The epsilon-delta definition of a limit and the challenge-and-response way of reading it, proving limits for linear and quadratic functions with the backwards scratch work and forwards proof, the epsilon-delta forms of one-sided and infinite limits, and using the definition to prove a limit law.
Subject: Calculus I · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Calculus I · Chapter 2 — Limits
The Precise Definition of a Limit
Objectives
Five outcomes. The first is a definition that took two thousand years to find, and the rest are what it makes possible.
OpenStax Calculus Volume 1, §2.5 The Precise Definition of a Limit §2.5, pp. 169-181 — the section these objectives are drawn from
Warm-up
Every section of this chapter has used the phrase gets arbitrarily close. Section 2.1 identified that phrase as exactly what the ancient Greeks lacked.
Discussion prompt
What would it take to make gets arbitrarily close to L into a statement you could actually verify or refute?
Hint: Turn the vague word 'close' into a number someone else gets to choose.
Answer:
The trouble with 'close' is that it has no scale. Close to 5 might mean within 1, or within a millionth, and the claim must hold for every such reading.
So let an adversary name the tolerance. If for every tolerance they name, however small, you can name a distance within which your inputs guarantee that tolerance, then no reading of 'close' can defeat you.
\[ \text{for every } \varepsilon > 0 \text{ there exists } \delta > 0 \;\ldots \]
That is the whole definition, and the two quantifiers are in that order for a reason: the tolerance comes first, and the response is allowed to depend on it.
Concept
The limit is L means: for every positive tolerance epsilon, there is a positive distance delta such that every input within delta of a, but not equal to a, produces an output within epsilon of L. The order of the two quantifiers is the whole content.
the epsilon-delta definition — The limit of f at a is L if for every epsilon greater than zero there exists delta greater than zero such that whenever the distance from x to a is positive and less than delta, the distance from f of x to L is less than epsilon.
\[ \forall \varepsilon > 0 \; \exists \delta > 0: \; 0 < |x - a| < \delta \;\Rightarrow\; |f(x) - L| < \varepsilon \]
Read it as a game. The challenger names epsilon; you must respond with a delta that works. If you have a rule producing a working delta for every possible epsilon, the limit is proved — and if some epsilon defeats every delta, it is disproved.
Figure (svg): The epsilon band around the limit and the delta band around the point, with the curve threading the box
OpenStax Calculus Volume 1, §2.5 The Precise Definition of a Limit §2.5, pp. 169-173
Section
Section 1
Concept
Epsilon is named first, by someone trying to defeat the claim. Delta is your reply and may depend on epsilon. The claim survives exactly when you have a reply for every possible challenge.
epsilon and delta — Epsilon is the tolerance on the output, chosen first and arbitrarily small. Delta is the distance on the input, chosen in response and permitted to depend on epsilon.
\[ 0 < |x-a| < \delta \;\Longrightarrow\; |f(x) - L| < \varepsilon \]
The strict inequality zero is less than the distance is what excludes the point itself, exactly as Section 2.2 required. Every feature of the informal definition is present here in a form that can be checked.
Figure (svg): The challenge-and-response structure, with epsilon shrinking and delta shrinking in reply
OpenStax Calculus Volume 1, §2.5 The Precise Definition of a Limit §2.5, pp. 169-174 — the formal definition
Picture it
A horizontal band of half-height epsilon, and the vertical band that fits inside it.
Figure (svg): The epsilon band around the limit and the delta band around the point, with the curve threading the box
The green band is the challenge and the yellow band is the answer. The requirement is that the curve, restricted to the yellow band, never leaves the green one — and that a yellow band exists however thin the green one is made.
Worked example
Example 2.35. Concrete epsilons first, to see the pattern.
\[ \text{For } f(x) = 2x + 1 \text{ at } a = 2 \text{ with } L = 5, \text{ find } \delta \text{ for } \varepsilon = 1, 0.1, 0.01. \]
Write what must be achieved
Why: The output within epsilon of 5.
\[ | (2 x + 1) - 5 | < \varepsilon \]
Simplify the left side
Why: Two x minus 4, which factors.
\[ 2 | x - 2 | < \varepsilon \]
Divide to isolate the input distance
Why: This is the condition on x.
\[ | x - 2 | < \varepsilon / 2 \]
Read off delta for each challenge
Why: Half of each epsilon.
\[ 0.5, 0.05, 0.005 \]
Notice the pattern
Why: It is always half.
\[ \delta = \varepsilon / 2\text{ always} \]
Figure (svg): The solution to Worked example answering three challenges shown as a ladder of expressions, one row per legal move
\[ \delta = \tfrac{\varepsilon}{2} \]
Verify: test the middle one explicitly
Why: With epsilon 0.1 and delta 0.05, take any x within 0.05 of 2, say 2.03. Then f of x is 6.06, and the distance from 5 is 1.06 — wait, that is 2 times 2.03 plus 1, which is 6.06, and 6.06 minus 5 is 1.06. That exceeds 0.1, so recompute: f of 2.03 is 6.06? No: 2 times 2.03 is 4.06, plus 1 is 5.06. The distance from 5 is 0.06, which is under 0.1. The arithmetic slip is instructive — always recompute rather than trusting the first reading, since these proofs live entirely on careful inequalities.
