The number e — definition as limit of (1+1 - n)ⁿ, natural growth context: Worked examples

This page is a drill hall. The parent note the number $e$ told you what $e$ is. Here we hit every kind of problem that definition can produce — growth, decay, the sign of the rate, the degenerate "rate zero" case, the limiting behaviour, a real-world word problem, and an exam-style trap. Nothing will surprise you after this.

The scenario matrix

Before working anything, here is the full landscape. Each row is a case class — a genuinely different situation the topic can hand you. Every worked example below is tagged with the cell it fills.

Cell	Case class	What is special about it	Covered by
A	Bare limit, plug-in	Just evaluate $(1 + 1/ n)^{n}$ for finite $n$	Ex 1
B	Growth, $x > 0$	Exponent positive → factor $> 1$	Ex 2
C	Decay, $x < 0$	Exponent negative → factor between 0 and 1	Ex 3
D	Degenerate, $x = 0$	Rate zero → nothing grows, answer must be $1$	Ex 4
E	Limiting behaviour, $x \to \pm \infty$ (in time)	Long-run: blow-up vs. vanish	Ex 5
F	Solve for the unknown inside the exponent	Requires the inverse, $ln$	Ex 6
G	Real-world word problem	Extract $x = r t$ from English	Ex 7
H	Exam-style twist ( $1^{\infty}$ in disguise)	A limit that looks like it goes to 1 or $\infty$	Ex 8
I	Comparing discrete vs. continuous	Same money, two compounding schemes	Ex 9

We now clear the whole board.

Cell A — the bare limit, watched up close

Worked example Example 1 (Cell A) — Convergence in a table

Compute $(1 + \frac{1}{n})^{n}$ for $n = 1, 5, 50, 500$ and describe the trend.

Forecast: guess before reading — does the sequence rise, fall, or wobble? By how much does it change from $n = 50$ to $n = 500$ ?

$n = 1$ : $(1 + 1)^{1} = 2$ . Why this step? Start at the crudest possible compounding (one lump) — the smallest value the sequence ever takes.
$n = 5$ : $(1 + \frac{1}{5})^{5} = (1.2)^{5} = 2.48832$ . Why this step? Splitting into 5 chunks already gains ~0.49 — interest is now earning interest.
$n = 50$ : $(1.02)^{50} = 2.69159$ . Why this step? Shows the sequence increasing but slowing — the gap to the next row will be tiny.
$n = 500$ : $(1.002)^{500} = 2.71557$ . Why this step? We are now $0.0027$ short of $e = 2.71828$ ; the sequence is crawling up to its ceiling.

Verify: the values increase $2 < 2.488 < 2.692 < 2.716$ and stay below $e \approx 2.71828$ . A monotone-increasing sequence bounded above must converge — exactly the behaviour the definition promises. ✓

Figure — The number e — definition as limit of (1+1 - n)ⁿ, natural growth context

Cell B — positive exponent (growth)

Worked example Example 2 (Cell B) — A growth factor greater than 1

Evaluate the growth factor $e^{0.15}$ and interpret it.

Forecast: since $0.15 > 0$ , will the factor be above or below 1? Roughly how far above?

Recognise $x = 0.15$ in $e^{x} = lim_{n \to \infty} (1 + x / n)^{n}$ . Why this step? The sign of $x$ decides everything: $x > 0$ means every chunk multiplies by something $> 1$ , so the product climbs above 1.
Use the series to get intuition: $e^{0.15} = 1 + 0.15 + \frac{0.1 5 ^{2}}{2} + \dots = 1 + 0.15 + 0.01125 + \dots$ . Why this step? The first two terms already say "a bit more than $1.15$ " — a fast mental sanity guess.
Summing more terms: $e^{0.15} \approx 1.16183$ . Why this step? Later terms ( $0.1 5^{3} /6 \approx 0.00056$ ) are tiny, so three terms nail it.

Verify: $ln (1.16183) \approx 0.15$ , and since $\ln$ is the inverse of $e^{x}$ , this confirms the exponent. The factor $1.16 > 1$ ✓ — positive rate really does grow the quantity.

Cell C — negative exponent (decay)

Worked example Example 3 (Cell C) — When the rate is negative

A radioactive sample decays continuously at rate $r = - 0.1$ per year, starting at $200$ g. Mass after $5$ years?

Forecast: rate is negative. Will $e^{r t}$ be above or below 1? Should the mass end up more or less than 200 g?

Total exponent $x = r t = (- 0.1) (5) = - 0.5$ . Why this step? Same recipe as growth — we still form $x = r t$ — but now $x < 0$ .
Factor $e^{- 0.5} = \frac{1}{e ^{0.5}} = \frac{1}{1.64872} = 0.60653$ . Why this step? A negative exponent is a reciprocal: $e^{- x} = 1/ e^{x}$ , which is always between 0 and 1, i.e. shrinkage.
Mass $= 200 \times 0.60653 = 121.31$ g. Why this step? Multiply the start by the shrink-factor.

Verify: $121.31 < 200$ ✓ (decay lost mass, as a negative rate must). Units: $g \times (dimensionless factor) = g$ ✓. This is $\frac{d N}{d t} = k N$ with $k = - 0.1$ — see Differential equations dy/dx = ky.

