Real analysis — rigorous epsilon-delta, metric spaces: Worked examples

The scenario matrix

Every $ε$ – $δ$ / metric problem is one of these cells. We label the cells with the tags Cell A … Cell I (these tags are only names for rows of this table — they are not used for anything else on the page). We will hit each one at least once below.

Cell	Case class	What makes it different	Example
Cell A	Linear $f$ (constant slope)	multiplier $M$ is already constant — no caging	Ex 1
Cell B	Quadratic / polynomial (variable slope)	must cap $∥ x - a ∥ < 1$ to bound the multiplier	Ex 2
Cell C	Root / reciprocal-type (algebra needed)	rationalise or invert to expose $∥ x - a ∥$	Ex 3
Cell D	Degenerate limit point ( $a = 0$ )	signs on both sides; punctured $0 < ∥ x ∥$	Ex 4
Cell E	Non-uniform continuity	show no single $δ$ works (negation)	Ex 5
Cell F	Sequence limit in $R$ ( $N$ instead of $δ$ )	discrete index, "for large $n$ "	Ex 6
Cell G	Non-Euclidean metric (taxicab / discrete)	distance axioms, unusual balls	Ex 7
Cell H	Real-world word problem	translate tolerance → $ε$ , setting → $δ$	Ex 8
Cell I	Exam twist (limit does not exist)	prove non-existence by contradiction	Ex 9

Intuition The single recipe behind almost every cell

Most cells (A, B, C, D, H) follow the exact same five-move loop. Read it once now; each example below is just this loop specialised. The flow uses plain step names (not the cell tags), so nothing is being reused:

graph LR
  S1["pick epsilon"] --> S2["factor out distance"]
  S2 --> S3["bound the multiplier M"]
  S3 --> S4["set delta from epsilon over M"]
  S4 --> S5["verify implication holds"]

Ex 1 skips step "bound M" (M already constant); Ex 2 is the full loop; Ex 3 does step "factor" via rationalising; Ex 8 is the same loop dressed as a word problem. Cells E, F, G, I break the loop on purpose and are handled by their own methods.

Cell A — linear, constant slope

Worked example

Ex 1. Prove $x \to 5 lim (4 x + 2) = 22$ . Forecast: the slope is $4$ . Guess: does the output error grow $4 \times$ faster than the input error, so $δ = ε /4$ ?

Write the target gap. $∣ f (x) - L ∣ = ∣ (4 x + 2) - 22∣ = ∣4 x - 20∣$ . Why this step? The whole game is about the size of $∣ f (x) - L ∣$ ; we simplify it first.
Factor out $∣ x - 5∣$ . $∣4 x - 20∣ = 4 ∣ x - 5∣$ . Why? $δ$ only controls $∣ x - 5∣$ , so we must express everything in terms of it. The leftover multiplier is the constant $M = 4$ — no caging needed (Cell A's defining feature).
Force it below $ε$ . We want $4∣ x - 5∣ < ε$ , i.e. $∣ x - 5∣ < ε /4$ . Why? Divide the desired inequality by the multiplier $4$ .
Choose $δ = ε /4$ . Why no $min$ ? There was no preliminary restriction ( $M$ was global), so nothing else to intersect.

Verify: if $0 < ∣ x - 5∣ < ε /4$ then $∣ f (x) - 22∣ = 4∣ x - 5∣ < 4 \cdot \frac{ε}{4} = ε .$ ✅ Take $ε = 0.1 \Rightarrow δ = 0.025$ ; at $x = 5.02$ , $∣ f - 22∣ = 4 (0.02) = 0.08 < 0.1$ . ✔

Cell B — quadratic, variable slope (caging)

Worked example

Ex 2. Prove $x \to 2 lim x^{2} = 4$ . Forecast: near $x = 2$ the slope of $x^{2}$ is about $2 x = 4$ , so guess a $δ$ roughly $ε /4$ — but the slope changes, so we'll need to cap it.

