$ exactly, since the line has slope $2$). This dependency of $\delta$ on $\varepsilon$ is the entire content of "$\delta$ as a function of $\varepsilon$" that every epsilon-delta proof must construct explicitly.

Example 1 (a genuinely nonlinear, asymmetric case). Show directly from the definition that

Here $c = 3$, $L = 3$, and $f(x) = x^2 - 2x$; note $L$ does not equal $c$ and the function is not affine, so we cannot reuse the previous theorem verbatim. Scratch work. We estimate

We want to bound $|x+1|$ by a constant, valid once $x$ is restricted near $3$. Restrict attention to $x$ with $|x - 3| < 1$, i.e. $2 < x < 4$. Then $3 < x + 1 < 5$, so $|x+1| < 5$. Under this restriction,

We want $5|x-3| < \varepsilon$, i.e. $|x-3| < \varepsilon/5$. Combining the two requirements $|x-3|<1$ and $|x-3|<\varepsilon/5$, we should take $\delta$ to be the minimum of these two bounds. Formal proof. Let $\varepsilon > 0$ be given. Define

This is a positive number since both $1 > 0$ and $\varepsilon/5 > 0$. Suppose $0 < |x - 3| < \delta$. Since $\delta \le 1$, we have $|x-3| < 1$, so $2 < x < 4$, hence $3 < x+1 < 5$ and in particular $|x+1| < 5$. Also since $\delta \le \varepsilon/5$, we have $|x-3| < \varepsilon/5$. Therefore

This holds for every $x$ satisfying $0<|x-3|<\delta$, so by definition $\lim_{x\to 3}(x^2-2x) = 3$. $\blacksquare$ Notice the shape of the argument: the "trivial" restriction $\delta \le 1$ is used purely to tame the varying factor $|x+1|$ into a fixed numerical bound, a technique ("bound one factor, then control the other") that recurs constantly in analysis.

Example 2 (a limit not at a "nice" point of the domain, with domain restrictions). Let

Show that $\displaystyle\lim_{x \to 2} f(x) = 1$. Here $c = 2 \in D$ is a limit point of $D$ (indeed every point of $\mathbb{R}$ is a limit point of $D$, since $D$ omits only the single point $1$). Note $c \ne 1$, so $f$ is defined at $c$, though the definition does not require this. Scratch work.

We need a lower bound on $|x-1|$ valid near $x=2$. Restrict to $|x - 2| < \tfrac12$, i.e. $\tfrac32 < x < \tfrac52$; then $\tfrac12 < x - 1 < \tfrac32$, so $|x - 1| > \tfrac12$, giving $\dfrac{1}{|x-1|} < 2$. Under this restriction,

We want $2|x-2| < \varepsilon$, i.e. $|x-2| < \varepsilon/2$. Formal proof. Let $\varepsilon > 0$. Set

Suppose $x \in D$ and $0 < |x-2| < \delta$. Since $\delta \le \tfrac12$, we get $\tfrac32 < x < \tfrac52$, hence $\tfrac12 < x-1 < \tfrac32$; in particular $x-1 \ne 0$ (consistent with $x \in D$) and $|x-1| > \tfrac12$, so $\frac{1}{|x-1|} < 2$. Since also $\delta \le \varepsilon/2$, we have $|x-2| < \varepsilon/2$. Combining,

Since $\varepsilon>0$ was arbitrary and $\delta$ was produced explicitly from it, $\lim_{x\to 2} f(x) = 1$. $\blacksquare$