OpenStax Calculus Volume 1, §2.5 The Precise Definition of a Limit §2.5, pp. 172-173
Fill the middle
The linear function from the worked example, solved for the input distance.
Fill in the blanks
|(2x+1) - 5| < \varepsilon \iff 2|x-2| < \varepsilon \iff |x-2| < \varepsilon/2
Why: The condition on the input is that its distance from 2 be under epsilon over 2, so that is the delta to choose. Because every step is an equivalence, the chain reverses and the choice provably works.
Worked example
Checkpoint 2.35. Swapping them destroys the definition.
\[ \text{What would it mean if } \delta \text{ had to be chosen before } \varepsilon? \]
Write the swapped statement
Why: There exists delta such that for every epsilon the implication holds.
Ask what a single delta would have to achieve
Why: It must work for every epsilon at once, however small.
Deduce the consequence
Why: The output distance must be smaller than every positive number.
\[ | f(x) - L | = 0 \]
Interpret
Why: The function would have to be exactly L throughout the delta band.
Figure (svg): The solution to Worked example why the order of quantifiers matters shown as a ladder of expressions, one row per legal move
\[ \exists\delta \forall\varepsilon \;\Longrightarrow\; f \equiv L \text{ near } a \]
Verify: check that the intended examples fail the swapped version
Why: The function 2x plus 1 is not constantly 5 anywhere near 2, so it would fail the swapped definition despite plainly having the limit 5. The swapped statement is a much stronger and quite useless condition. That the order matters this much is why the definition is written with the universal quantifier first, and it is worth reading the two quantifiers aloud in order every time.
OpenStax Calculus Volume 1, §2.5 The Precise Definition of a Limit §2.5, pp. 173-174
Trap
\[ \varepsilon = 1 \Rightarrow \delta = 0.5; \quad \varepsilon = 0.1 \Rightarrow \delta = 0.05 \]
Conclude the limit is proved
Why: The student checks a few challenges and stops.
\[ \text{so the limit is } 5 \quad \text{(not yet proved)} \]
The definition demands every positive epsilon, of which there are infinitely many. Checking three of them establishes nothing about the rest.
\[ \text{given any } \varepsilon > 0, \text{ take } \delta = \tfrac{\varepsilon}{2} \]
Produce a RULE that answers every challenge at once
Why: The proof is the formula for delta in terms of epsilon, not a list of cases.
Concrete epsilons are useful scratch work for spotting the pattern, and Example 2.35 does exactly that. But the proof is the general rule, and the write-up must begin with 'let epsilon be any positive number' rather than with a particular value.
Ranking
Who moves when.
Put in order
Why: The order is the definition. Epsilon is fixed before delta is chosen, which is what lets delta depend on it; and x is chosen after delta, which is why delta must work for every admissible x rather than for one you pick.
Two truths and a lie
All three are about the definition.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and it has the relationship backwards. A delta that works for a given epsilon also works for every LARGER epsilon, since a larger tolerance is easier to meet. Smaller epsilons generally demand smaller deltas — that is precisely why delta is written as a function of epsilon.
Prediction
Commit before reasoning.
Predict first
What exactly constitutes a proof that a limit equals L?
Correct: A rule producing a working delta for every positive epsilon.
\[ \text{proof} = \text{a function } \delta(\varepsilon) \text{ that always works} \]
Why: The definition quantifies over all positive epsilon, so the proof must too — which means a formula rather than a list. A table is evidence only, as Section 2.2's oscillating example showed, and checking finitely many epsilons leaves infinitely many unchecked. Continuity is a stronger statement that presupposes the limit rather than establishing it. The deliverable is always the formula for delta.
Section
Section 2
Concept
To find delta, start from the inequality you must achieve and solve it for the distance from x to a. Since every step is reversible, reading the chain in the other direction is the proof.
backwards scratch work — The technique of solving the desired output inequality for the input distance in order to discover delta. It is not itself the proof; the proof is the same chain of equivalences written in the forward direction.
\[ |f(x)-L| < \varepsilon \iff \cdots \iff |x-a| < g(\varepsilon) \;\Longrightarrow\; \delta = g(\varepsilon) \]
For a linear function the chain is a single division and delta comes out as epsilon divided by the absolute value of the slope. A steeper line needs a smaller delta, which is exactly what the picture shows.
Figure (svg): The algebra of a linear epsilon-delta proof, run backwards to find delta and forwards to verify
OpenStax Calculus Volume 1, §2.5 The Precise Definition of a Limit §2.5, pp. 174-177 — proving limits from the definition
Picture it
The chain of equivalences, and which direction each use runs.
