Compounding frequency versus compounding scaling.

The reason we transition from a limit where \(n\) tends to infinity to raising \(e\) to the power of \(x\) comes down to a beautiful mathematical bridge: compounding frequency versus compounding scaling.

When you calculate \(e\) normally, you are looking at what happens to a base unit of \(1\) growing at a \(100\%\) interest rate over infinitely small intervals (\(n \to \infty\)). When you raise it to the power of \(x\), you are altering the total time or the internal rate of that growth engine.

Here is the exact mathematical breakdown of how \(e^{x}\) naturally emerges from that infinity limit, structured so you can easily translate it into a high-retention Remotion animation.

The Mathematical Proof (The “Why”)

The Mathematical Proof (The “Why”)

The standard definition of \(e\) when \(n\) tends to infinity is:
\(e=\lim _{n\rightarrow \infty }\left(1+\frac{1}{n}\right)^{n}\)

Now, imagine instead of growing at a \(100\%\) rate, your system grows at a rate of \(x\) (for example, a \(200\%\) rate, where \(x = 2\)). The formula for continuous compounding at a rate of \(x\) becomes:
\(\text{Growth}=\lim _{n\rightarrow \infty }\left(1+\frac{x}{n}\right)^{n}\)

To see why this equals \(e^{x}\), we use a clever algebraic substitution. Let’s define a new variable, \(m\), such that:
\(m=\frac{n}{x}\quad \implies \quad n=mx\)

As \(n\) approaches infinity (\(\infty \)), \(m\) must also approach infinity because \(x\) is just a constant number. Now, let’s substitute \(m\) back into our growth equation: [1, 2]

\(\lim _{n\rightarrow \infty }\left(1+\frac{x}{n}\right)^{n}=\lim _{m\rightarrow \infty }\left(1+\frac{1}{m}\right)^{mx}\)

Using the laws of exponents, we can pull the power of \(x\) outside the main brackets:
\(\lim _{m\rightarrow \infty }\left[\left(1+\frac{1}{m}\right)^{m}\right]^{x}\)

Look closely at the expression inside the brackets. It is the exact definition of \(e\) that we started with, just using the letter \(m\) instead of \(n\):
\(\left[\lim _{m\rightarrow \infty }\left(1+\frac{1}{m}\right)^{m}\right]^{x}=e^{x}\)

The Story Hook: Raising \(e\) to the \(x\) isn’t an arbitrary rule we invented later. The math proves that changing the growth rate inside the infinity limit is fractionally identical to taking the final infinite baseline (\(e\)) and scaling it exponentially by \(x\).