Lesson 18: Limits and Continuity (Introduction)

The limit is calculus' founding idea: the value a function approaches as its input approaches a point — whether or not it ever arrives. China met it early: Liu Hui cut the circle with polygons of 3,072 sides, writing "the finer the cut, the smaller the loss; cut again and again until it cannot be cut, and it coincides with the circle" — a limit argument in words. Zhuangzi's stick, halved daily and never exhausted, is an infinite series wearing philosophical clothes. This lesson builds limits rigorously: the intuitive definition, one-sided and infinite limits, continuity, the squeeze theorem, and the ε-δ definition that makes "approaches" precise — the gateway to Lesson 19's derivatives.

The limit concept

lim(x→a) f(x) = L means: as x comes arbitrarily close to a (without ever equalling a), f(x) comes arbitrarily close to L. The value f(a) is irrelevant — it may differ from L or not exist. For f(x) = (x² − 1)/(x − 1) at a = 1: the function is undefined at 1, yet for x near 1 it equals x + 1 (the factor (x − 1) cancels because x ≠ 1), so the limit is 2. That cancellation, legal only because the limit never touches the point, is the whole secret of differential calculus.

One-sided limits: |x|/x approaches −1 from the left and +1 from the right at 0; the two-sided limit exists only when both one-sided limits agree, so it does not exist here. Infinite limits and limits at infinity: 1/x² → ∞ as x → 0; (3x² + 1)/(2x² − x) → 3/2 as x → ∞ (divide top and bottom by x²: leading coefficients rule). Indeterminate forms 0/0, ∞/∞, ∞ − ∞ are not answers but instructions: factor, rationalise, or divide by the highest power. The squeeze theorem: if g ≤ f ≤ h near a and both bounds tend to L, so does f — the proof that lim(x→0) sin x/x = 1 stands on it.

Continuity at a requires all three: f(a) defined, lim(x→a) f(x) exists, and they are equal. Failures classify as removable (a hole or a misdefined point), jump (one-sided limits disagree), or infinite (a vertical asymptote). Continuity's payoff is the Intermediate Value Theorem: a continuous function on [a, b] takes every value between f(a) and f(b) — existence of roots without finding them.

Worked examples

Worked example 1 — factor and cancel. lim(x→2) (x² − 4)/(x − 2): direct substitution gives 0/0 (indeterminate, not "undefined"). Factor: (x − 2)(x + 2)/(x − 2) = x + 2 for x ≠ 2. Limit: 4.

Worked example 2 — rationalise. lim(x→0) (√(1 + x) − 1)/x. Multiply top and bottom by the conjugate √(1 + x) + 1: numerator becomes (1 + x) − 1 = x; cancel x (legal, x ≠ 0): 1/(√(1 + x) + 1) → 1/(1 + 1) = 1/2. (Lesson 19 will recognise this as the derivative of √x at x = 1.)

Worked example 3 — squeeze. lim(x→0) x² sin(1/x) = 0. Since −1 ≤ sin(anything) ≤ 1, we have −x² ≤ x² sin(1/x) ≤ x², and both bounds → 0. By the squeeze theorem the limit is 0 — even though sin(1/x) itself oscillates infinitely fast and has no limit. Bounding beats computing when the function misbehaves.

Worked example 4 — ε-δ, gently. Prove lim(x→3) 2x = 6. The definition: for every ε > 0 there is δ > 0 such that 0 < |x − 3| < δ implies |f(x) − 6| < ε. Compute: |2x − 6| = 2|x − 3|. We need 2|x − 3| < ε, i.e. |x − 3| < ε/2. Choose δ = ε/2; then the implication holds by construction. ∎ Liu Hui's phrasing maps exactly: "for any demanded precision ε, cut finely enough (δ)". For x² the same proof needs one preliminary bound: if δ ≤ 1 then |x + 2| < 5 near x = 3... in general, bound the extra factor first, then choose δ = min(1, ε/bound).

Worked example 5 — classify a discontinuity. f(x) = (x² − 1)/(x − 1) for x ≠ 1, with f(1) = 3 defined separately. The limit at 1 is 2 (above), but f(1) = 3: limit ≠ value, so f is discontinuous at 1 — removably: redefining f(1) = 2 alone repairs it. Compare |x|/x at 0 (jump, unrepairable: the two sides disagree) and 1/x² at 0 (infinite). Three species; name them on sight.

Challenge problem — Gaokao standard

Problem. Prove that x³ − 2x − 5 = 0 has a root between 2 and 3, and locate it within an interval of length 1/8 by bisection.

Solution. f(x) = x³ − 2x − 5 is a polynomial, hence continuous everywhere. f(2) = 8 − 4 − 5 = −1 < 0 and f(3) = 27 − 6 − 5 = 16 > 0. By the Intermediate Value Theorem, a root lies in (2, 3). Bisect: f(2.5) = 15.625 − 5 − 5 = 5.625 > 0, so the root is in (2, 2.5). f(2.25) = 11.39 − 4.5 − 5 = 1.89 > 0: root in (2, 2.25). f(2.125) = 9.596 − 4.25 − 5 = 0.346 > 0: root in (2, 2.125), an interval of length 0.125 = 1/8. (The root is ≈ 2.094.) Existence by IVT, location by bisection — the theory and the algorithm in one problem, and the pattern behind every numerical root-finder in Lesson 19.