Figure (svg): The algebra of a linear epsilon-delta proof, run backwards to find delta and forwards to verify
Every line is an if-and-only-if, which is what makes the search legitimate. When a step is only an implication rather than an equivalence — as happens for the quadratic — the direction has to be watched carefully.
Worked example
Example 2.37. Scratch work, then the write-up.
\[ \text{Prove that } \lim_{x \to 2}(3x - 1) = 5. \]
Write the target inequality
Why: Output within epsilon of 5.
\[ | (3 x - 1) - 5 | < \varepsilon \]
Simplify
Why: Three x minus 6, which factors.
\[ 3 | x - 2 | < \varepsilon \]
Solve for the input distance
Why: Divide by 3.
\[ | x - 2 | < \varepsilon / 3 \]
Declare the choice
Why: This is the rule.
\[ \text{let } \delta = \varepsilon / 3 \]
Write the forward proof
Why: Assume the input condition and derive the output one.
\[ 0 < | x - 2 | < \varepsilon / 3\text{ gives } 3 | x - 2 | < \varepsilon \]
Figure (svg): The solution to Worked example a linear proof in full shown as a ladder of expressions, one row per legal move
\[ \delta = \tfrac{\varepsilon}{3} \]
Verify: run the forward chain explicitly
Why: Suppose the distance from x to 2 is positive and under epsilon over 3. Multiplying by 3 gives 3 times that distance under epsilon. But 3 times the distance from x to 2 is exactly the distance from 3x minus 1 to 5, since the difference simplifies to 3x minus 6. So the output condition holds, for every x satisfying the input condition — which is what the definition demanded. The proof is three lines and every step is reversible.
OpenStax Calculus Volume 1, §2.5 The Precise Definition of a Limit §2.5, pp. 175-176
Fill the middle
The linear proof from the worked example, with slope 3.
Fill in the blanks
3|x-2| < \varepsilon \iff |x - 2| < \varepsilon/3
Why: Delta is epsilon over 3. For any linear function the answer is epsilon divided by the absolute value of the slope, which is why steeper lines need tighter input bands.
Worked example
Checkpoint 2.37. A steeper line needs a tighter delta.
\[ \text{Find } \delta \text{ for } \lim_{x \to 1}(10x + 3) = 13 \text{ and compare with the previous slope.} \]
Write the target
Why: Output within epsilon of 13.
\[ | (10 x + 3) - 13 | < \varepsilon \]
Simplify and factor
Why: Ten x minus 10.
\[ 10 | x - 1 | < \varepsilon \]
Solve
Why: Divide by 10.
\[ | x - 1 | < \varepsilon / 10 \]
Compare with the slope-3 case
Why: Delta is epsilon over the slope in both.
\[ \delta = \varepsilon / | m | \]
Figure (svg): The solution to Worked example the slope decides the ratio shown as a ladder of expressions, one row per legal move
\[ \delta = \frac{\varepsilon}{|m|} \]
Verify: check the picture agrees
Why: A steeper line converts a given input wobble into a bigger output wobble, so to keep the output inside a fixed band the input band must be narrower — and epsilon over 10 is indeed much smaller than epsilon over 3. The general formula also predicts trouble for a slope of zero, and correctly so: a constant function needs no restriction at all, since the output is exactly L for every input and any delta works.
OpenStax Calculus Volume 1, §2.5 The Precise Definition of a Limit §2.5, pp. 176-177
Error analysis
A student proves a limit and produces a delta.
Annotate
On: \( |x^2 - 9| = |x-3||x+3| < \varepsilon \;\Longrightarrow\; \text{take } \delta = \frac{\varepsilon}{|x+3|} \)
This is the single most common error in epsilon-delta proofs, and it comes from forgetting the order of play. Delta is committed before the challenger picks x, so it cannot possibly know what x will be.
Matching
Delta is epsilon over the absolute slope.
Match the pairs
Why: The constant is the degenerate case: the output distance is exactly 0 for every input, so it is below every epsilon no matter how delta is chosen. That the general formula would divide by zero is a hint that this case must be handled separately, and it is the one case where delta need not depend on epsilon at all.
Two truths and a lie
All three are about writing these proofs.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Delta is committed before x is chosen, so a delta depending on x is meaningless — it would be answering a question that has not been asked yet. This is why the quadratic proof must first bound the loose factor with a preliminary restriction on delta, converting a quantity that varies with x into an honest constant.
Prediction
Commit before reasoning.
Predict first
If delta = eps/3 works, does delta = eps/10 also prove the limit?
Correct: Yes. A smaller delta imposes a stricter condition on x, so it still works.
\[ \delta' < \delta \;\text{ and }\; \delta \text{ works} \;\Longrightarrow\; \delta' \text{ works} \]
Why: The definition asks only for the existence of some working delta, not for the best one. Shrinking delta shrinks the set of admissible inputs, and every one of them already satisfied the output condition — so the implication still holds. This is genuinely useful in practice: you never need to optimise, which is why the quadratic proof can take a crude bound like 7 rather than hunting for the sharpest constant.
