Calculus I · Patterns That Go Somewhere

The many faces of $e$

The sequence $\left(1+\frac1n\right)^n$ is the most important $1^\infty$ fight in all of mathematics, and it has a whole family. What if we replace that $1$ in the numerator with some other number $x$?

$$\boxed{\;\left(1+\frac{x}{n}\right)^n \longrightarrow e^{\,x}\;}$$

This single fact is the bridge from this chapter to Chapter 5. It says $e$ is not a one-off curiosity: it spawns an entire function $e^x$, and that function is born from compounding. Set $x=1$ and recover the original $\to e$. Set $x=-1$: $\left(1-\frac1n\right)^n\to e^{-1}=\frac1e\approx0.368$. Set $x=2$: $\left(1+\frac2n\right)^n\to e^2$. Try a few on a calculator and watch them land on powers of $e$.

◆ Layer 2 — Why $\left(1+\frac{x}{n}\right)^n\to e^x$ (one honest loan)

Take the logarithm, which turns the frightening exponent into a product (Chapter 5 makes $\ln$ rigorous; here we borrow only that $\ln$ pulls a power down into a multiplier): $$\ln\left(1+\frac{x}{n}\right)^n = n\,\ln\!\left(1+\frac{x}{n}\right).$$ Let $t=\frac{x}{n}\to0$, so $n=\frac{x}{t}$, and the right side becomes $x\cdot\dfrac{\ln(1+t)}{t}$. One more famous limit closes it: $\dfrac{\ln(1+t)}{t}\to1$ as $t\to0$ (this is the slope of $\ln$ at $1$, proven in Chapter 5 — the loan). So the logarithm of our sequence $\to x\cdot1=x$, which means the sequence itself $\to e^x$. $\blacksquare$ Notice it is the same $\frac{\text{small}}{\text{small}}\to1$ shape that turned up in Fight one. The limits that fight back are almost always derivatives wearing a costume — you’ll see this confirmed in Chapters 2 and 3.

The race decides everything: three sequences, three fates

The form $1^\infty$ is indeterminate precisely because the answer hinges on a race — how fast the base sinks toward $1$ versus how fast the exponent climbs to $\infty$. Watch three sequences that all look like “$1^\infty$” land in three different worlds:

$$\left(1+\frac1n\right)^{n}\to e,\qquad \left(1+\frac{1}{n^2}\right)^{n}\to 1,\qquad \left(1+\frac1n\right)^{n^2}\to \infty.$$

In the first, base and exponent are matched and balance at $e$. In the second, the base rushes to $1$ too fast (like $\frac1{n^2}$) for a mere $n$ in the exponent to matter — its logarithm runs like $n\cdot\frac1{n^2}=\frac1n\to0$, so the limit is $e^0=1$. In the third, the exponent $n^2$ is too strong for the base’s gentle approach — its logarithm runs like $n^2\cdot\frac1n=n\to\infty$, so it explodes.

Same form, three answers, decided entirely by who wins. Here’s the gauge that settles any of them at a glance: for a base near $1$, look at $(\text{exponent})\times(\text{base}-1)$; whatever number $k$ that approaches, the limit is $e^{k}$. Run the three cases through it and out pop $e^1=e$, then $e^0=1$, then $e^\infty=\infty$. That is the entire secret of $1^\infty$ forms, and you now hold it.

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