4.1.9 · D5Calculus I — Limits & Derivatives
Question bank — Epsilon-delta definition of a limit — formal proofs
Reminder of the machine you are testing everything against:
True or false — justify
The value of that works for a proof is the only correct choice
False. Any smaller positive also works — if a band of width keeps outputs inside, so does any narrower band. The game asks you to find one , not the largest.
Once you find a that works for a given , a smaller will fail
False. Shrinking only restricts to points even closer to , whose outputs were already inside the -band. Smaller is always safe; larger is what risks failing.
If then
False. The strict excludes entirely, so the limit says nothing about — may be undefined at or defined to some unrelated value. Equality is the extra condition called Continuity.
" gets closer and closer to " is enough to guarantee
False. Getting closer is not the same as getting arbitrarily close: could approach steadily and still "get closer" from far away. The definition demands you beat every , not merely decrease the gap.
is allowed to depend on
True — and it usually must. The quantifier order "" means is chosen after seeing , so writing or is exactly correct.
is allowed to depend on
False. You commit to before the adversary picks inside the -band, so an -dependent is not a legal move — it answers a question that hasn't been asked yet.
If for a fixed some works, the limit exists
False. One succeeding proves nothing; the limit exists only if you can produce a working for every . Failing even one (as with at ) kills the limit.
If no single can satisfy the definition, then has no limit at
True. The limit, when it exists, is unique — so "no works" is precisely the negation, meaning the limit does not exist.
The definition requires to be defined on all of
False (in spirit). It only constrains with ; need not be defined at , and we only ever require defined on the punctured neighbourhood we actually use.
Making smaller forces smaller
Usually true but not logically required. Smaller means a tighter output band, which typically needs tighter input control; but for a constant function any works for every , so no shrinking is forced.
Spot the error
"Proof that : choose ." — what's wrong?
contains , but must be a single number fixed before is picked. Fix: bound via a preliminary , then use .
"Let . I'll find an close to with close to , done." — why isn't this a proof?
Finding one good is trivial and meaningless. You must guarantee every in the punctured -band lands in the -band — a universal statement over , not an existence one.
"Since , and , and , we get " — the preliminary step is missing. Why does that break it?
The bound is only valid after imposing ; without the factor could be huge (e.g. ), and would be false.
"Choose so both bands match" — what confusion is this?
It treats and as the same tolerance. They live on opposite sides: measures outputs, measures inputs, and is a response to , not equal to it in general.
" because is undefined at ." — why is that reasoning invalid?
Being undefined at is irrelevant (the limit ignores ). The real reason the limit fails is that just left of the function is and just right it is , so no single band of half-width under can catch both.
"To negate the limit, I keep and just add 'not' at the end." — what's the actual negation?
Negation flips the quantifiers: with and . You must produce a bad and then defeat every .
"I proved , so the limit holds." — is enough?
The definition demands strict . Usually harmless (run the argument with , or note the inequality was strict earlier), but a bare at the boundary is technically not what was required — state or fix the constant.
Why questions
Why is the inequality on the input written rather than just ?
The extra removes the single point , so the limit describes the approach, not the value at the point. This is what lets limits detect removable holes.
Why must be strictly positive, never zero?
would leave the punctured interval empty, making the implication vacuously true for any — the definition would be useless. A positive forces you to actually control a neighbourhood of real inputs.
Why do we always try to rewrite as (factor)?
Because is the quantity directly controls. Exposing it as an explicit factor converts "input is close" into "output is close," which is the whole conversion the proof needs.
Why does a constant leftover factor make the proof easy but a variable factor need extra work?
A constant gives a clean . A variable factor depends on , so you must first trap it under a fixed bound (via a preliminary restriction) before it can play the role of a constant.
Why take instead of adding the two conditions?
You need both conditions to hold simultaneously, and guarantees is small enough to satisfy the tighter of the two. Adding them would give a larger that could violate one condition.
Why does the quantifier order (ε before δ) matter so much?
It sets who moves first: the adversary reveals the target band , then you respond with . Swapping to "" would (wrongly) demand one that beats every tolerance at once.
Why is the natural choice to disprove ?
The two one-sided values and are distance apart, so any band of half-width cannot contain both. Choosing is the largest bound that still forces the contradiction.
Why can a smaller never hurt a valid proof?
Shrinking only keeps a subset of the previously-allowed inputs, all of which already produced outputs inside the -band. The implication stays true on any smaller punctured interval.
Edge cases
For a constant function , what works and why?
Any works, because automatically for every . This is the degenerate case where need not depend on at all.
Does require to be defined at ?
No. Only the punctured neighbourhood matters, so may be undefined; e.g. despite the hole at .
If left and right approaches give different values, what does the two-sided definition say?
No single can satisfy it, so the two-sided limit does not exist — this is exactly why One-sided limits are defined separately, splitting into and .
Can a limit exist even if is wildly discontinuous at (a single misplaced point)?
Yes. Redefining at the single point changes nothing, because the definition never inspects ; the limit sees only the surrounding punctured interval.
What breaks if a function oscillates without settling near , like at ?
For every punctured interval around contains points where hits both and , so no traps the outputs in one -band and the limit fails to exist.
How does the epsilon–delta idea change for Uniform continuity?
The same – game is played, but must work for all points simultaneously — it may depend on but not on the location .
How does the definition of a limit hide inside the Definition of the derivative?
The derivative is , an ordinary – limit of the difference quotient — same challenge–response game with the quotient playing the role of .
What is the discrete cousin of in Sequences and their limits?
The index cutoff : instead of " within of " you demand "," and (like ) is chosen in response to .
Recall One-line self-test
If someone hands you a candidate limit and a proof, ask three questions in order: (1) Is a single number free of ? (2) Does it come after and only depend on ? (3) Does the conclusion give strict for every in the punctured band? Any "no" is a trap sprung.
Connections
- Epsilon-delta definition of a limit — formal proofs — the parent recipe these traps stress-test.
- Continuity — the trap lives exactly on the border between limit and continuity.
- One-sided limits — resolves the "different left/right values" edge case.
- Definition of the derivative — an – limit in disguise.
- Uniform continuity — same game, stricter rule.
- Sequences and their limits — swap for .