Common mistakes

• "0/0 = the limit does not exist." 0/0 is a form, not a verdict (Worked example 1: the limit is 4). Indeterminate means "work required".
• Confusing limit with value. f(1) = 3 while the limit is 2 (Worked example 5). The limit never inspects the point itself.
• ∞ − ∞ treated as 0. √(x² + 1) − x → 0 only after rationalising (it equals 1/(√(x² + 1) + x)); other ∞ − ∞ forms tend elsewhere. The form is indeterminate; each case is computed, not assumed.
• Ignoring one-sided limits. Declaring a limit at a jump (like |x|/x at 0) because "both sides approach something" — they must approach the same something.
• Cancelling (x − a) without justification. It is legal precisely because the limit considers x ≠ a. Stating that reason is worth marks; the limit's definition is doing real work.

Exam Experience

Multiple choice. lim(x→2) (x² − 4)/(x − 2) equals: A. 0 B. 2 C. 4 D. does not exist Answer: C. Factor and cancel (valid for x ≠ 2), leaving x + 2 → 4. Option D is the 0/0 trap.

Fill in the blank. lim(x→∞) (5x² + 3)/(2x² − x) = ____. Answer: 5/2. Divide numerator and denominator by x²: (5 + 3/x²)/(2 − 1/x) → 5/2.

Solution question (10 marks). (1) Compute lim(x→1) (x³ − 1)/(x − 1). (2) Using ε-δ, prove that lim(x→1) (x² + x + 1) = 3. Marking scheme. (1) Factor: x³ − 1 = (x − 1)(x² + x + 1) — 2 marks. Cancel (x ≠ 1 in the limit) and substitute: 1 + 1 + 1 = 3 — 2 marks. [4 marks] (2) Given ε > 0. Estimate |(x² + x + 1) − 3| = |x² + x − 2| = |x − 1| · |x + 2| — 2 marks. Restrict δ ≤ 1, so |x − 1| < 1 gives 0 < x < 2 and |x + 2| < 4 — 2 marks. Choose δ = min(1, ε/4): then |x − 1| < δ implies |x² + x − 2| < 4 × (ε/4) = ε — 2 marks, completing the proof. [6 marks] Total: 10 marks. Part (2)'s structure — estimate, bound the extra factor, choose δ — is the universal ε-δ template; learn it once, apply it everywhere.

Going deeper — two famous limits, and why maxima exist

The trigonometric limit. lim(x→0) (sin x)/x = 1 (x in radians — the statement is false in degrees, which is one reason calculus adopts radians). The squeeze proof compares three areas on the unit circle for 0 < x < π/2: the inner triangle (area (1/2) sin x cos x... in its standard form: (1/2) sin x < (1/2) x < (1/2) tan x, dividing by (1/2) sin x gives 1 < x/sin x < 1/cos x, and inverting squeezes sin x/x between cos x and 1, both → 1. ∎ Consequence: lim(x→0) (sin 3x)/x = 3 — substitute u = 3x and use sin u/u → 1: (sin 3x)/x = 3 × (sin 3x)/(3x) = 3 × sin u/u → 3. Nearly every trigonometric derivative traces back to this one limit (Lesson 19's (sin x)′ = cos x is proved through it).

The number e. lim(n→∞) (1 + 1/n)ⁿ = e ≈ 2.71828: the compound-interest limit — Lesson 13's geometric growth with the compounding made continuous. It pairs with the exponential's self-derivative (eˣ)′ = eˣ to anchor all growth models. Existence of extrema (Extreme Value Theorem): a continuous function on a closed bounded interval [a, b] attains a maximum and a minimum. The engine is Bolzano–Weierstrass: every bounded sequence has a convergent subsequence — proved by repeatedly bisecting the interval and keeping the half containing infinitely many terms (completeness of the reals does the rest). This theorem is the silent licence for Lesson 19's "set f′ = 0 to find the maximum": it guarantees there is a maximum, attained either inside (where Fermat forces f′ = 0 or undefined) or at the endpoints. Rigor is not decoration; it is the reason the algorithm works.

Second look — two limits worked to the end

Limit 1. lim(x→∞) (√(x² + 1) − x). The form is ∞ − ∞ (indeterminate — Lesson 18's warning). Rationalise with the conjugate: (√(x² + 1) − x) × (√(x² + 1) + x)/(√(x² + 1) + x) = [(x² + 1) − x²]/(√(x² + 1) + x) = 1/(√(x² + 1) + x). As x → ∞ the denominator → ∞, so the limit is 0. (Rate check: the gap behaves like 1/(2x) — the curve y = √(x² + 1) approaches its asymptote y = x with a gap that halves as x doubles.)

Limit 2. lim(x→0) (tan x)/x. Write tan x = sin x/cos x: the quotient is (sin x/x) × (1/cos x). By the extension's famous limit, sin x/x → 1; and 1/cos x → 1/1 = 1. Product of limits: 1. The limit laws license splitting into factors exactly when each factor's limit exists and the denominator's is nonzero — both checked. Two morals: ∞ − ∞ and 0/0 are starting signals for the conjugate and factorisation tools, and complicated limits are chains of famous ones. Keep the famous five at your fingertips: sin x/x → 1, (1 + 1/n)ⁿ → e, x² sin(1/x) → 0 by squeeze, rational functions at ∞ by leading terms, and √(x² + 1) − x → 0 by conjugation.