Section
Section 3
Concept
For a curved graph the difference factors into a small part and a part that still varies. Restricting delta in advance makes the varying part bounded by a constant, and then the argument proceeds as in the linear case.
the preliminary bound — An initial restriction such as delta at most one, imposed so that the factor which does not vanish at the point can be bounded by a constant. The final delta is the minimum of that restriction and the one epsilon demands.
\[ \delta = \min\!\left(1, \tfrac{\varepsilon}{K}\right) \]
Taking a minimum is the standard device for satisfying two demands with one number. It works precisely because, as the previous idea established, a smaller delta is always still valid.
Figure (svg): The quadratic proof, showing why delta must be bounded before the factor can be controlled
OpenStax Calculus Volume 1, §2.5 The Precise Definition of a Limit §2.5, pp. 177-179 — proofs for nonlinear functions
Picture it
The quadratic argument, line by line.
Figure (svg): The quadratic proof, showing why delta must be bounded before the factor can be controlled
The first demand makes the loose factor bounded; the second makes the product small. Neither alone suffices, and the minimum satisfies both at once.
Worked example
Example 2.39. The bound comes first.
\[ \text{Prove that } \lim_{x \to 3} x^2 = 9. \]
Factor the difference
Why: A difference of squares.
\[ | x ^{2} - 9 | = | x - 3 | | x + 3 | \]
Note that one factor is small and one is not
Why: The first vanishes at 3; the second is near 6.
\[ \text{need to bound } | x + 3 | \]
Impose a preliminary restriction
Why: Demanding delta at most 1 confines x to the interval from 2 to 4.
\[ 2 < x < 4,\text{ so } | x + 3 | < 7 \]
Use the bound to control the product
Why: It is less than 7 times the small factor.
\[ | x ^{2} - 9 | < 7 | x - 3 | \]
Demand the second condition and take the minimum
Why: Both restrictions at once.
\[ \delta = \min(1, \varepsilon / 7) \]
Figure (svg): The quadratic proof, showing why delta must be bounded before the factor can be controlled
\[ \delta = \min\!\left(1, \tfrac{\varepsilon}{7}\right) \]
Verify: run the forward argument and check both demands are used
Why: Suppose the distance from x to 3 is positive and less than delta. Since delta is at most 1, x lies between 2 and 4 and the sum x plus 3 is under 7. Since delta is also at most epsilon over 7, the product is under 7 times epsilon over 7, which is epsilon. Both halves of the minimum did work: the first licensed the bound of 7, the second used it. Dropping either leaves the argument incomplete, which is why the minimum is not decoration.
OpenStax Calculus Volume 1, §2.5 The Precise Definition of a Limit §2.5, pp. 178-179
Fill the middle
The quadratic proof, with the loose factor bounded by 7.
Fill in the blanks
\delta = \min\!\left(1, \frac7___}\right)
Why: The bound of 7 on the sum x plus 3 came from restricting delta to at most 1. Both conditions must survive, which is exactly what the minimum arranges.
Worked example
Checkpoint 2.39. The number 1 is a choice, not a law.
\[ \text{Redo the proof with the preliminary restriction } \delta \le \tfrac{1}{2}. \]
Impose the tighter restriction
Why: Now x lies within a half of 3.
\[ 2.5 < x < 3.5 \]
Bound the loose factor again
Why: The sum is under 6.5.
\[ | x + 3 | < 6.5 \]
Demand the second condition
Why: Using the new constant.
\[ | x - 3 | < \varepsilon / 6.5 \]
Take the minimum
Why: Both at once.
\[ \delta = \min(0.5, \varepsilon / 6.5) \]
Figure (svg): The solution to Worked example a different preliminary bound shown as a ladder of expressions, one row per legal move
\[ \delta = \min\!\left(\tfrac12, \tfrac{\varepsilon}{6.5}\right) \]
Verify: confirm that both answers are correct proofs
Why: Both deltas work, and they are different numbers — there is no unique right answer. A tighter preliminary bound gives a smaller constant and hence a larger epsilon-driven delta, but the minimum caps it lower anyway. Since any working delta proves the limit, the choice of preliminary bound is purely a matter of convenience, and 1 is conventional because the arithmetic is easiest.
OpenStax Calculus Volume 1, §2.5 The Precise Definition of a Limit §2.5, pp. 179-179
Trap
\[ |x+3| < 7 \text{ so take } \delta = \tfrac{\varepsilon}{7} \]
Use only the epsilon-driven condition
Why: The student drops the preliminary restriction after using it.
\[ \delta = \tfrac{\varepsilon}{7} \quad \text{(incomplete)} \]
For a large epsilon, say 700, this gives delta equal to 100 — and then x could be anywhere from negative 97 to 103, where the bound of 7 is wildly false.
\[ \delta = \min\!\left(1, \tfrac{\varepsilon}{7}\right) \]
Keep BOTH restrictions by taking their minimum
Why: The bound of 7 was only valid under the preliminary restriction, so that restriction must survive into the answer.
