7 min readConcept sheetShort answer; 5-mark set of limits

Limits of functions and sequences

A limit is the value f(x) approaches as x gets close to a number. For a sequence, ask what the terms approach as n grows without bound.

Notationlim (x → a) f(x) = L

First trySubstitute x = a

0/0 formFactorise and cancel, then substitute

At infinityk/x → 0 as x → ∞

Limit rules

Constantlim k = k

Sum / differencelim [f ± g] = lim f ± lim g

Constant multiplelim kf = k lim f

Productlim [f × g] = lim f × lim g

Quotientlim [f ÷ g] = lim f ÷ lim g, if lim g ≠ 0

Polynomiallim (x → a) p(x) = p(a): just substitute

Infinitylim (x → ∞) k/xⁿ = 0 for n > 0

How to find a limit as x → a

Direct substitution and the 0/0 form

Problem

Find (a) lim (x → 2) (3x² − 4x + 1), (b) lim (x → 3) (x² − 9)/(x − 3), (c) lim (x → 2) (x² − 5x + 6)/(x − 2).

Working

(a) A polynomial: substitute. 3(4) − 4(2) + 1 = 12 − 8 + 1 = 5.
(b) Substituting gives 0/0, so factorise: (x² − 9) = (x + 3)(x − 3).
(x + 3)(x − 3)/(x − 3) = x + 3 for x ≠ 3. Limit = 3 + 3 = 6.
(c) x² − 5x + 6 = (x − 2)(x − 3), so the fraction is x − 3 for x ≠ 2.
Limit = 2 − 3 = −1.

Answer

(a) 5 (b) 6 (c) −1

Limits at infinity

Problem

Find (a) lim (x → ∞) (4x² + 3x)/(2x² − 5), (b) the limit of the sequence aₙ = (3n + 1)/(n + 2), (c) the long-run average cost AC = (5,000 + 20x)/x as output x → ∞.

Working

(a) Divide every term by x², the highest power: (4 + 3/x)/(2 − 5/x²).
As x → ∞, 3/x → 0 and 5/x² → 0, so the limit is 4/2 = 2.
(b) Divide by n: (3 + 1/n)/(1 + 2/n) → 3/1 = 3.
(c) AC = 20 + 5,000/x. At x = 100, AC = 70; at x = 1,000, AC = 25; at x = 10,000, AC = 20.5.
As x → ∞, 5,000/x → 0, so AC → 20.

Answer

(a) 2 (b) 3 (c) Rs. 20: average cost falls towards the variable cost per unit.

A famous sequence: (1 + 1/n)ⁿ

n	1	10	100	1,000	n → ∞
(1 + 1/n)ⁿ	2.0000	2.5937	2.7048	2.7169	e = 2.71828…

In Sri Lankan business

As a Sri Lankan garment factory's output grows, its fixed cost is spread over more units, so average cost per unit 20 + 5,000/x falls towards the variable cost of Rs. 20. The limit shows why large orders lower unit cost.

Exam angle

Short answer: “Evaluate lim (x → 2) (x² − 4)/(x − 2).” Factorise: (x + 2)(x − 2)/(x − 2) = x + 2, so the limit is 4.

Short answer: “Find lim (x → ∞) (6x + 1)/(2x − 3).” Divide by x: (6 + 1/x)/(2 − 3/x) → 6/2 = 3.

Structured question: “Evaluate three limits: a polynomial, a 0/0 form and a limit at infinity, and interpret AC = 20 + 5,000/x as x → ∞.” Substitute; factorise and cancel; divide by the highest power. AC → Rs. 20.

Remember it

Substitute first. If you get 0/0, factorise and cancel. If x → ∞, divide by the highest power. A number over infinity is 0.