Worked examples — Epsilon-delta definition of a limit — formal proofs

If any symbol below feels unearned, revisit Epsilon-delta definition of a limit — formal proofs first — but I re-explain each tool the moment it appears.

---

The scenario matrix

An $\epsilon$–$\delta$ problem is fully described by what kind of factor appears when you rewrite $|f(x)-L|$ as $(\text{factor})\cdot |x-a|$, plus what weirdness lives near $a$. Here is every cell:

#	Cell (what makes it different)	The core difficulty	Worked in
A	Constant function	No $	x-a
B	Linear, factor is a constant	Clean $\delta=\varepsilon/	M
C	Quadratic/polynomial, factor varies	Must bound the factor first	Ex 3
D	Quotient — denominator can shrink to $0$	Bound the denominator away from 0	Ex 4
E	Square root — factor lives under a root	Rationalise to expose $	x-a
F	Limit is FALSE (oscillation)	Negate the definition	Ex 6
G	One-sided limit / value at $a$ irrelevant	Only one side of $a$ counts	Ex 7
H	Word problem (real tolerance)	Translate "within" into $\epsilon ,\delta$	Ex 8
I	Exam twist: $\delta$ depending on $\epsilon$ and the point	Uniform vs pointwise	Ex 9

Notation reminder (each symbol earned here so nothing below is a surprise):

Read the picture below before any algebra. Figure s01 is the mental image behind every example on this page: the horizontal green strip is the $\delta$-band around the target input $a$; the horizontal-you-look-up-into blue strip is the $\epsilon$-band around the claimed limit $L$. A limit exists when — no matter how thin the blue band the adversary paints — you can always shrink the green band so that the whole curve, wherever it crosses the green strip, stays trapped inside the blue strip. Keep this image in mind: every proof below is just computing how narrow the green strip must be.

---

Example 1 — Constant function (Cell A)

Forecast: Guess before reading — what will $\delta$ be? (A trick: does $\delta$ even need to depend on $\epsilon$ here?)

1. Compute the output distance. $|f(x)-L|=|7-7|=0.$ Why this step? Every proof starts by writing the vertical gap. Here $f$ never moves, so the gap is $0$ for every $x$.
2. Compare to $\epsilon$. Since $0<\epsilon$ always, $|f(x)-L|=0<\epsilon$ automatically. Why this step? We need $|f(x)-L|<\epsilon$. It holds no matter what — the input distance never entered.
3. Choose $\delta$. Any positive number works. Take $\delta =1$. Why this step? The game demands some $\delta >0$; the value is irrelevant because step 2 never used $\delta$.

Proof. Let $\epsilon >0$. Choose $\delta =1$. If $0<|x-5|<1$, then $|7-7|=0<\epsilon$. $■$

Look at figure s01: picture a flat orange line instead of a sloped one. A flat line has output distance $0$ everywhere, so it sits inside any blue band no matter how thin — that's why $\delta$ is free. This is the degenerate corner of the whole matrix.

Verify: $|7-7|=0$, and $0<\epsilon$ for every allowed $\epsilon >0$. ✓

---

Example 2 — Linear, constant factor (Cell B)

Forecast: The slope is $4$. Guess the form of $\delta$ before reading.

1. Output distance. $|(4x+3)-(-1)|=|4x+4|=4|x+1|.$ Why this step? We rewrite the gap as $(\text{factor})\cdot |x-a|$ with $a=-1$, so $|x-a|=|x+1|$. This is the pivot move (parent note): turn an output bound into an input bound.
2. Identify the factor. The factor is the constant $4$ — it does not depend on $x$. Why this step? A constant factor means no pre-restriction is needed; $\delta$ is a clean fraction of $\epsilon$.
3. Solve for $\delta$. We want $4|x+1|<\epsilon$, i.e. $|x+1|<\epsilon /4$. So take $\delta =\epsilon /4$. Why this step? Choosing $\delta =\epsilon /4$ makes the last inequality automatic.