Target gap and factor. $∣ x^{2} - 4∣ = ∣ x - 2∣ ∣ x + 2∣$ . Why? $x^{2} - 4 = (x - 2) (x + 2)$ splits off the controllable $∣ x - 2∣$ . The leftover $∣ x + 2∣$ is not constant — that's Cell B.
Cage $∣ x + 2∣$ with a preliminary restriction $∣ x - 2∣ < 1$ . Then $1 < x < 3 \Rightarrow 3 < x + 2 < 5 \Rightarrow ∣ x + 2∣ < 5$ . Why cap first? $∣ x + 2∣$ grows without bound over all of $R$ ; we can only bound it by staying near $2$ .
Combine. With that cap, $∣ x^{2} - 4∣ < 5 ∣ x - 2∣$ . To beat $ε$ : need $∣ x - 2∣ < ε /5$ . Why? Now the multiplier is a fixed $M = 5$ ; divide as in Cell A.
Choose $δ = min (1, ε /5)$ . Why $min$ ? Both requirements — $∣ x - 2∣ < 1$ (to keep the cage valid) and $∣ x - 2∣ < ε /5$ — must hold at once.

Figure below. The lavender curve is $f (x) = x^{2}$ . The butter-yellow horizontal band is the target tube $∣ f - 4∣ < ε$ ; the two coral dashed lines at $x = 1$ and $x = 3$ mark the cap $∣ x - 2∣ < 1$ ; the mint vertical band is the actual $δ$ -band of half-width $min (1, ε /5)$ . Notice the mint band sits inside both the coral cap and the region whose curve stays in the yellow tube — that visual "inside both" is exactly why the answer is a $min$ .

Verify: for $ε = 1$ , $δ = min (1, 0.2) = 0.2$ . At $x = 2.2$ : $∣ x^{2} - 4∣ = ∣4.84 - 4∣ = 0.84 < 1$ . ✔ For $ε = 25$ , $δ = min (1, 5) = 1$ — the cap $1$ takes over, exactly why we needed the $min$ .

Cell C — root type (rationalise to expose distance)

Worked example

Ex 3. Prove $x \to 9 lim x = 3$ . Forecast: $x$ is gentle (slope $\frac{1}{2 9} = \frac{1}{6}$ ) near $9$ , so guess a generous $δ$ , roughly $6 ε$ .

Domain guard first. $x$ needs $x \geq 0$ . We will therefore build the restriction $δ \leq 9$ into the choice from the start, so that $0 < ∣ x - 9∣ < δ$ forces $x > 0$ . Why up front? If we let $δ$ exceed $9$ , the punctured ball around $9$ could reach $x \leq 0$ where $x$ is undefined — the proof would talk about points that don't exist. Guarding now keeps every later line legal.
Target gap. $∣ x - 3∣$ . There's no obvious $∣ x - 9∣$ yet. Why worry? $δ$ controls $∣ x - 9∣$ , not $∣ x - 3∣$ . We must manufacture $∣ x - 9∣$ .
Rationalise. Multiply by $\frac{x + 3}{x + 3}$ : $∣ x - 3∣ = \frac{∣ ( x - 3 ) ( x + 3 ) ∣}{x + 3} = \frac{∣ x - 9∣}{x + 3} .$ Why this trick? $(x - 3) (x + 3) = x - 9$ — the difference of squares turns the messy root into the plain $∣ x - 9∣$ we can control.
Bound the denominator from below. By step 1 we have $x \geq 0$ , so $x \geq 0$ and $x + 3 \geq 3$ , hence $∣ x - 3∣ = \frac{∣ x - 9∣}{x + 3} \leq \frac{∣ x - 9∣}{3} .$ Why lower-bound the denominator? A bigger denominator makes the fraction smaller; using the smallest denominator ( $3$ ) gives the safest (largest) upper bound.
Force below $ε$ and choose $δ$ . Want $\frac{∣ x - 9∣}{3} < ε$ , i.e. $∣ x - 9∣ < 3 ε$ . Combined with the domain guard $δ \leq 9$ , the airtight choice is $δ = min (9, 3 ε) .$ Why $min$ ? We need $∣ x - 9∣ < 3 ε$ and $∣ x - 9∣ \leq 9$ (to keep $x > 0$ ) simultaneously — exactly the situation a $min$ resolves.

