The number e: growth added in smaller and smaller steps

The problem

A plant grows by 100% in a year: size 1 becomes 2. Now suppose the growth is added in two halves, 50% each time, with each half building on the last. You get 1.5 × 1.5 = 2.25. Split it into 12 monthly steps and you reach 2.613.

Daily steps give 2.7146. Smaller steps always give a bit more, but the total never runs away. It closes in on one special number, 2.71828…, called e.

lim(n→∞) (1 + 1/n)ⁿ = e ≈ 2.718281. Equivalently, lim(x→0) (1 + x)^(1/x) = e. Also lim(x→0) (aˣ − 1)/x = ln a, so lim(x→0) (eˣ − 1)/x = 1.

Key terms

e as a series

Expanding (1 + 1/n)ⁿ by the binomial theorem and letting n → ∞ gives 1 + 1 + 1/2! + 1/3! + 1/4! + …, which adds up to 2.718281…

(1 + x)^(1/x) → e

Put n = 1/x. As n → ∞, x → 0, so the same limit becomes lim(x→0) (1 + x)^(1/x) = e.

Matching the exponent

In (1 + k/n)^(cn), put m = n/k. Then it becomes [(1 + 1/m)^m]^(kc), which tends to e^(kc).

(aˣ − 1)/x → ln a

Put aˣ − 1 = y, so x = logₐ(1 + y) and y → 0 as x → 0. The limit becomes 1/logₐe = ln a. With a = e it is 1.

Short questions with model answers

1. Q1. Express lim(n→+∞) (1 + 3/n)^(2n) in terms of e.

   • m = n/3 ⇒ 3/n = 1/m, 2n = 6m
   • [(1 + 1/m)^m]⁶ → e⁶

   e⁶

2. Q2. Express lim(h→0) (1 + 2h)^(1/h) in terms of e.

   • m = 2h ⇒ 1/h = 2/m
   • [(1 + m)^(1/m)]² → e²

   e²

3. Q3. Find lim(x→0) (2ˣ − 1)/x and check it numerically at x = 0.001.

   • lim(x→0) (aˣ − 1)/x = ln a ⇒ ln 2 = 0.693147
   • (2^0.001 − 1)/0.001 = 0.693387

   ln 2 ≈ 0.6931

Common mistakes

• ✗ Saying (1 + 1/n)ⁿ → 1 because 1 + 1/n → 1 and 1 to any power is 1.

  ✓ The base gets closer to 1 while the power grows without end; the two effects balance at e ≈ 2.718, not 1.

• ✗ Answering e³ or e² for lim (1 + 3/n)^(2n).

  ✓ Both numbers multiply: the 3 inside and the 2 outside give e^(3 × 2) = e⁶.

• ✗ Writing lim(x→0) (aˣ − 1)/x = a.

  ✓ It is ln a, the natural logarithm of a. For a = 2 that is 0.693, not 2.

MCQs

1. 1. lim(n→+∞) (1 + 2/n)ⁿ equals:

   1. (a) e
   2. (b) 2e
   3. (c) e²
   4. (d) 1
   Show answer

   (c) With m = n/2 it becomes [(1 + 1/m)^m]² → e².

2. 2. lim(n→+∞) (1 − 1/n)ⁿ equals:

   1. (a) e
   2. (b) e⁻¹
   3. (c) 0
   4. (d) 1
   Show answer

   (b) Put m = −n: the expression becomes [(1 + 1/m)^m]⁻¹ → e⁻¹.

3. 3. lim(x→0) (eˣ − 1)/x equals:

   1. (a) 0
   2. (b) 1
   3. (c) e
   4. (d) ∞
   Show answer

   (b) ln e = 1, from the result lim (aˣ − 1)/x = ln a.

4. 4. lim(x→0) (1 + 3x)^(2/x) equals:

   1. (a) e⁵
   2. (b) e⁶
   3. (c) e^(2/3)
   4. (d) e³
   Show answer

   (b) With m = 3x, 2/x = 6/m, so [(1 + m)^(1/m)]⁶ → e⁶.

5. 5. lim(x→−∞) eˣ equals:

   1. (a) −∞
   2. (b) 0
   3. (c) 1
   4. (d) e
   Show answer

   (b) eˣ = 1/e⁻ˣ, and e⁻ˣ → ∞ as x → −∞, so eˣ → 0.

Quick revision

• e = lim(n→∞) (1 + 1/n)ⁿ = lim(x→0) (1 + x)^(1/x) ≈ 2.718281.
• (1 + k/n)^(cn) → e^(kc): substitute m = n/k to match the pattern.
• lim(x→0) (aˣ − 1)/x = ln a and lim(x→0) (eˣ − 1)/x = 1.