Proof. Let $\epsilon >0$. Choose $\delta =\epsilon /4$. If $0<|x+1|<\delta$, then $|(4x+3)-(-1)|=4|x+1|<4\delta =4\cdot \frac{\epsilon }{4}=\epsilon .■$

Look at figure s02: the orange line is steep (slope $4$). Watch how a blue $\epsilon$-band of height $0.4$ only lets through a green $\delta$-band of width $0.1$ — the steeper the line, the narrower the input window it forces. That ratio $\frac{\text{width}}{\text{height}}=\frac{1}{4}$ is exactly $\delta /\epsilon$; the slope is the amplifier.

Verify: Pick $\epsilon =0.1\Rightarrow \delta =0.025$. Take $x=-1+0.02$ (inside $\delta$). Then $4x+3=4(-0.98)+3=-0.92$, and $|-0.92-(-1)|=0.08<0.1$. ✓

---

Example 3 — Quadratic, variable factor (Cell C)

Forecast: The factor will not be constant. Predict which extra step appears.

1. Factor the difference. $|x^2-9|=|x-3||x+3|.$ Why this step? $a=3$, so we want $|x-3|$ out front; the leftover factor is $|x+3|$.
2. Notice the danger. $|x+3|$ grows with $x$ — not a constant. We cannot set $\delta =\epsilon /|x+3|$ (parent note's steel-man mistake). Why this step? $\delta$ must be a fixed number chosen before $x$; it may not contain $x$.
3. Pre-restrict. Impose $|x-3|<1$. Then $2<x<4$, so $5<x+3<7$, giving $|x+3|<7$. Why this step? Trapping $x$ near $3$ caps the variable factor by the constant $7$.
4. Combine both conditions. From $|x^2-9|<7|x-3|$ we want $7|x-3|<\epsilon$, i.e. $|x-3|<\epsilon /7$. We need $|x-3|<1$ and $|x-3|<\epsilon /7$, so $\delta =\min \left(1,\frac{\epsilon }{7}\right).$ Why this step? The $\min$ guarantees both the cap (step 3) and the $\epsilon$-condition hold at once.

Proof. Let $\epsilon >0$. Choose $\delta =\min (1,\epsilon /7)$. If $0<|x-3|<\delta$: since $\delta \le 1$, $|x+3|<7$; since $\delta \le \epsilon /7$, $|x^2-9|=|x-3||x+3|<7|x-3|<7\cdot \frac{\epsilon }{7}=\epsilon .■$

Verify: $\epsilon =0.7\Rightarrow \delta =\min (1,0.1)=0.1$. Take $x=3.1$: $x^2=9.61$, gap $=0.61<0.7$. ✓ And at the boundary $x=3.1$, $|x+3|=6.1<7$, confirming the cap.

---

Example 4 — Quotient, denominator near zero (Cell D)

Forecast: The new danger is division. What could make $\left|\frac{1}{x}-\frac{1}{2}\right|$ blow up? Guess where you must forbid $x$.

1. Output distance. $\left|\frac{1}{x}-\frac{1}{2}\right|=\left|\frac{2-x}{2x}\right|=\frac{|x-2|}{2|x|}.$ Why this step? Combine over a common denominator so $|x-2|=|x-a|$ appears on top — the controllable distance.
2. Spot the danger. The factor is $\frac{1}{2|x|}$. If $x$ drifts toward $0$, this explodes. We must keep $x$ away from $0$. Why this step? A denominator shrinking to $0$ makes the factor unbounded — worse than Cell C, where the factor merely grew.
3. Pre-restrict. Impose $|x-2|<1$, so $1<x<3$, hence $|x|>1$, giving $\frac{1}{2|x|}<\frac{1}{2}.$ Why this step? Trapping $x$ in $(1,3)$ keeps it a safe distance from $0$, capping the factor by $\frac{1}{2}$.
4. Combine. Now $\left|\frac{1}{x}-\frac{1}{2}\right|<\frac{1}{2}|x-2|$. Want $\frac{1}{2}|x-2|<\epsilon$, i.e. $|x-2|<2\epsilon$. So $\delta =\min (1,2\epsilon ).$ Why this step? $\min$ enforces both "stay away from $0$" and the $\epsilon$-condition.