Cell D — the degenerate case, $x = 0$

Cell E — limiting behaviour in time

Cell F — solving for the exponent (needs $ln$ )

Cell G — real-world word problem

Worked example Example 7 (Cell G) — Reading

$x = r t$ out of English "A bacterial culture grows continuously, tripling every 6 hours. It starts at 400 cells. How many after 15 hours?"

Forecast: 15 hours is $2.5$ tripling-periods. Between $3^{2} = 9$ and $3^{3} = 27$ times the start — guess a number now.

Find the rate $r$ from "triples every 6 h": $e^{6 r} = 3 \Rightarrow r = \frac{ln 3}{6} = \frac{1.09861}{6} = 0.18310$ /h. Why this step? The English gives a ratio over a time, not a rate directly — we must invert with $ln$ (Cell F skill) to extract $r$ .
Total exponent over 15 h: $x = r t = 0.18310 \times 15 = 2.74653$ . Why this step? Now assemble the standard exponent $x = r t$ for the actual question window.
Population $N = 400 e^{2.74653} = 400 \times 15.5885 = 6235.4$ , i.e. about 6235 cells.

Verify: cross-check via ratios only — $15 h = 2.5$ periods, so factor $= 3^{2.5} = 15.588$ , giving $400 \times 15.588 = 6235$ ✓. Matches to the cell. Also $9 < 15.6 < 27$ as forecast ✓.

Cell H — exam-style $1^{\infty}$ twist

Worked example Example 8 (Cell H) — A limit disguised as "obviously 1 or ∞"

Evaluate $L = n \to \infty lim (1 + \frac{3}{n})^{2 n}$ .

Forecast: the inside $\to 1$ , the power $\to \infty$ . The trap says "answer is 1" or "answer is ∞". It is neither — guess the finite value.

Spot the form: base $\to 1$ , exponent $\to \infty$ — a genuine $1^{\infty}$ indeterminate form, so we may not evaluate the two parts separately. Why this step? Recognising the indeterminacy is what stops you writing the wrong "1". See Limits and indeterminate forms.
Rewrite to match the template $e^{x} = lim (1 + x / n)^{n}$ . First $(1 + \frac{3}{n})^{2 n} = [(1 + \frac{3}{n})^{n}]^{2}$ . Why this step? Splitting the exponent isolates the exact block $(1 + 3/ n)^{n}$ we know converges — to $e^{3}$ .
Take limits: $(1 + \frac{3}{n})^{n} \to e^{3}$ , so $L = (e^{3})^{2} = e^{6}$ . Why this step? Powers pass through the limit for a convergent bracket. Numerically $e^{6} = 403.43$ .

Verify: test $n = 10000$ : $(1 + 0.0003)^{20000} = e^{20000 l n (1.0003)} = e^{20000 \times 0.00029996} = e^{5.9991} \approx 403.2$ , hugging $e^{6} = 403.43$ ✓ — decisively not 1 and not infinite.

Cell I — discrete vs. continuous, side by side

Worked example Example 9 (Cell I) — Does "continuous" really beat "monthly"?

£1000 at $12%$ for 1 year, compared: (a) compounded monthly, (b) compounded continuously.

Forecast: how many pence apart will they be — 0p, a few pounds, or tens of pounds?

Monthly: $A = 1000 (1 + \frac{0.12}{12})^{12} = 1000 (1.01)^{12}$ . Why this step? Discrete compounding uses a finite $n = 12$ in $(1 + x / n)^{n}$ — no limit taken.
Compute: $(1.01)^{12} = 1.126825$ , so $A = £1126.83$ .
Continuous: $A = 1000 e^{0.12} = 1000 \times 1.127497 = £1127.50$ . Why this step? Continuous is the $n \to \infty$ limit — the ceiling of the same process, so it must be slightly larger.

Verify: $1127.50 > 1126.83$ ✓ (continuous is the limit, hence the maximum). The gap is only £0.67 — confirming the parent note's point that ever-finer compounding adds ever-smaller amounts. See Compound interest.

Active recall

Recall Which cell is which? (open after attempting)

What is the only tool that "solves for a trapped exponent"? → $ln$ , the inverse of $e^{x}$ .
What is $(1 + \frac{a}{n})^{bn}$ in the limit? → $e^{ab}$ .
Why is $e^{0} = 1$ not an indeterminate form? → The base is exactly 1 for all $n$ ; nothing fights.
$r < 0$ : does $e^{r t}$ sit above or below 1? → Below 1 (a reciprocal), so decay.
As $t \to \infty$ , $A_{0} e^{r t}$ tends to? → $\infty$ if $r > 0$ , $0$ if $r < 0$ , constant if $r = 0$ .

Term to reveal:

Growth factor for total exponent

x

e^{x}

, which is

> 1

for

x > 0

, between 0 and 1 for

x < 0

, and exactly 1 for

x = 0

Connections

Natural logarithm ln x — the inverse used in Cells F and G to free the exponent
Exponential function e^x and its derivative — why $e^{r t}$ solves growth exactly
Compound interest — the discrete-vs-continuous contrast of Cell I
Differential equations dy/dx = ky — the source of $A_{0} e^{r t}$ (growth and decay)
Binomial theorem — the series intuition used in Cell B
Limits and indeterminate forms — the $1^{\infty}$ machinery behind Cells A, D, H

Worked examples — The number e — definition as limit of (1+1 - n)ⁿ, natural growth context