Limits: where a function is heading

The problem

How do you find the area of a circle with only straight lines? Draw a square inside a unit circle: area 2. An octagon: 2.828. A 16-sided polygon: 3.061. Keep doubling the sides and the area creeps towards 3.142, which is π.

No polygon ever equals the circle, yet their areas head to exactly one number. That number is a limit. Calculus is built on this idea.

Suppose f(x) approaches a specific number L as x approaches a from both the left and the right. Then L is the limit of f(x) as x → a: lim(x→a) f(x) = L. f need not be defined at a itself.

Key terms

x → 0

x takes values like 1, ½, ¼, ⅛ … getting as small as we please. x → 0 means x is very close to zero but not actually zero; x = 0 means it is zero.

x → a

x gets as close as we please to a from both the left and the right, but x − a ≠ 0.

x → ∞

x takes values like 1, 10, 100, 1000 … growing as large as we please. ∞ is not a number; it describes this unending increase.

Theorems on limits

Let lim f = L and lim g = M. Then the limit of a sum, difference, multiple, product or power is the sum, difference, multiple, product or power of the limits. For a quotient it is L/M, provided M ≠ 0.

Short questions with model answers

1. Q1. Inscribed polygons in a circle of radius 1 have areas 2 (4 sides), 2√2 (8 sides) and 3.061 (16 sides). What is the limit of the area as the number of sides n → ∞?

   • 2 → 2.828 → 3.061 → …
   • Area of circle = πr² = π(1)² ≈ 3.142

   lim(n→∞) area = π ≈ 3.142

2. Q2. Evaluate lim(x→2) (3x + 4)/(x + 3) using the theorems.

   • lim(3x + 4) = 10, lim(x + 3) = 5
   • 10 ÷ 5 = 2

   2

3. Q3. Evaluate lim(x→−2) (2x³ + 5x)/(3x − 2).

   • 2(−2)³ + 5(−2) = −16 − 10 = −26
   • 3(−2) − 2 = −8
   • −26 ÷ −8 = 13/4 = 3.25

   13/4

Common mistakes

• ✗ Treating x → a as if it meant x = a.

  ✓ x → a means x is as close to a as we please but never equal to it. The limit is about the approach, not the value at a.

• ✗ Using the quotient theorem when the denominator's limit is zero.

  ✓ Theorem 5 needs M ≠ 0. If substitution gives 0/0, simplify first by factorising or rationalising (1.5).

• ✗ Saying the limit cannot exist because f(a) is undefined.

  ✓ The definition says f need not be defined at a. (x² − 1)/(x − 1) is undefined at 1, yet its values close in on 2.

MCQs

1. 1. lim(x→3) (2x + 4) equals:

   1. (a) 6
   2. (b) 7
   3. (c) 10
   4. (d) 12
   Show answer

   (c) Substitute x = 3: 2(3) + 4 = 10.

2. 2. lim(x→1) (3x² − 2x + 4) equals:

   1. (a) 3
   2. (b) 5
   3. (c) 9
   4. (d) 1
   Show answer

   (b) 3 − 2 + 4 = 5: for a polynomial, lim P(x) = P(c).

3. 3. lim(x→3) √(x² + x + 4) equals:

   1. (a) 4
   2. (b) 16
   3. (c) √13
   4. (d) 5
   Show answer

   (a) 9 + 3 + 4 = 16, and √16 = 4.

4. 4. The quotient theorem lim f/g = L/M holds provided:

   1. (a) L ≠ 0
   2. (b) M ≠ 0
   3. (c) L = M
   4. (d) a ≠ 0
   Show answer

   (b) The limit of the denominator, M, must not be zero.

5. 5. “x → 0” means:

   1. (a) x is zero
   2. (b) x is very close to zero but not zero
   3. (c) x is negative
   4. (d) x is infinite
   Show answer

   (b) It describes an unending approach to zero; x itself is never 0.

Quick revision

• lim(x→a) f(x) = L means f(x) closes in on L as x approaches a from both sides; f(a) need not exist.
• Six theorems: limits of sums, differences, multiples, products, quotients (M ≠ 0) and powers.
• For a polynomial, lim(x→c) P(x) = P(c): just substitute.