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

Recall the objects in play, so no symbol here is unearned:

$n$ = how many times we chop a growth period into equal compounding chunks.
$(1 + \frac{1}{n})^{n}$ = value after compounding $n$ times at 100% total rate.
$e = lim_{n \to \infty} (1 + \frac{1}{n})^{n} \approx 2.71828$ — the ceiling that value creeps toward.
$e^{x} = lim_{n \to \infty} (1 + \frac{x}{n})^{n}$ — same idea at a general rate $x$ .
The series $e = \sum_{k = 0}^{\infty} \frac{1}{k !} = 1 + 1 + \frac{1}{2 !} + \frac{1}{3 !} + \dots$

If any of those feel shaky, revisit the parent note first.

True or false — justify

The limit

lim_{n \to \infty} (1 + \frac{1}{n})^{n}

equals

1

because the inside tends to

1

False. This is a

1^{\infty}

indeterminate form: the base sinks toward

1

while the exponent blows up. The two effects fight, and the balance point is

e

, not

1

The sequence

(1 + \frac{1}{n})^{n}

eventually overshoots

e

and comes back down.

False. It is strictly increasing yet bounded above by

3

, so it climbs toward

e

from below and never exceeds it.

Since

e

is irrational, the series

1 + 1 + \frac{1}{2 !} + \frac{1}{3 !} + \dots

must be infinite in value.

False. Irrational just means "no exact fraction/no repeating decimal." The sum is finite (about

2.718

); irrationality is about the decimal pattern, not the size.

e = 2.718

exactly, so calculators are just being fussy with the extra digits.

False.

e

is irrational (in fact transcendental); its decimal never terminates or repeats.

2.718

is a rounding, not the value.

Compounding £1 at 100% "infinitely often" gives infinite money.

False. Each extra split adds a smaller increment (the

k

-th correction is

\frac{1}{k !}

, crushed by factorials), so the total is bounded — it lands at

e \approx 2.72

The formula

e^{x} = lim_{n \to \infty} (1 + \frac{x}{n})^{n}

only makes sense for positive

x

False. It holds for any real

x

. For

x < 0

it models continuous decay; e.g.

e^{- 1} = lim (1 - \frac{1}{n})^{n} \approx 0.368

Plugging

x = 0

into

(1 + \frac{x}{n})^{n}

correctly gives

e^{0} = 1

True. With

x = 0

the base is exactly

1

for every

n

, so

1^{n} = 1

throughout, and the limit is

1

— no indeterminacy here because the base never actually approaches

1

, it is

1

Doubling the compounding frequency

n

roughly doubles the final amount.

False. Going from

n = 100

n = 1000

moves the value only from

2.705

2.717

. The gains shrink toward zero because we're already near the ceiling

e

Spot the error

(1 + \frac{1}{n})^{n} = 1 + \frac{n}{n} = 1 + 1 = 2

, so the limit is

2

The exponent does not distribute over a sum — that's repeated multiplication, not a single addition. Expanding honestly (binomial theorem) leaves many positive terms beyond

2

, pushing the value up to

e \approx 2.72

"As

n \to \infty

the base

\to 1

, and the power

\to \infty

; a big power of

1

is still

1

, so limit

= 1

The base is never actually

1

during the limit — it's

1 + \frac{1}{n}

, slightly above

1

. A number slightly above

1

raised to a huge power can grow; you cannot evaluate base and exponent separately in a

1^{\infty}

form.

e^{r t}

for £1000 at 5% over 3 years uses

r = 5

, so the exponent is

5 \times 3 = 15

The rate must be a decimal fraction:

r = 0.05

, giving exponent

0.15

, not

15

. Using

5

inflates the answer astronomically. Percentages become fractions before entering

e

"In the series derivation,

\frac{n ( n - 1 ) \dots ( n - k + 1 )}{n ^{k}} \to 0

because the numerator has fewer factors than

n^{k}

It has exactly

k

factors, matching the

k

powers of

n

in the denominator. Written as

1 \cdot (1 - \frac{1}{n}) (1 - \frac{2}{n}) \dots

, it tends to

1

, not

0

— that's why term

k

becomes

\frac{1}{k !}

"Compounding continuously means adding the interest infinitely many times, so use

lim (1 + \frac{x}{n}) \cdot n

Continuous compounding is repeated multiplication — the

n

is an exponent, not a factor. The correct object is

(1 + \frac{x}{n})^{n}

; interest earns interest, which multiplication captures and

\times n

does not.