Verify: $ε = 0.3 \Rightarrow δ = min (9, 0.9) = 0.9$ . At $x = 9.9$ : $∣ 9.9 - 3∣ \approx ∣3.1464 - 3∣ = 0.1464 < 0.3$ . ✔ Our guaranteed bound was $\frac{∣ x - 9∣}{3} = \frac{0.9}{3} = 0.3$ , and indeed the true error $0.1464$ sits below it. ✔

Cell D — degenerate limit point $a = 0$ (both signs)

Worked example

Ex 4. Prove $x \to 0 lim x^{2} sin \frac{1}{x} = 0$ (squeeze, via $ε$ – $δ$ ). Forecast: $sin (1/ x)$ oscillates wildly as $x \to 0$ from both sides, yet $x^{2}$ crushes it. Guess $δ = ε$ .

Target gap. $∣ x^{2} sin \frac{1}{x} - 0∣ = x^{2} ∣ sin \frac{1}{x} ∣$ . Why? $L = 0$ , so the gap is just the magnitude of the function.
Bound the wild part. $∣ sin θ ∣ \leq 1$ for all real $θ$ , so $x^{2} ∣ sin \frac{1}{x} ∣ \leq x^{2} = ∣ x ∣^{2} .$ Why is this the key move? We can never control $sin (1/ x)$ directly, but we don't need to — we only need an upper bound, and $1$ is a universal one. This handles both signs of $x$ at once since $x^{2} \geq 0$ regardless.
Force below $ε$ . Want $∣ x ∣^{2} < ε$ , i.e. $∣ x ∣ < ε$ . Why square root? We solve $∣ x ∣^{2} < ε$ for $∣ x ∣$ ; both sides positive so the root is monotone.
Choose $δ = ε$ , and note $0 < ∣ x - 0∣ < δ$ (punctured — we never use $x = 0$ , where $sin (1/ x)$ is undefined anyway; this is exactly why the definition punctures the point).

Verify: $ε = 0.04 \Rightarrow δ = 0.2$ . Worst case at any $x$ with $∣ x ∣ < 0.2$ : $∣ x^{2} sin \frac{1}{x} ∣ \leq x^{2} < 0.04$ . Check $x = - 0.15$ : $x^{2} = 0.0225 < 0.04$ regardless of the sine value. ✔ (Negative $x$ works identically — the degenerate two-sided case is fully covered.)

Cell E — non-uniform continuity (the failure case)

Worked example

Ex 5. Show $f (x) = \frac{1}{x}$ is not uniformly continuous on $(0, 1)$ . Forecast: as $x \to 0^{+}$ the graph gets vertically steep — guess that no single $δ$ can serve all points, because near $0$ tiny input moves cause huge output jumps.

Write the full definition, then negate it. Uniform continuity is $\forall ε > 0 \exists δ > 0 \forall x, y : ∣ x - y ∣ < δ ⟹ ∣ f (x) - f (y) ∣ < ε .$ Its logical negation (what we must prove) is $\exists ε > 0 \forall δ > 0 \exists x, y : ∣ x - y ∣ < δ and ∣ f (x) - f (y) ∣ \geq ε .$ Why write both? We must chase the exact negation: we get to choose one $ε$ , but must then defeat every $δ$ the opponent offers by producing bad points $x, y$ .
Fix $ε = 1$ . Now, given an arbitrary $δ > 0$ , pick two points close together but near $0$ : $x = \frac{1}{n}, y = \frac{1}{n + 1}, with n large.$ Why these? Both live in $(0, 1)$ , and their distance $∣ x - y ∣ = \frac{1}{n ( n + 1 )} \to 0$ — so we can make $∣ x - y ∣ < δ$ for large $n$ .
Compute the output gap. $∣ f (x) - f (y) ∣ = ∣ n - (n + 1) ∣ = 1$ . Why the punchline? Reciprocals of consecutive $1/ n$ differ by exactly $1$ — no matter how tiny $∣ x - y ∣$ , the output gap stays $\geq ε = 1$ .
Conclude. For each $δ$ we found $x, y$ with $∣ x - y ∣ < δ$ yet $∣ f (x) - f (y) ∣ = 1 \geq 1 = ε$ . So no $δ$ works ⇒ not uniformly continuous. ∎

Figure below. The lavender curve is $f (x) = 1/ x$ on $(0, 1)$ . Two coral dots sit at $x = \frac{1}{3}$ and $y = \frac{1}{4}$ ; the mint dotted verticals drop to the axis to show how close together the inputs are (short horizontal green arrow, "tiny $∣ x - y ∣$ "), while the coral vertical arrow on the right measures the output gap, which is a full unit high. Two inputs a hair apart, outputs a whole unit apart — that is the failure.

