MATH1961 · Mathematics 1a (advanced)
Limits and Continuity
This chapter holds the single most important statement in the calculus stream: the precise ε–δ definition of a limit. Mainstream calculus says “f(x) gets close to L as x gets close to a” — a feeling, not mathematics. MATH1961 replaces it with a challenge–response game with a precise winning condition, and asks you to prove limits, not just apply laws. You learn to read the definition as a game (a sceptic names ε, you answer with δ), run the standard play to prove a limit from the definition, use the limit laws and the squeeze theorem to compute, and — the recurring rigour trap — prove a limit does not exist by exhibiting two sequences with different limits. Continuity is then defined as ‘limit equals value’, and the completeness of ℝ is cashed in to prove the Intermediate and Extreme Value Theorems.
What this chapter covers
- 012.1 The ε–δ definition of a limit — reading it as a game
- 02How to PROVE a limit from the definition — the standard backward-then-forward play
- 032.2 Limit laws, the squeeze (sandwich) law & the fundamental limit
- 04Limit DNE by the two-sequence method — the rigour trap
- 05Continuity, and the IVT / EVT proved via completeness
Worked example: prove a limit does not exist (two sequences)
- +1State the strategy. To prove a limit fails to exist, exhibit two sequences xn → 0 (with xn ≠ 0) along which f(xn) tends to different values. One sequence is never enough.
- +1First sequence. Let xn = 1/(nπ). Then xn → 0 and sin(1/xn) = sin(nπ) = 0 for every n, so f(xn) → 0.
- +1Second sequence. Let yn = 1/(2πn + π/2). Then yn → 0 and sin(1/yn) = sin(2πn + π/2) = 1 for every n, so f(yn) → 1.
- +1Conclude. Two sequences both approach 0 but give limits 0 and 1. If limx→0 f(x) existed it would have to equal both, a contradiction — so the limit does not exist. ■
Key terms
- Epsilon–delta definition
- limx→a f(x) = L means: for every ε > 0 there exists δ > 0 such that 0 < |x − a| < δ implies |f(x) − L| < ε. Read as a game: a sceptic names a tolerance ε, you respond with a window δ (chosen after ε, usually depending on it) so every x in that window lands in the ε-band. The strict |x − a| > 0 means the value at a itself is ignored.
- Squeeze (sandwich) theorem
- If g(x) ≤ f(x) ≤ h(x) near a and both g and h tend to the same limit L, then f also tends to L. It is the tool for awkward oscillating limits — the standard application proves the fundamental limit (sin x)/x → 1 as x → 0.
- Limit does not exist (DNE)
- A limit fails to exist when the function approaches different values along different routes to the point. The rigorous proof exhibits two sequences xn, yn → a with f(xn) and f(yn) tending to different limits. Using only one sequence is the recurring exam slip.
- Continuity
- f is continuous at a when limx→a f(x) = f(a): the limit exists, the value exists, and they agree. A removable discontinuity is a point where the limit exists but differs from (or is missing) the value — the limit never looks at f(a).
- Intermediate Value Theorem (IVT)
- If f is continuous on [a, b] and N lies between f(a) and f(b), then f(c) = N for some c in (a, b). In MATH1961 it is proved from the completeness of ℝ (via a supremum), not assumed — which is why the completeness chapter comes first.
Limits and Continuity FAQ
What is the actual method to prove a limit from the ε–δ definition?
Four steps. (1) Take ε > 0 arbitrary. (2) Work backwards from the target |f(x) − L| < ε, bounding it by something simple times |x − a|. (3) Read off a δ (in terms of ε) that makes the bound < ε. (4) Verify forwards: assume 0 < |x − a| < δ and show |f(x) − L| < ε. The marks are in doing backward-design then a clean forward check, with δ chosen after ε.
Why is the quantifier order in the ε–δ definition so important?
Because it is for-all ε then there-exists δ — δ is chosen after, and usually depends on, ε. Reversing it (one δ that works for every ε) describes something false. The whole definition lives in that order, and partial credit on the exam is attached to writing it correctly.
How do I prove a limit does NOT exist?
Find two sequences approaching the point along which the function tends to different limits. For sin(1/x) at 0, one sequence gives constant 0 and another gives constant 1; since a genuine limit would have to equal both, it cannot exist. One sequence is never enough — that single-sequence shortcut is a marked error.
Why does MATH1961 prove the IVT and EVT instead of just stating them?
Because they are not obvious — they are exactly where the completeness of ℝ is used. The IVT is proved by taking a supremum of the set where f stays below the target value; the EVT uses boundedness plus a sup/inf argument. A mainstream student quotes ‘continuous, so IVT’; an Advanced student is expected to build the completeness argument.
Exam move
Memorise the ε–δ definition cold, with the quantifiers in order, and be able to write it without looking. Then drill the four-step proof play on a ladder of examples (linear, then quadratic, then a square root needing rationalisation) until the backward-design step is automatic. Keep a one-page list of the limit laws and the fundamental limit (sin x)/x → 1, and rehearse the two-sequence DNE template on sin(1/x) and (−1)-type oscillations. For continuity and the IVT/EVT, learn the proof via completeness, not just the statement — that is the Advanced premium. State the theorem, name where completeness enters, then build the supremum argument.
