Epsilon-delta definition of a limit — formal proofs

WHY do we even need this?

The casual phrase "$f(x)$ gets close to $L$ as $x$ gets close to $a$" is too vague to prove anything:

• "Close" — how close?
• Does $f$ have to reach $L$? (No — limits ignore the point $a$ itself.)
• Could $f(x)$ wobble forever near $a$? (Then no limit exists, but the vague phrase can't catch that.)

The epsilon–delta definition replaces fuzzy words with a game with measurable tolerances, so a limit becomes something you can actually win an argument about.

WHAT is the definition?

Decoding each symbol:

• $|f(x)-L|<\varepsilon$ → the output is inside a band of half-width $\varepsilon$ around $L$.
• $0<|x-a|<\delta$ → the input is inside a band of half-width $\delta$ around $a$. The strict $0<$ excludes $x=a$ (a limit doesn't care what happens at $a$).
• The quantifier order matters: $\varepsilon$ is given to you first (the challenge), then you find $\delta$ (your reply). Your $\delta$ is allowed to depend on $\varepsilon$.

HOW to write a proof (the universal recipe)

Every $\varepsilon$–$\delta$ proof has two halves:

1. Scratchwork (find $\delta$): Start from the goal $|f(x)-L|<\varepsilon$ and work backwards, factoring out $|x-a|$, until you see what $\delta$ must be.
2. Clean proof (forwards): "Let $\varepsilon>0$. Choose $\delta = \dots$. Then assume $0<|x-a|<\delta$ and derive $|f(x)-L|<\varepsilon$."

Example 1 — a linear function (the cleanest case)

Claim: $\displaystyle\lim_{x\to 3}(2x+1)=7.$

Scratchwork. Here $f(x)-L = (2x+1)-7 = 2x-6 = 2(x-3)$. So $|f(x)-L| = 2|x-3|.$ Why this step? We factored out $|x-3|=|x-a|$, the controllable input distance. To make $2|x-3|<\varepsilon$ we need $|x-3|<\varepsilon/2$. So take $\delta=\varepsilon/2$.

Proof. Let $\varepsilon>0$. Choose $\delta=\dfrac{\varepsilon}{2}$. Suppose $0<|x-3|<\delta$. Then $|(2x+1)-7| = 2|x-3| < 2\delta = 2\cdot\frac{\varepsilon}{2}=\varepsilon.\qquad\blacksquare$ Why this works: the factor was the constant $2$, so $\delta$ is a clean fraction of $\varepsilon$.

Example 2 — a quadratic (factor must be bounded first)

Claim: $\displaystyle\lim_{x\to 2}x^2=4.$

Scratchwork. $|x^2-4| = |x-2|\,|x+2|$. The factor $|x+2|$ is not constant — it depends on $x$. Why is this a problem? We can't pick $\delta=\varepsilon/|x+2|$ because $\delta$ must be a fixed number, not depend on $x$.

Fix — pre-restrict $x$. First impose a preliminary restriction $|x-2|<1$. Then $1<x<3$, so $|x+2| = x+2 < 5.$ Why this step? Once $x$ is near $2$, the factor $|x+2|$ can't exceed $5$ — we bounded the troublesome factor by a constant. Now $|x^2-4| = |x-2|\,|x+2| < 5\,|x-2|.$ To get $5|x-2|<\varepsilon$ we need $|x-2|<\varepsilon/5$. We need both $|x-2|<1$ and $|x-2|<\varepsilon/5$, so take the smaller: $\boxed{\delta = \min\!\left(1,\ \tfrac{\varepsilon}{5}\right).}$

Proof. Let $\varepsilon>0$. Choose $\delta=\min(1,\varepsilon/5)$. Suppose $0<|x-2|<\delta$. Since $\delta\le1$, we have $|x+2|<5$. Since $\delta\le\varepsilon/5$, $|x^2-4| = |x-2|\,|x+2| < 5|x-2| < 5\cdot\frac{\varepsilon}{5}=\varepsilon.\qquad\blacksquare$

Example 3 — proving a limit is FALSE (negation)

Claim: $\displaystyle\lim_{x\to 0} \frac{|x|}{x}$ does not exist.

To negate the definition, swap the quantifiers:

Here $f(x)=1$ for $x>0$ and $f(x)=-1$ for $x<0$. Suppose someone claims the limit is $L$. Take $\varepsilon=1$. For any $\delta>0$, the interval $(-\delta,\delta)$ contains both a point $x_+>0$ (where $f=1$) and $x_-<0$ (where $f=-1$). These differ by $2$, so they can't both be within $1$ of $L$. Hence no $L$ works. Why $\varepsilon=1$? It's small enough that the $\varepsilon$-band can't straddle both $+1$ and $-1$ simultaneously.

Active recall

Connections

• Limit Laws (sum, product, quotient) — each law is proved with $\varepsilon$–$\delta$.
• Continuity — $f$ continuous at $a$ means $\lim_{x\to a}f(x)=f(a)$ (same definition, with $L=f(a)$).
• One-sided limits — replace $0<|x-a|<\delta$ with $a<x<a+\delta$ or $a-\delta<x<a$.
• Definition of the derivative — a derivative is an $\varepsilon$–$\delta$ limit of difference quotients.
• Uniform continuity — same game, but $\delta$ must work for all $a$ at once.
• Sequences and their limits — discrete analogue with $N$ instead of $\delta$.

Concept Map

Hinglish (regional understanding)