Verify: $n = 100$ : $x = 0.01$ , $y = \frac{1}{101} \approx 0.009901$ , $∣ x - y ∣ \approx 9.9 \times 1 0^{- 5}$ (tiny), but $∣ f (x) - f (y) ∣ = ∣100 - 101∣ = 1$ . ✔ (Contrast: on a closed bounded set this failure cannot happen.)

Cell F — sequence limit ( $N$ instead of $δ$ )

Cell G — a non-Euclidean metric (taxicab vs discrete)

Worked example

Ex 7. In $R^{2}$ with the taxicab metric $d_{1} (x, y) = ∣ x_{1} - y_{1} ∣ + ∣ x_{2} - y_{2} ∣$ , describe the open ball $B (0, 1)$ ; and describe $B (0, 1)$ under the discrete metric. Forecast: taxicab "circles" are usually drawn as diamonds — guess a square rotated $4 5^{\circ}$ . For the discrete metric, guess something strange (only the centre?).

Taxicab ball — read the defining inequality. $B (0, 1) = {(x_{1}, x_{2}) : ∣ x_{1} ∣ + ∣ x_{2} ∣ < 1}$ . Why a diamond? The boundary $∣ x_{1} ∣ + ∣ x_{2} ∣ = 1$ is four straight segments; e.g. in the first quadrant it is the line $x_{1} + x_{2} = 1$ . Four such segments enclose the unit square turned $4 5^{\circ}$ — a diamond. (We take the triangle inequality for $d_{1}$ as already known from the parent note; here we only need the shape.)
Discrete ball — read its defining inequality. With $d (x, y) = 1$ if $x \neq = y$ , else $0$ : $B (0, 1) = {x : d (x, 0) < 1} = {x : d (x, 0) = 0} = {0} .$ Why only the centre? The only distance strictly less than $1$ is $0$ , achieved solely at $x = 0$ . (A radius $r > 1$ would instead give the whole space.)

Figure below (two panels). Left: the mint diamond is the taxicab $B (0, 1)$ ; the interior coral dot $(0.5, 0.4)$ has $d_{1} = 0.9 < 1$ so it is inside, while $(0.7, 0.7)$ has $d_{1} = 1.4$ so it is outside. Right: the discrete $B (0, 1)$ is a single lavender dot at the origin; every other point (hollow coral circles) is at distance exactly $1$ , hence excluded. Same radius, wildly different balls — the metric decides the geometry.

Verify: point $(0.5, 0.4)$ : $d_{1} = 0.9 < 1$ ⇒ inside diamond ✔; point $(0.7, 0.7)$ : $d_{1} = 1.4 \neq < 1$ ⇒ outside ✔. Discrete: $d ((0.5, 0.4), 0) = 1 \neq < 1$ ⇒ not in discrete $B (0, 1)$ , only $0$ is. ✔

Cell H — real-world word problem

Worked example

Ex 8. A machine cuts rods to a target length by setting a dial $x$ (cm). The finished length is $L (x) = 3 x$ (cm). A customer's tolerance is $\pm 0.06$ cm around the target $30$ cm. How precisely ( $δ$ ) must the dial be set at $x = 10$ ? Forecast: the "gain" is $3 \times$ , so the dial tolerance should be $3 \times$ tighter than the output tolerance: guess $0.02$ cm.