Proof. Let $\epsilon >0$. Choose $\delta =\min (1,2\epsilon )$. If $0<|x-2|<\delta$: since $\delta \le 1$, $|x|>1$ so $\frac{1}{2|x|}<\frac{1}{2}$; since $\delta \le 2\epsilon$, $\left|\frac{1}{x}-\frac{1}{2}\right|=\frac{|x-2|}{2|x|}<\frac{|x-2|}{2}<\frac{2\epsilon }{2}=\epsilon .■$

Look at figure s03: the orange $\frac{1}{x}$ curve plunges toward infinity as $x\to 0$. The red-shaded strip on the left is the forbidden zone — if the green $\delta$-band ever reached into it, the output would escape the blue band no matter how small $\epsilon$ is. Notice the curve is gentle near $x=2$ (small slope $\approx -\frac{1}{4}$), which is why a modest $\delta$ suffices there.

Verify: $\epsilon =0.05\Rightarrow \delta =\min (1,0.1)=0.1$. Take $x=2.1$: $\frac{1}{x}=0.47619$, gap $=|0.47619-0.5|=0.02381<0.05$. ✓

---

Example 5 — Square root, rationalise (Cell E)

Forecast: How do you pull $|x-4|$ out from under a square root? (Hint: what "undoes" $a-b$ times $a+b$?)

1. Output distance. $|\sqrt{x}-2|.$ The variable is stuck under a root, so $|x-4|$ isn't visible yet. Why this step? We must expose $|x-a|$ to use the pivot.
2. Rationalise. Multiply top and bottom by the conjugate $\sqrt{x}+2$: $|\sqrt{x}-2|=\frac{|\sqrt{x}-2|(\sqrt{x}+2)}{\sqrt{x}+2}=\frac{|x-4|}{\sqrt{x}+2}.$ Why this tool and not another? The identity $(\sqrt{x}-2)(\sqrt{x}+2)=x-4$ is the only move that turns a root difference into the plain difference $x-4$ — exactly the distance we can control.
3. Pre-restrict to keep the root defined. Impose $|x-4|<4$, which forces $0<x<8$, so $x>0$ and $\sqrt{x}$ is a real number. Then $\sqrt{x}\ge 0$, so $\sqrt{x}+2\ge 2$, giving the factor $\frac{1}{\sqrt{x}+2}\le \frac{1}{2}.$ Why this step? Unlike Cells C and D, the danger here isn't a growing/exploding factor — it's that $\sqrt{x}$ stops existing for $x<0$. The restriction $|x-4|<4$ fences $x$ into the region where the square root is defined, and as a bonus already caps the factor by the constant $\frac{1}{2}$.
4. Choose $\delta$. Then $|\sqrt{x}-2|\le \frac{1}{2}|x-4|$. Want $\frac{1}{2}|x-4|<\epsilon$, i.e. $|x-4|<2\epsilon$. Combining with step 3's requirement $|x-4|<4$: $\delta =\min (4,2\epsilon ).$ Why the $\min$? It enforces both "$\sqrt{x}$ exists" (from $|x-4|<4$) and the $\epsilon$-condition at once.

Proof. Let $\epsilon >0$. Choose $\delta =\min (4,2\epsilon )$. If $0<|x-4|<\delta$: since $\delta \le 4$, $x>0$ so $\sqrt{x}$ exists and $\sqrt{x}+2\ge 2$; hence $|\sqrt{x}-2|=\frac{|x-4|}{\sqrt{x}+2}\le \frac{|x-4|}{2}<\frac{2\epsilon }{2}=\epsilon .■$

Verify: $\epsilon =0.1\Rightarrow \delta =\min (4,0.2)=0.2$. Take $x=4.2$: $\sqrt{4.2}=2.04939$, gap $=0.04939<0.1$. ✓

---

Example 6 — Proving a limit is FALSE (Cell F)

Forecast: To disprove, we negate the definition. Which quantifier flips first?