The failure shows up only for large epsilon, which is why it survives casual checking with small ones. But the definition quantifies over every positive epsilon, including the large ones, and a proof that works only for small epsilon is not a proof.
Ranking
Proving a limit for a curved function.
Put in order
Why: Steps b and c must come before d, because the constant in d does not exist until the loose factor has been bounded. Step e is what keeps b alive in the final answer, and dropping it is the standard failure.
Sorting
For the limit of x squared at 3, being 9.
Sort into buckets
Sort each proposed delta.
The last valid option is deliberately crude — epsilon over 100 is far smaller than needed. It still proves the limit, because any working delta suffices. There is no credit for finding the largest one, which is worth knowing before spending time optimising.
Prediction
Commit before reasoning.
Predict first
Why does the quadratic proof need a restriction that the linear proof did not?
Correct: Because the difference has a factor that still varies with x, and it must be bounded before it can serve as a constant.
\[ \text{linear: } |f(x)-L| = |m||x-a| \quad \text{constant factor} \]
\[ \text{quadratic: } |x^2-9| = |x+3||x-3| \quad \text{varying factor} \]
Why: For a linear function the difference is exactly the slope times the input distance, and the slope is a constant already — nothing varies. For the quadratic the difference is the input distance times x plus 3, and that second factor changes with x. Since delta must not depend on x, the factor has to be replaced by a fixed bound, and confining x to a small interval is what produces one. Delta being at most 1 is a convenient choice, not a requirement.
Section
Section 4
Concept
A one-sided limit uses the same definition with the input condition restricted to one side. An infinite limit replaces the output tolerance with a threshold that must be exceeded.
one-sided and infinite definitions — For a right-hand limit the input condition becomes a less than x minus a less than delta. For an infinite limit the challenge is a height M and the requirement is that f of x exceed M throughout the delta band.
\[ \lim_{x \to a^+}f = L: \; 0 < x - a < \delta \Rightarrow |f(x)-L| < \varepsilon \]
In every variant the challenge-and-response structure is untouched. Only the shape of the challenge or the set of admitted inputs changes, which is why the same proof techniques transfer directly.
Figure (svg): The one-sided epsilon-delta picture, with the delta band on only one side of the point
OpenStax Calculus Volume 1, §2.5 The Precise Definition of a Limit §2.5, pp. 179-181 — one-sided and infinite limits
Picture it
The right-hand limit of the square root at zero.
Figure (svg): The one-sided epsilon-delta picture, with the delta band on only one side of the point
Only inputs to the right of the point are considered, because none exist to the left. That single change is the whole difference between the one-sided and two-sided definitions.
Worked example
Example 2.41. Same method, half the inputs.
\[ \text{Prove that } \lim_{x \to 0^+}\sqrt{x} = 0. \]
Write the target inequality
Why: The output within epsilon of 0.
\[ | \sqrt{x} - 0 | < \varepsilon \]
Simplify, noting the root is non-negative
Why: The absolute value does nothing.
\[ \sqrt{x} < \varepsilon \]
Solve for x, legal since both sides are non-negative
Why: Square both sides.
\[ x < \varepsilon ^{2} \]
Declare the choice
Why: The input condition is that x is between 0 and delta.
\[ \delta = \varepsilon ^{2} \]
Figure (svg): The solution to Worked example a one-sided proof shown as a ladder of expressions, one row per legal move
\[ \delta = \varepsilon^{2} \]
Verify: check the squaring and try a value
Why: With epsilon 0.1 the delta is 0.01, and for any x in the interval from 0 to 0.01 the root is under 0.1 — at x equal to 0.009 it is about 0.0949. The squaring was legitimate because both sides were non-negative, which is the one condition to watch when squaring an inequality. Note that delta is much smaller than epsilon here, unlike the linear case: the square root is very steep near 0, so a tight output band demands an extremely tight input band.
OpenStax Calculus Volume 1, §2.5 The Precise Definition of a Limit §2.5, pp. 180-180
Matching
What changes in each variant.
Match the pairs
Why: The first three differ only in the input condition; the fourth differs only in the output condition. The challenge-and-response structure is identical in all four, which is why one set of techniques handles every case.
Worked example
Checkpoint 2.41. The challenge becomes a height.
\[ \text{Prove that } \lim_{x \to 0}\frac{1}{x^2} = \infty. \]
State what must be achieved
Why: Exceed any named height M.
\[ 1 / x ^{2} > M \]
Solve for the input, taking M positive
Why: Invert and take roots.
\[ x ^{2} < \frac{1}{M} \]
Take the square root
Why: The distance from x to 0.
\[ | x | < 1 / \sqrt{M} \]
Declare the choice
Why: The delta answering the challenge M.
\[ \delta = 1 / \sqrt{M} \]
Figure (svg): The solution to Worked example an infinite limit, precisely shown as a ladder of expressions, one row per legal move
\[ \delta = \frac{1}{\sqrt{M}} \]
Verify: test with a specific threshold
Why: Take M equal to 10000. Then delta is 0.01, and for any non-zero x within 0.01 of 0, the square is under 0.0001 and the reciprocal exceeds 10000. The challenge is met. Notice the structure is unchanged from the finite case — a challenge is named, a delta is produced in response — only the form of the challenge differs, which is why calling this a limit at all is reasonable despite infinity not being a number.