Translate. Output target $L = 30$ at dial $x = 10$ . "Within tolerance" means $∣ L (x) - 30∣ < ε$ with $ε = 0.06$ cm. Dial precision is our $δ$ : $∣ x - 10∣ < δ$ . Why $ε$ – $δ$ at all? "Get the length right enough by setting the dial precisely enough" is literally the challenge–response game.
Factor. $∣3 x - 30∣ = 3∣ x - 10∣$ (Cell A structure — constant gain $M = 3$ ).
Solve. $3∣ x - 10∣ < 0.06 \Rightarrow ∣ x - 10∣ < 0.02$ . So $δ = 0.02$ cm. Why divide by 3? The machine amplifies dial error threefold, so we must shrink the input error threefold.

Verify (units): dial off by $0.02$ cm ⇒ length off by $3 \times 0.02 = 0.06$ cm $= ε$ , exactly the boundary. A dial within $0.02$ cm keeps length strictly within $0.06$ cm. ✔ Units: cm·(dimensionless gain)=cm ✔.

Cell I — exam twist: prove a limit does NOT exist

Worked example

Ex 9. Show $x \to 0 lim \frac{∣ x ∣}{x}$ does not exist. Forecast: $∣ x ∣/ x$ is $+ 1$ for $x > 0$ and $- 1$ for $x < 0$ . Guess: the two sides disagree, so no single $L$ can win the game.

Assume a limit $L$ exists and challenge with $ε = \frac{1}{2}$ . By the definition there must be a $δ > 0$ so that $0 < ∣ x ∣ < δ \Rightarrow ∣ f (x) - L ∣ < \frac{1}{2}$ . Why contradiction? To disprove $\forall ε \exists δ$ , we show one $ε$ that no $δ$ can satisfy; assuming an $L$ lets us derive an impossibility.
Any $δ > 0$ admits points on both sides. Pick $x_{+} = δ /2 > 0$ and $x_{-} = - δ /2 < 0$ ; both satisfy $0 < ∣ x ∣ < δ$ , so both must obey the $\frac{1}{2}$ -bound. Why both signs? A punctured ball around $0$ always contains left and right points — that is the degenerate two-sided nature of $a = 0$ .
Evaluate at each side. Since $f (x_{+}) = + 1$ and $f (x_{-}) = - 1$ , the $\frac{1}{2}$ -bound forces $∣1 - L ∣ < \frac{1}{2} and ∣ - 1 - L ∣ < \frac{1}{2} .$ Why both must hold? Both $x_{+}$ and $x_{-}$ lie in the punctured $δ$ -ball, so the definition applies to each.
Derive the contradiction (triangle inequality). These say $L$ is within $\frac{1}{2}$ of $+ 1$ and within $\frac{1}{2}$ of $- 1$ at once — but $+ 1$ and $- 1$ are a full $2$ apart: $2 = ∣1 - (- 1) ∣ = ∣ (1 - L) + (L - (- 1)) ∣ \leq ∣1 - L ∣ + ∣ L + 1∣ < \frac{1}{2} + \frac{1}{2} = 1,$ i.e. $2 < 1$ — impossible. Why the triangle inequality? It is the one tool that fuses " $L$ near $+ 1$ " and " $L$ near $- 1$ " into a single false statement, exactly the distance axiom powering the theory.
Conclude. The assumption "a limit $L$ exists" led to $2 < 1$ . So no such $L$ exists ⇒ the limit does not exist. ∎

Verify: at $x = + 0.001$ : value $= + 1$ ; at $x = - 0.001$ : value $= - 1$ ; the two one-sided values differ by $2$ , unbeatable by any $ε < 2$ (in particular $ε = \frac{1}{2}$ ). ✔

Recall Self-test: match cell to method

Which technique caps $∣ x - a ∣ < 1$ first ::: The variable-slope polynomial case (Cell B), to bound the non-constant multiplier How do you prove a limit does NOT exist ::: Find one $ε$ and two approaches giving separated values; use the triangle inequality for a contradiction (Cell I) The discrete-metric ball $B (0, 1)$ is ::: just ${0}$ — only distance $< 1$ is $0$ Sequence limit uses which threshold ::: an index $N = ⌈ 1/ ε ⌉$ instead of a $δ$ Why does Ex 3 need $δ = min (9, 3 ε)$ not just $3 ε$ ::: the cap $9$ keeps the punctured ball inside the domain $x \geq 0$ where $x$ is defined