Why questions

Why do we need a limit at all — why not just pick a very large fixed

n

No fixed

n

ever reaches the exact value; each finite

n

still undershoots. The limit names the precise ceiling that all those approximations approach, which is a specific irrational number.

Why do the correction terms

\frac{1}{k !}

guarantee the total stays finite?

Factorials grow faster than any geometric ratio, so from some point on

\frac{1}{k !}

is smaller than

\frac{1}{2 ^{k}}

. Since

\sum \frac{1}{2 ^{k}}

is finite, the smaller-termed series is finite too — bounded below

3

Why does the number

e

show up in bacteria and money and radioactive decay alike?

All obey "rate of change is proportional to current amount," i.e.

\frac{d y}{d x} = k y

. Its solution

y = y_{0} e^{k x}

comes directly from the continuous-compounding limit, so

e

is the universal base of self-feeding growth. See Differential equations dy/dx = ky.

Why is

e

the "natural" base rather than, say,

10

2

Because

\frac{d}{d x} e^{x} = e^{x}

: the exponential with base

e

is its own rate of change, so growth described by

e

needs no messy conversion factor. Any other base

b

drags along an extra

ln b

. See Exponential function e^x and its derivative.

Why does the binomial theorem turn the limit into a clean sum of

\frac{1}{k !}

Expanding

(1 + \frac{1}{n})^{n}

gives terms

(k n) \frac{1}{n ^{k}}

; the

n

-dependent parts each

\to 1

n \to \infty

, leaving just

\frac{1}{k !}

per term. So the messy

n

vanishes and a fixed series remains. See Binomial theorem.

Why can't we just say "the base is near

1

, so approximate

(1 + \frac{1}{n})^{n} \approx 1

Because it's an indeterminate form

1^{\infty}

: the tiny excess

\frac{1}{n}

gets amplified

n

times, and

n \cdot \frac{1}{n} = 1

worth of "growth" survives — enough to build

e

, not

1

Edge cases

What does

(1 + \frac{x}{n})^{n}

tend to when

x = 1

Exactly

e

. Setting

x = 1

recovers the basic definition

lim (1 + \frac{1}{n})^{n}

— the

x

-version is just the general case, and

x = 1

is the special one.

What happens to

e^{x}

x \to - \infty

It tends to

0

(but never reaches it). Negative

x

means continuous decay; the larger the negative exponent, the closer to zero the value, staying strictly positive.

(1 - \frac{1}{n})^{n}

also convergent, and to what?

Yes — it's the

x = - 1

case, converging to

e^{- 1} \approx 0.368

. The base dips just below

1

and the same balancing act gives a value below

1

, not

0

What is

e^{0}

, and is it a

1^{\infty}

trap?

e^{0} = 1

, and it is not a trap: with

x = 0

the base

(1 + \frac{0}{n})

is exactly

1

for all

n

, so nothing indeterminate happens — it's a genuine, boring

1

If interest is

0%

(rate

r = 0

) compounded continuously, what factor multiplies your money?

The factor is

e^{0} = 1

— money is unchanged. Zero rate means no growth, and the formula correctly returns "multiply by

1

What is

lim_{n \to \infty} (1 + \frac{1}{n})

without the exponent

n

Just

1

. Stripping the power kills the compounding; only the exponent-times-base interplay produces

e

. This isolates why the exponent is the whole story.

Does the value of

(1 + \frac{1}{n})^{n}

ever equal

e

for a finite whole number

n

No. Every finite

n

gives a rational number strictly less than

e

; since

e

is irrational, no finite term can hit it exactly.

e

is only the limit, never a member of the sequence.

Connections

Natural logarithm ln x — undoes $e^{x}$ ; the "which power gives this?" question.
Exponential function e^x and its derivative — why $e$ is the natural base.
Compound interest — where the limit is born.
Differential equations dy/dx = ky — the universal growth law behind $e$ .
Binomial theorem — turns the limit into the $\sum \frac{1}{k !}$ series.
Limits and indeterminate forms — the $1^{\infty}$ case in full.