Po-Shen Loh

A cute proof that makes $e$ natural

Posted Apr 16, 2025; last updated Apr 17, 2025

For the full article covering many properties of $e$ , including history and comparison with existing methods of teaching: PDF from arXiv. A video explanation will be posted here shortly.

This webpage pulls out the part of the article which uses Pre-Calculus language to explain what is so natural about $e$ , while intuitively connecting the following two important properties:

The slope of the tangent line to $y = e^x$ at the point $(x, e^x)$ is just $e^x$ . (In Calculus language: $e^x$ is its own derivative.)
The expression $\big(1 + \frac 1n \big)^n$ approaches $e$ as $n$ grows.

Key conceptual starting point

Geometrically, there really is only one exponential function curve shape, because all exponential function curves $y = a^x$ (with positive real bases $a$ ) are just horizontal stretches of each other. This is exactly like how all ellipses are just stretches of each other (and for the same reason).

For example, $y = 8^x$ , stretched horizontally by a factor of $6$ , is $y = 8^{x/6} = (\sqrt{2})^x$ .

All exponentials are stretches of each other

Geometrically, since stretching is a continuous process, exactly one of these horizontally stretched exponential curves has the property that its tangent line at its $y$ -intercept has the particularly nice and natural slope of $1$ .

We define $\boldsymbol e$ to be the unique positive real base corresponding to that curve.

Easy $e$ approximation

Let's find a number whose exponential curve has tangent slope $1$ at the $y$ -axis. For this, we take the curve $y = 3^x$ and estimate what factor to horizontally stretch it. To start, we must estimate the slope of the tangent line to $y = 3^x$ at its $y$ -intercept $A(0, 3^0)$ . But how? Does that need Calculus? No! Algebra is enough!

Consider a very-nearby point on the curve: $B(h, 3^h)$ , where $h$ is tiny but not zero. The slope of line $AB$ is $\frac{3^h - 3^0}{h - 0}$ Use $h = 0.0001$ to approximate that tangent slope: $\frac{3^{0.0001} - 1}{0.0001 - 0} \approx 1.09867$ Thus a horizontal stretch by a factor of $\approx 1.09867$ will make the tangent slope $\approx 1$ . So $3^{x/1.09867} = (3^{1/1.09867})^x$ has a tangent slope of $\approx 1$ .

Therefore, $3^{1/1.09867} \approx 2.71814$ is close to $e$ . This is pretty good, because actually $e \approx 2.718281828459045$ .

Beautiful tangent slopes everywhere

The same method derives the slope of the tangent line to $y = e^x$ at any point $P(x, e^x)$ . Consider a very-nearby point on the curve: $Q(x+h, e^{x+h})$ , where $h$ is tiny but not zero. The slope of line $PQ$ is $\frac{e^{x+h} - e^x}{(x+h) - x} = e^x \cdot \left( \frac{e^h - e^0}{h - 0} \right).$

The bracket is the slope of the line through $(0, e^0)$ and $(h, e^h)$ , so as $h$ shrinks, the bracket becomes the slope of the tangent to $y = e^x$ at the $y$ -intercept. That miraculously cleans up to just $1$ by our definition of $e$ . (And that is precisely why we built the definition this way.)

So, the slope of the tangent at $P(x, e^x)$ is just $e^x$ .

Rephrased in Calculus language: $e^x$ is its own derivative. This is perhaps the single most important property of $e$ , because all of the Calculus facts stemming from $e$ can be deduced from this fact.

Compound interest limit

Pre-Calculus usually teaches a different definition of $e$ , as the limit of the expression $\big(1 + \frac 1 n \big)^n$ which arises from continuously compounded interest. To reconcile the approaches, we now visually prove that $\big(1 + \frac 1 n \big)^n$ approaches the same number we defined.

Since $\log_b x$ is the inverse function of $b^x$ for any base $b$ , using our base $e$ we get $\begin{align*} \left(1 + \frac 1 n\right)^n &= \left(e^{\log_e \left(1 + \frac 1 n\right)}\right)^n \\ &= e^{n \log_e \left(1 + \frac 1 n\right)} \end{align*}$ We used base $b = e$ (instead of, say, $10$ ) because it now conveniently suffices to show that the expression in the exponent tends to $1$ as $n$ grows. That expression rearranges into a slope calculation! $n \log_e \left(1 + \frac 1 n\right) = \frac{\log_e \left(1 + \frac 1 n\right)}{\frac{1}{n}}$ $= \frac{\log_e \left(1 + \frac 1 n\right) - \log_e (1)}{\left(1 + \frac 1 n\right) - (1)}$ That's the slope of the line through the point $(1, 0)$ on the curve $y = \log_e x$ and another point very nearby on the curve. As $n$ grows, that tends to the slope of the tangent line at $(1, 0)$ . We are done as soon as we prove that slope is $1$ (which is also a natural objective to seek).

To that end, since $\log_e x$ is the inverse function of $e^x$ , their graphs are reflections over the line $y = x$ .

Both of the following lines have slope $1$ :

the tangent line to $y = e^x$ through $(0, 1)$ by definition of $e$ ; and
the line $y = x$ .

So, they are parallel, making this nice reflection:

Therefore, the slope of the tangent line to $y = \log_e x$ at $(1, 0)$ is indeed $1$ , completing the proof!