You've seen $e$ used in exponential functions and natural logarithms, but where does this specific number come from? Why isn't it a nice round number like 2 or 3?
The number $e$ emerges naturally from calculus. It is defined as a limit, and that limit has a beautiful connection to compound interest and rates of change.
The key formulas:
$$\boxed{e = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n \approx 2.71828...}$$
$$\boxed{e = \lim_{x \to 0}(1 + x)^{1/x}}$$
These two formulas say the same thing in different notation. The first uses discrete steps ($n$ increasing through integers), while the second uses a continuous variable ($x$ approaching 0).
| Formula | Description |
|---|---|
| $e = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n$ | Discrete form (for positive integers $n$) |
| $e = \lim_{x \to 0}(1 + x)^{1/x}$ | Continuous form |
| $e \approx 2.71828182845...$ | Numerical value (irrational) |
Numerical evidence:
| $n$ | $\left(1 + \frac{1}{n}\right)^n$ |
|---|---|
| 1 | 2.000 |
| 10 | 2.594 |
| 100 | 2.705 |
| 1,000 | 2.717 |
| 10,000 | 2.7181 |
| 100,000 | 2.71827 |
| $\to \infty$ | $\to e \approx 2.71828...$ |
We can derive this from what we know about $\ln x$.
Starting point: We know $\frac{d}{dx}(\ln x) = \frac{1}{x}$, so $f'(1) = 1$ when $f(x) = \ln x$.
Step 1. Use the definition of derivative at $x = 1$: $$f'(1) = \lim_{h \to 0}\frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0}\frac{\ln(1 + h) - \ln 1}{h}$$
Step 2. Since $\ln 1 = 0$: $$f'(1) = \lim_{h \to 0}\frac{\ln(1 + h)}{h} = \lim_{h \to 0}\frac{1}{h}\ln(1 + h) = \lim_{h \to 0}\ln(1 + h)^{1/h}$$
Step 3. Since $f'(1) = 1$: $$\lim_{h \to 0}\ln(1 + h)^{1/h} = 1$$
Step 4. Apply the exponential function to both sides (using continuity): $$\lim_{h \to 0}(1 + h)^{1/h} = e^1 = e$$
This gives us the continuous form. Substituting $h = \frac{1}{n}$ (so $h \to 0$ as $n \to \infty$) gives the discrete form.
Imagine you invest \$1 at 100% annual interest. How much do you have after one year?
Annual compounding (once per year): $$A = 1 \cdot (1 + 1)^1 = \$2$$
Semi-annual compounding (twice per year, 50% each time): $$A = 1 \cdot \left(1 + \frac{1}{2}\right)^2 = (1.5)^2 = \$2.25$$
Quarterly compounding (4 times per year, 25% each time): $$A = 1 \cdot \left(1 + \frac{1}{4}\right)^4 = (1.25)^4 \approx \$2.44$$
Monthly compounding: $$A = 1 \cdot \left(1 + \frac{1}{12}\right)^{12} \approx \$2.61$$
Daily compounding: $$A = 1 \cdot \left(1 + \frac{1}{365}\right)^{365} \approx \$2.714$$
Continuous compounding (infinitely many times): $$A = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e \approx \$2.718$$
No matter how frequently you compound, you can never exceed $e$ dollars. This is why $e$ is called the "natural" base for exponential growth.
For principal $P$, annual rate $r$, over time $t$:
$$A = Pe^{rt}$$
This formula appears throughout science and engineering whenever something grows (or decays) continuously at a rate proportional to its current value.
Calculate $\left(1 + \frac{1}{n}\right)^n$ for $n = 2$, $n = 5$, and $n = 10$. Verify that the values approach $e$.
You invest \$1000 at 5% annual interest, compounded continuously. How much do you have after 10 years?
Evaluate $\lim_{n \to \infty}\left(1 + \frac{2}{n}\right)^n$.
Show that $\lim_{x \to 0}\frac{\ln(1 + x)}{x} = 1$.
Prove that $\lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^n = e^x$ for any real number $x$.
Explain why $e$ is the "natural" base for exponential functions. Specifically, why is $e$ the unique base $b$ for which $\frac{d}{dx}(b^x) = b^x$?
| Mistake | Why It's Wrong | Correct Approach |
|---|---|---|
| Thinking $(1 + \frac{1}{n})^n \to \infty$ | The base approaches 1 as $n \to \infty$ | The limit is finite: $e \approx 2.718$ |
| Using $e \approx 3$ | This is too imprecise for most calculations | Use $e \approx 2.718$ or leave as $e$ |
| Writing $\left(1 + \frac{1}{n}\right)^n = 1 + \frac{n}{n} = 2$ | Cannot distribute exponent over sum | Use the limit or binomial expansion |
| Confusing $\left(1 + \frac{1}{n}\right)^n$ with $\left(1 + n\right)^{1/n}$ | These are different expressions! | The first $\to e$; the second $\to 1$ |
$e$ as "natural growth":
Imagine a quantity that grows continuously at a rate equal to its current size. After one unit of time, it will have grown by a factor of exactly $e$.
The limit captures this: As you compound more and more frequently (larger $n$), you approach but never exceed $e$.
Looking back:
Looking ahead:
Historical note: The number $e$ was discovered by Jacob Bernoulli while studying compound interest in 1683. Euler later named it and established it as a fundamental mathematical constant.
| Previous | Up | Next |
|---|---|---|
| Integrals Yielding ln | Section 6.4 | Logarithmic Differentiation |
Last updated: 2026-01-23