Answer the forecast first — the quantifiers swap order. The original definition reads "$\forall \epsilon \exists \delta \dots$": for every challenge $\epsilon$ there exists a reply $\delta$ that wins. To say the limit is not $L$, we negate that whole sentence. Negation turns each $\forall$ into $\exists$ and each $\exists$ into $\forall$, in the same order: the leading $\forall \epsilon$ becomes $\exists \epsilon$ (we now get to pick one clever $\epsilon$), and the following $\exists \delta$ becomes $\forall \delta$ (it must fail for every $\delta$ the claimant offers). The inner implication $P\Rightarrow Q$ negates to $P\text{ and not }Q$, producing $\exists x$ with $0<|x-a|<\delta$ and $|f(x)-L|\ge \epsilon$. So the flip is: the outer $\forall$ becomes $\exists$ first, then the inner $\exists$ becomes $\forall$.

1. Suppose a limit $L$ exists. We derive a contradiction for every $L$. Why this step? Negation must defeat all candidate values, not one.
2. Pick $\epsilon =1$. Why $1$? The two output values $+1$ and $-1$ are $2$ apart. A band of half-width $1$ around any single $L$ has total width $2$ and cannot contain both $+1$ and $-1$ strictly inside.
3. Take any $\delta >0$. The interval $(-\delta ,\delta )$ (minus $0$) always contains a point $x_{+}=\delta /2>0$ and $x_{-}=-\delta /2<0$. Why this step? We must produce a bad $x$ for each $\delta$ the claimant might offer — this is exactly the flipped "$\forall \delta$".
4. Contradiction. $f(x_{+})=1$ and $f(x_{-})=-1$. At least one of $|1-L|$ and $|-1-L|$ is $\ge 1$, because if both were $<1$ then $|1-(-1)|\le |1-L|+|-1-L|<2$ — but $|1-(-1)|=2$, impossible. Why this step? The triangle inequality forces one of the two gaps to hit $\epsilon =1$.

So for $\epsilon =1$, every $\delta$ fails. The limit does not exist. $■$

Look at figure s04: the orange graph jumps from $-1$ to $+1$ at $x=0$. The blue band of half-width $1$ is drawn around the most generous choice $L=0$; even then the two open circles (the values just left and right of $0$) poke out of the band. No vertical slide of the band can swallow both — that visual impossibility is the proof.

Verify: With $L$ arbitrary, $\max (|1-L|,|-1-L|)\ge 1$ for all real $L$. Numeric spot-check at the "best" $L=0$: $|1-0|=1\ge 1$. ✓

---

Example 7 — One-sided limit (Cell G)

Forecast: The $+$ means "$x$ approaches from the right only." How does the input condition change?

1. Rewrite the input band. Instead of $0<|x-0|<\delta$, a right-hand limit uses $0<x<\delta$ (only positive $x$, see One-sided limits). Why this step? $\sqrt{x}$ isn't real for $x<0$, so only the right side is meaningful.
2. Output distance. $|\sqrt{x}-0|=\sqrt{x}.$ Why this step? $L=0$ makes the gap simply $\sqrt{x}$.
3. Solve. Want $\sqrt{x}<\epsilon$. Since $x>0$, square both sides: $x<{\epsilon }^2$. So take $\delta ={\epsilon }^2$. Why squaring is legal here: both sides are non-negative, so squaring preserves the inequality — this is the tool that "undoes" the root going the other direction.

Proof. Let $\epsilon >0$. Choose $\delta ={\epsilon }^2$. If $0<x<\delta ={\epsilon }^2$, then $\sqrt{x}<\sqrt{{\epsilon }^2}=\epsilon$. $■$

Verify: $\epsilon =0.3\Rightarrow \delta =0.09$. Take $x=0.08$: $\sqrt{0.08}=0.28284<0.3$. ✓ Note $\delta ={\epsilon }^2$ shrinks faster than $\epsilon$ — the curve is very steep near $0$.

---

Example 8 — Word problem, real tolerance (Cell H)

Forecast: Which quantity is $\epsilon$? Which is $\delta$? Guess the window before computing.