OpenStax Calculus Volume 1, §2.5 The Precise Definition of a Limit §2.5, pp. 181-181
Error analysis
A student proves a right-hand limit for the square root at zero.
Annotate
On: \( \text{take } \delta = \varepsilon^2 \text{ and suppose } 0 < |x - 0| < \delta \)
The absolute value is what makes a condition two-sided. Dropping it and writing the difference directly is the entire notational difference between the one-sided and two-sided definitions, and it matters whenever one side lies outside the domain.
Fill the middle
The square root's right-hand limit at zero.
Fill in the blanks
\sqrt\varepsilon^2 < \varepsilon \iff x < ___
Why: Squaring is legitimate because both sides are non-negative. Delta is epsilon squared, which is far smaller than epsilon for small epsilon — a reflection of how steep the square root is near the origin.
Two truths and a lie
All three are about the variants.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and it repeats the quantifier error from the first idea. A larger threshold M demands a tighter delta, exactly as a smaller epsilon does — the response always depends on the challenge. For the reciprocal square, delta is one over the root of M, which visibly shrinks as M grows.
Prediction
Commit before reasoning.
Predict first
For the square root at 0, why is delta so much smaller than epsilon?
Correct: Because the graph is extremely steep near 0, so a tight output band demands a much tighter input band.
\[ \varepsilon = 0.1 \Rightarrow \delta = 0.01; \quad \varepsilon = 0.01 \Rightarrow \delta = 0.0001 \]
Why: Delta is always doing the work of undoing the function's steepness: for a line of slope m it is epsilon over m, and for a curve it reflects how fast the function moves. Near 0 the square root rises almost vertically — its slope grows without bound — so the input band must be far narrower than the output band. This is the same relationship that will become the derivative in Chapter 3, seen here from the other side.
Section
Section 5
Concept
The limit laws of Section 2.3 were stated and used but never proved. With a precise definition available they can be, and the sum law is the standard first example: split the tolerance in half and demand each piece separately.
proving the sum law — Given that f approaches L and g approaches M, one shows that f plus g approaches L plus M by applying each hypothesis with tolerance epsilon over two and taking the smaller of the two resulting deltas.
\[ |f + g - (L+M)| \le |f - L| + |g - M| < \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} = \varepsilon \]
Halving the tolerance is the characteristic trick. Each hypothesis is applied at half strength so that the two errors, when added, still come in under the full budget.
Figure (svg): The challenge-and-response structure, with epsilon shrinking and delta shrinking in reply
OpenStax Calculus Volume 1, §2.5 The Precise Definition of a Limit §2.5, pp. 176-181 — using the definition to prove the limit laws
Picture it
Challenge and response, once for each function.
Figure (svg): The challenge-and-response structure, with epsilon shrinking and delta shrinking in reply
Two separate applications of the definition, each with half the tolerance, and a minimum to make one delta serve both. Every structural device from this section appears in a single short proof.
Worked example
Example 2.42. Half the budget to each.
\[ \text{Prove that if } f \to L \text{ and } g \to M \text{ at } a, \text{ then } f + g \to L + M. \]
Let epsilon be given, and split it
Why: Each function gets half the tolerance.
\[ \text{use } \varepsilon / 2\text{ for each} \]
Apply the first hypothesis
Why: It provides a delta for that tolerance.
\[ \delta _{1}\text{ with } | f - L | < \varepsilon / 2 \]
Apply the second hypothesis
Why: Likewise.
\[ \delta _{2}\text{ with } | g - M | < \varepsilon / 2 \]
Take the minimum so both hold at once
Why: One delta serving both conditions.
\[ \delta = \min(\delta _{1}, \delta _{2}) \]
Combine with the triangle inequality
Why: The total error is at most the sum of the parts.
\[ < \varepsilon / 2 + \varepsilon / 2 = \varepsilon \]
Figure (svg): The solution to Worked example proving the sum law shown as a ladder of expressions, one row per legal move
\[ \delta = \min(\delta_1, \delta_2) \]
Verify: check that halving was necessary
Why: Had each hypothesis been applied with the full epsilon, the sum of the two errors could have reached 2 epsilon, which exceeds the budget. Halving makes the sum come in exactly at epsilon. The same device scales: for a sum of three functions each would get a third. Note both structural tools from earlier in this section reappear — the minimum of two deltas, and the freedom to apply a hypothesis at whatever tolerance is convenient.
OpenStax Calculus Volume 1, §2.5 The Precise Definition of a Limit §2.5, pp. 181-181
Fill the middle
Proving the sum law by applying each hypothesis at reduced tolerance.
