Single-Variable-Calculus · Unit 6 · Video 4 · Interactive Practice

Why (1 + 1/n)ⁿ Approaches e

IKey Formulas

Formula	Name	What it says
$\ln\big[(1 + \tfrac{1}{n})^n\big] = n \ln(1 + \tfrac{1}{n})$	Take the log	The moving exponent comes down as a factor: a form $\infty \cdot 0$
$n \ln(1 + \tfrac{1}{n}) = \dfrac{\ln(1 + \Delta x) - \ln 1}{\Delta x}$ , $\ \Delta x = \tfrac{1}{n}$	Difference quotient	The difference quotient of $\ln$ at $x_0 = 1$ ; $n \to \infty$ is $\Delta x \to 0$
$\dfrac{d}{dx}\ln x \Big\|_{x = 1} = \dfrac{1}{x}\Big\|_{x = 1} = 1$	The limit of the log	So $n \ln(1 + \tfrac{1}{n}) \to 1$
$(1 + \tfrac{1}{n})^n = e^{\,n \ln(1 + 1/n)} \to e^1 = e$	Undo the log	$e^x$ is continuous, so its output follows its input

Key Insight: A base sinking to $1$ against an exponent climbing to $\infty$ has no automatic answer. Taking the log turns the contest into a derivative in disguise — the slope of $\ln x$ at $x = 1$ — and that slope is exactly $1$ .

IIVisualization 1 — Neither 1 Nor Infinity

As $n$ grows the base $1 + \tfrac{1}{n}$ sinks to $1$ , yet $(1 + \tfrac{1}{n})^n$ levels off at $e$ .

n 100

IIIVisualization 2 — The Secant That Becomes the Tangent

$n \ln(1 + \tfrac{1}{n})$ is the slope of a secant of $y = \ln x$ from $x = 1$ to $x = 1 + \Delta x$ .

Δx 1.500

💡 The secant slope is $\ln$ of $(1 + \Delta x)^{1/\Delta x}$ , so the readout's last value is $(1 + \tfrac{1}{n})^n$ itself — as the slope tends to $1$ , it tends to $e^1$ .

IVVisualization 3 — Computing e by Hand

How close does $(1 + \tfrac{1}{n})^n$ get to $e \approx 2.71828$ for a given $n$ ?

n Your guess 2.500

💡 Every finite $n$ gives only an approximation: $e$ is irrational, so its decimal expansion never ends and never repeats.

VQuiz Questions

Problem 1 · Change the Variable

Given: $\Delta x = \tfrac{1}{n}$ — rewrite $n \ln\!\big(1 + \tfrac{1}{n}\big)$ in terms of $\Delta x$ .

✅ Correct! Inside the log

\tfrac{1}{n}

becomes

\Delta x

, and the factor

n

becomes

\tfrac{1}{\Delta x}

. As

n \to \infty

\Delta x \to 0

— the derivative's limit.

❌ The factor in front is upside down.

n

is the reciprocal of

\Delta x = \tfrac{1}{n}

, so it becomes

\tfrac{1}{\Delta x}

, not

\Delta x

❌ The $1$ inside the log went missing. Only the

\tfrac{1}{n}

is replaced:

\ln\!\big(1 + \tfrac{1}{n}\big)

becomes

\ln(1 + \Delta x)

❌ Not quite. A factor multiplying a log stays outside the log; replace

n

\tfrac{1}{\Delta x}

and

\tfrac{1}{n}

\Delta x

separately.

Show solution

Rewrite each piece. Inside the log, $\tfrac{1}{n} = \Delta x$ . The factor in front is $n = \tfrac{1}{\Delta x}$ . So

n \ln\!\Big(1 + \frac{1}{n}\Big) = \frac{1}{\Delta x}\,\ln(1 + \Delta x) = \frac{\ln(1 + \Delta x)}{\Delta x}.

The value is unchanged — only the variable is new: $n \to \infty$ is the same as $\Delta x \to 0$ .

Problem 2 · Match the Pieces

Given: $\dfrac{\ln(1 + \Delta x) - \ln 1}{\Delta x}$ , compared with the definition $f'(x_0) = \displaystyle\lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}$ .

Which function and base point is this the difference quotient of?

What is its limit as $\Delta x \to 0$ ?

✅ Correct! The limit is the slope of

\ln x

x = 1

, which is

\tfrac{1}{1} = 1

— so

n \ln(1 + \tfrac{1}{n}) \to 1

❌ $x_0$ is where the function is evaluated, not where $\Delta x$ goes. Here

f(x_0) = \ln 1

, so

x_0 = 1

;

\ln 0

is not even defined.

❌ That is $\ln 1$ , not the slope at $1$ . The height of

\ln x

x = 1

0

, but the limit of the quotient is the derivative there,

\tfrac{1}{x}

x = 1

❌ The quotient does not blow up. Numerator and denominator both shrink to

0

together; their ratio tends to the slope

\tfrac{1}{x}\big|_{x=1} = 1

❌ Match term by term.

f(x_0 + \Delta x)

lines up with

\ln(1 + \Delta x)

and

f(x_0)

with

\ln 1

❌ Use the derivative of the natural log.