1. Translate. "Within $0.02$ L of $8.5$" is $|V(t)-8.5|<\epsilon$ with $\epsilon =0.02$. The timing window is the $\delta$ we must find. Why this step? The output tolerance is the demand; the input (time) tolerance is our control knob.
2. Output distance. $|V(t)-8.5|=|2t+0.5-8.5|=|2t-8|=2|t-4|.$ Why this step? Pivot: factor out $|t-4|=|t-a|$ with $a=4$.
3. Identify the factor and solve. The factor is the constant $2$ (litres per second). Want $2|t-4|<0.02$, i.e. $|t-4|<0.01$. So $\delta =0.01$ s. Why this step? Constant factor means $\delta =\epsilon /2=0.02/2=0.01$ — the same clean division as Cell B.
4. State the answer as a window. The volume stays in spec whenever the timer is within $\pm 0.01$ s of $4$ s — a total window of $0.02$ s. Why this step? $|t-4|<0.01$ is exactly the interval $(3.99,4.01)$; its full width is $2\delta =0.02$ s.

Proof. Let $\epsilon =0.02$. Choose $\delta =0.01$. If $0<|t-4|<0.01$, then $|V(t)-8.5|=2|t-4|<2\cdot 0.01=0.02=\epsilon .■$

Verify: At $t=4.005$ (inside the window): $V=2(4.005)+0.5=8.51$, gap $=|8.51-8.5|=0.01<0.02$. ✓ At the boundary $t=4.01$: $V=8.52$, gap $=0.02$ — exactly $\epsilon$, so strictly inside requires $|t-4|<0.01$, confirming the strict window. Units: rate $2$ L/s $\times$ $0.01$ s $=0.02$ L — dimensionally consistent.

---

Example 9 — Exam twist: pointwise vs uniform (Cell I)

Forecast: In Ex 3 we got $\delta =\min (1,\epsilon /7)$ at $a=3$. Guess what replaces the $7$ for a general $a$.

1. General factor. $|x^2-a^2|=|x-a||x+a|.$ With pre-restriction $|x-a|<1$, we get $|x+a|<|2a|+1$. Why this step? Same Cell-C machinery, but keep $a$ symbolic: $|x-a|<1$ gives $|x|<|a|+1$, so $|x+a|\le |x|+|a|<|2a|+1$.
2. General $\delta$. $\delta =\min \left(1,\frac{\epsilon }{|2a|+1}\right).$ Why this step? Replace the constant $7$ (which was $2\cdot 3+1$) by the point-dependent cap $|2a|+1$. Notice $\delta$ now depends on both $\epsilon$ and the point $a$.
3. Read off the trend. As $a\to \infty$, the denominator $|2a|+1\to \infty$, so $\delta \to 0$. Why this matters: the same $\epsilon$ needs an ever-smaller $\delta$ far out. No single positive $\delta$ works for all $a$ simultaneously — that's exactly why $x^2$ is continuous everywhere but not uniformly continuous on $\mathbb{R}$.

Look at figure s05: the blue curve plots the largest usable $\delta$ against the point $a$ (for a fixed $\epsilon$). At $a=3$ (red dot) it sits at $0.1$; at $a=10$ (orange dot) it has dropped to $0.033$. The curve slides toward $0$ forever — no horizontal floor exists, which is the visual meaning of "no single $\delta$ works for all $a$."

Verify (matches Ex 3): At $a=3$, $|2a|+1=7$ — recovers $\delta =\min (1,\epsilon /7)$. ✓ At $a=10$, denominator $=21$, so for $\epsilon =0.7$, $\delta =\min (1,0.0333)=0.0333$ — over $3\times$ smaller than at $a=3$ (where it was $0.1$).

---

Coverage check

Recall Did we hit every matrix cell?

A::: Ex 1 constant. B::: Ex 2 linear. C::: Ex 3 quadratic. D::: Ex 4 quotient. E::: Ex 5 root. F::: Ex 6 non-existence. G::: Ex 7 one-sided. H::: Ex 8 word problem. I::: Ex 9 uniform-vs-pointwise. Every cell covered.

Active recall

Connections

• Epsilon-delta definition of a limit — formal proofs — the parent; this page is its worked-example tournament.
• Limit Laws (sum, product, quotient) — Ex 4's quotient handling is the seed of the quotient law.
• Continuity — every example here also proves $f$ continuous at $a$ (since $L=f(a)$ each time it's defined).
• One-sided limits — Ex 7 uses the one-sided input band.
• Uniform continuity — Ex 9 is the gateway.
• Definition of the derivative — same $\epsilon$–$\delta$ machinery on difference quotients.
• Sequences and their limits — swap $\delta$ for a threshold index $N$.