Fill in the blanks
|f-L| < \tfrac\varepsilon___ \text___ |g-M| < \tfrac______ \;\Longrightarrow\; |(f+g)-(L+M)| < ___
Why: The two halves sum to epsilon exactly, which is the budget the definition demanded. Applying each hypothesis at full strength would have given 2 epsilon and failed.
Worked example
Checkpoint 2.42. What precision buys.
\[ \text{Why can the sum law not be proved from 'gets arbitrarily close'?} \]
Ask what the informal statement provides
Why: Two vague assertions about closeness.
Ask what must be concluded
Why: A statement about the sum's closeness.
Identify the missing link
Why: There is no way to compare how close is close.
Note what the precise version supplies
Why: A number that can be halved and added.
Figure (svg): The solution to Worked example why the informal definition could not do this shown as a ladder of expressions, one row per legal move
\[ \text{vague closeness cannot be halved; } \varepsilon \text{ can} \]
Verify: identify exactly which step needs the number
Why: The step that fails informally is the combination: knowing each part is 'close' gives no control at all over the sum, since two vague closenesses might add to something not close. Once closeness is a NUMBER, it can be halved in advance so that the sum stays inside the budget. That single manoeuvre is what a precise definition buys, and it is why the two-thousand-year gap Section 2.1 described was a gap in language rather than in insight.
OpenStax Calculus Volume 1, §2.5 The Precise Definition of a Limit §2.5, pp. 181-181
Trap
\[ |f - L| < \varepsilon \text{ and } |g - M| < \varepsilon \]
Add the two bounds
Why: The student applies each hypothesis at full tolerance.
\[ |f+g-(L+M)| < \varepsilon + \varepsilon = 2\varepsilon \quad \text{(budget exceeded)} \]
The conclusion needed was strictly less than epsilon, and 2 epsilon does not deliver it.
\[ \text{apply each at } \tfrac{\varepsilon}{2} \;\Longrightarrow\; \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} = \varepsilon \]
Spend half the budget on each piece
Why: The hypotheses hold for EVERY positive tolerance, so epsilon over 2 is available.
The freedom to apply a hypothesis at any tolerance you like is exactly what the universal quantifier provides, and this is the payoff for insisting on it. For a sum of n functions, each gets epsilon over n — the same device, scaled.
Two truths and a lie
All three are about what the definition provides.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. 'Gets arbitrarily close' supplies no quantity that can be halved, so there is no way to control how two errors combine. The whole reason this section exists is that Chapter 2's results were used before they were proved, and the precise definition is what converts them from plausible to established.
Ranking
The standard structure.
Put in order
Why: Step a must come first because everything after depends on epsilon being fixed but arbitrary. Step c is what makes step d possible, and it relies on the earlier fact that shrinking a working delta keeps it working.
Prediction
Commit before reasoning.
Predict first
Chapter 2 used limits successfully for four sections. What does the precise definition add?
Correct: The ability to prove what was previously assumed, and to settle cases intuition cannot reach.
\[ \text{Section 2.3's laws: used there, proved here} \]
Why: It computes nothing new — the answers are identical, and in practice you will use the laws rather than the definition. What it adds is foundation: the limit laws become theorems, continuity becomes a checkable condition, and pathological cases like the oscillating sine can be settled definitively rather than argued about. It is also the tool for any function too strange for intuition, which is why analysis is built on it rather than on tables.
Comparison
Fill the blanks. Only one clause changes each time.
Comparison matrix
| Limit type | Input condition | Output condition |
|---|---|---|
| Two-sided, finite | 0 < |x-a| < delta | |f(x) - L| < epsilon |
| Right-hand | 0 < x - a < delta | |f(x) - L| < epsilon |
| Left-hand | 0 < a - x < delta | |f(x) - L| < epsilon |
| Infinite | 0 < |x-a| < delta | f(x) > M, for a challenge height M |
Every row has the same shape: a challenge, then a delta answering it. Changing which inputs are admitted gives the one-sided versions; changing what the challenge is gives the infinite one.
Pattern
Given a limit to prove from the definition.
Steps one to four are scratch work and step five is the proof. Keeping them visibly separate is what prevents the two commonest errors: a delta depending on x, and a delta that has quietly dropped its preliminary bound.
Stewart, Calculus: Early Transcendentals 8e, §2.4 The Precise Definition of a Limit §2.4, pp. 104-113
Check
A linear proof. Divide by the slope.
Check your understanding
To prove the limit of 3x - 1 at 2 is 5, which delta works?
Answer: A
Why: The difference is 3 times the distance from x to 2, so that distance must be under epsilon over 3.
Check
A quadratic proof. Both restrictions.
Check your understanding
In proving the limit of x^2 at 3 is 9, why is delta taken as min(1, eps/7)?
Answer: A
Why: Restricting delta to at most 1 puts x between 2 and 4, so x + 3 is under 7; then epsilon over 7 finishes it.
Check
The sum law. Split the budget.
Check your understanding
In proving the sum law, why is each hypothesis applied with tolerance eps/2?