\dfrac{d}{dx}\ln x = \dfrac{1}{x}

— evaluate it at the base point.

Show solution

Line up the pieces: $f(x_0 + \Delta x) \leftrightarrow \ln(1 + \Delta x)$ and $f(x_0) \leftrightarrow \ln 1$ . So $f = \ln$ and $x_0 = 1$ .

The limit of the difference quotient is the derivative at the base point:

\lim_{\Delta x \to 0} \frac{\ln(1 + \Delta x) - \ln 1}{\Delta x} = \frac{d}{dx}\ln x\,\Big|_{x=1} = \frac{1}{x}\,\Big|_{x=1} = 1.

Subtracting $\ln 1 = 0$ changed nothing about the value; it only made the difference quotient visible.

Problem 3 · A Base Just Below One

Given: $\displaystyle\lim_{n \to \infty} \Big(1 - \frac{1}{n}\Big)^n$ — find it by the same route: take the log, spot a derivative, undo the log.

What does the exponent $n \ln\!\big(1 - \tfrac{1}{n}\big)$ tend to?

So the limit is

✅ Correct! The exponent tends to

-1

, and

e^x

is continuous, so

\big(1 - \tfrac{1}{n}\big)^n \to e^{-1} = \tfrac{1}{e} \approx 0.368

❌ Track the sign of $\Delta x$ . With

\Delta x = -\tfrac{1}{n}

, the factor

n

-\tfrac{1}{\Delta x}

, so the difference quotient comes with a minus sign in front.

❌ That is the $\infty \cdot 0$ trap.

\ln\!\big(1 - \tfrac{1}{n}\big) \to 0

but

n \to \infty

; the product needs the derivative to settle it.

❌ A base tending to $1$ does not force the power to $1$ . The exponent grows at the same time — the log of the expression tends to

-1

, not

0

❌ Substitute $\Delta x = -\tfrac{1}{n}$ . Then

n \ln\!\big(1 - \tfrac{1}{n}\big) = -\dfrac{\ln(1 + \Delta x) - \ln 1}{\Delta x}

with

\Delta x \to 0

❌ Undo the log. The expression equals

e

raised to its own log, and that log tends to

-1

Show solution

Step 1 — Take the log. $\ln\big[(1 - \tfrac{1}{n})^n\big] = n \ln\!\big(1 - \tfrac{1}{n}\big)$ , a form $\infty \cdot 0$ .

Step 2 — Substitute. Let $\Delta x = -\tfrac{1}{n}$ , so $\Delta x \to 0$ as $n \to \infty$ and $n = -\tfrac{1}{\Delta x}$ :

n \ln\!\Big(1 - \frac{1}{n}\Big) = -\frac{\ln(1 + \Delta x) - \ln 1}{\Delta x} \;\longrightarrow\; -\frac{d}{dx}\ln x\,\Big|_{x=1} = -1.

Step 3 — Undo the log. $e^x$ is continuous, so

\Big(1 - \frac{1}{n}\Big)^n = e^{\,n \ln(1 - 1/n)} \;\longrightarrow\; e^{-1} = \frac{1}{e} \approx 0.368.

Problem 4 · A Different Base and Exponent

Given: $\displaystyle\lim_{n \to \infty} \Big(1 + \frac{2}{n}\Big)^{3n}$ — find the limit.

✅ Correct! The log is

3n \ln(1 + \tfrac{2}{n}) = 6 \cdot \dfrac{\ln(1 + \Delta x) - \ln 1}{\Delta x}

with

\Delta x = \tfrac{2}{n}

, which tends to

6 \cdot 1 = 6

❌ The $2$ and the $3$ multiply, not add. With

\Delta x = \tfrac{2}{n}

, the factor

3n

becomes

3 \cdot \tfrac{2}{\Delta x} = \tfrac{6}{\Delta x}

❌ The constants belong in the exponent. They multiply the log of the expression, so after undoing the log they sit in the power of

e

❌ Check the substitution.

\Delta x = \tfrac{2}{n}

means

n = \tfrac{2}{\Delta x}

, so

3n = \tfrac{6}{\Delta x}

— the

2

multiplies, it does not divide.

❌ Not quite. Take the log, set

\Delta x = \tfrac{2}{n}

, and write

3n

in terms of

\Delta x

Show solution

Step 1 — Take the log. $\ln\big[(1 + \tfrac{2}{n})^{3n}\big] = 3n \ln\!\big(1 + \tfrac{2}{n}\big)$ .

Step 2 — Substitute $\Delta x = \tfrac{2}{n}$ , so $n = \tfrac{2}{\Delta x}$ and $3n = \tfrac{6}{\Delta x}$ :

3n \ln\!\Big(1 + \frac{2}{n}\Big) = 6 \cdot \frac{\ln(1 + \Delta x) - \ln 1}{\Delta x} \;\longrightarrow\; 6 \cdot \frac{d}{dx}\ln x\,\Big|_{x=1} = 6.

Step 3 — Undo the log (continuity of $e^x$ ):

\Big(1 + \frac{2}{n}\Big)^{3n} = e^{\,3n \ln(1 + 2/n)} \;\longrightarrow\; e^6.

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