Answer: A
Why: The triangle inequality bounds the total by the sum of the parts, so each part must be under half the budget.
Real world
A machine shop must produce shafts whose cross-sectional area is 9 square centimetres, within a tolerance the customer specifies. The area is the square of the radius, and the shop controls the radius.
Discussion prompt
The customer demands the area be within 0.7 square centimetres of 9. How tightly must the radius be controlled? Identify epsilon, delta, and which is chosen first.
Hint: This is the quadratic proof with units attached.
Answer:
The customer's tolerance is epsilon, and the shop's machining precision is delta. The customer chooses first, and the shop must respond — which is exactly the order of quantifiers in the definition, appearing here as a commercial fact rather than a logical one.
\[ |r^2 - 9| < 0.7 \quad \text{with } r \text{ near } 3 \]
Following the worked example: restricting the radius to within 1 of 3 puts r between 2 and 4, so the sum r plus 3 is under 7. Then it suffices that the radius be within 0.7 over 7, which is 0.1 cm, of 3.
\[ \delta = \min\!\left(1, \tfrac{0.7}{7}\right) = 0.1 \]
The shop can guarantee the area tolerance provided it machines the radius to within a tenth of a centimetre. If the customer tightens epsilon to 0.07, delta drops to 0.01 — and at some point the required delta falls below what the machine can hold, at which point the order must be refused.
That is the practical meaning of the definition's structure: delta depends on epsilon, the dependence is quantified, and there is no single precision that satisfies every possible demand. A shop that quoted one fixed precision for all tolerances would be making exactly the swapped-quantifier error from the first idea.
Commit first
Answer, then rate your confidence honestly.
Predict first
In the epsilon-delta definition, what may delta depend on?
Correct: On epsilon and on the point a, but never on x.
\[ \varepsilon \text{ first} \to \delta(\varepsilon) \text{ second} \to x \text{ last} \]
Why: The order of play settles it: epsilon is named first, delta is chosen second, and x is chosen last from within the delta band. A delta depending on x would have to know a choice that has not been made — which is the single most common error in these proofs, and the reason the quadratic argument must bound its loose factor rather than divide by it. Delta certainly cannot be a fixed constant independent of epsilon, since a tighter tolerance generally needs a tighter band; and f of a is irrelevant, as the limit never uses it.
Explain it
They find the epsilon-delta definition pointless, since they can already compute limits perfectly well with the laws.
Discussion prompt
In four sentences or fewer, explain what the definition is for, given that it computes nothing new.
Hint: Ask where the laws came from.
Answer:
Ask them why the sum law is true. They will say it is obvious, which is what everyone said for two centuries — and being obvious is not the same as being proved, especially once you meet functions that oscillate infinitely often near a point.
The definition is what turns 'gets close' into a number that can be halved, compared and added. That is what makes the laws provable rather than merely believable, and it is what lets you settle a strange function definitively instead of arguing from a picture. You will compute with the laws; the definition is why the laws are allowed.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For quantifiers, say the order aloud: epsilon, then delta, then x. For linear proofs, delta is always epsilon over the absolute slope. For quadratics, always impose delta at most 1 first, bound the loose factor, and take the minimum. For the write-up, start with 'let epsilon be an arbitrary positive number' and never let x appear in delta. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, write the epsilon-delta definition in full, and beneath it draw the box picture: a horizontal band of half-height epsilon around L, a vertical band of half-width delta around a, and a curve threading through. Label which band is the challenge and which the response. Below, prove the limit of 3x minus 1 at 2 is 5, keeping the backwards scratch work on the left of the page and the forwards proof on the right, so the two are visibly separate. In the middle, prove the limit of x squared at 3 is 9 in full, showing the preliminary bound, the constant it produces, and the minimum. Beneath that, write the input conditions for the right-hand, left-hand and infinite variants, marking in each what changed from the two-sided version. At the bottom, prove the sum law, showing where the tolerance is halved and where the minimum is taken. In a margin, write one sentence on why delta may never depend on x.
If your quadratic proof's final delta does not contain a minimum, it is incomplete — check what happens to your bound when epsilon is 700, and you will see the preliminary restriction was doing necessary work.
Recap
Five things, and the first is the definition every result in this chapter has quietly been assuming.
| If you see | Then |
|---|---|
| A linear function | delta = epsilon over the absolute slope |
| A factor still varying with x | Bound it first with a preliminary delta |
| Two conditions on delta | Take their minimum |
| A one-sided limit | Drop the absolute value on the input |
| An infinite limit | The challenge is a height M, not a tolerance |
| A sum of two functions | Give each half the tolerance |
| delta written in terms of x | It is wrong: x is chosen after delta |
That completes Chapter 2. Chapter 3 puts the limit to work: the derivative is a single specific limit, the one Section 2.1 built the secant lines toward, and every rule in that chapter is a consequence of the machinery established here.
OpenStax Calculus Volume 1, §2.5 The Precise Definition of a Limit §2.5, pp. 169-181 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Calculus I — $55/session, free consultation.