# Intuitively, why are the two limit definitions of $e^x$ equivalent?

Intuitively, why does...

$$\lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^{xn}=\lim_{n \rightarrow \infty} \left(1+\frac{x}{n}\right)^{n}=e^x$$

Note, I'm not asking why $$e^x$$ is one of the two limits. I understand the first limit, or at least I think I do.

In terms of continuous growth however (I don't just want a mathematical proof - the more intuitive the answer is the better), I'd like to understand why the two limits are equivalent!

Why is letting some principal amount grow by $$\frac{1}{n}$$ times its current value $$xn$$ times equal to letting that principal grow by $$\frac{x}{n}$$ times its current value $$n$$ times if we allow $$n \rightarrow \infty$$?

Thanks!

• Think about integration, specifically integrating $1/x^{1+1/n}$ when $n$ is very large. Remember that the integral of $1/x = \ln{x}$. – bob.sacamento Jul 11 at 20:01

Forget that $$n$$ is an integer. For $$x > 0$$ we have

$$\lim_{y\to\infty} \left(1+\frac{x}{y}\right)^y = \lim_{y\to\infty} \left(1+\frac{1}{\frac{y}{x}}\right)^{x\cdot \frac{y}{x}}$$

If $$y \to \infty$$ then $$z:= \frac{y}{x} \to \infty$$ as well so change of variables gives that this is equal to $$\lim_{z\to\infty} \left(1+\frac1z\right)^{xz}$$ which is the desired expression.

In general $$\lim_{x\to \infty} \left( 1+\frac{a}{x} \right) ^{bx}=e^{ab}$$ and they are only based on the definition of the number $$e$$ $$e:=\lim_{n\to \infty} \left( 1+\frac{1}{n} \right)^n$$ Here is a simple proof: \begin{align} \lim_{x\to \infty} \left( 1+\frac{a}{x} \right) ^{bx} &= \lim_{x\to \infty} \left[ \left( 1+\frac{a}{x} \right)^{x/a} \right]^{ab} \\ &= \lim_{n\to \infty} \left[ \left( 1+\frac{1}{n} \right)^n \right]^{ab} \quad (\textrm{take } n=x/a) \\ &= \left[ \lim_{n\to \infty} \left( 1+\frac{1}{n} \right)^n \right]^{ab} \\ &= e^{ab}. \end{align}

Apply binomial theorem resp. the binomial series, then $$\left(1+\frac1n\right)^{nx}=1+x+\sum_{k=2}^{\infty}\frac{x(x-\frac1n)...(x-\frac{k-1}n)}{k!}$$ and $$\left(1+\frac xn\right)^{n}=1+x+\sum_{k=2}^{\infty}\frac{(1-\frac1n)...(1-\frac{k-1}n)}{k!}x^k$$ Both expansions converge in their coefficients to the same limit $$\frac{x^k}{k!}$$, the complicated part is to show that exchanging the limit in the series and in the coefficients is permissible.

Consider that for $$x>0$$, $$n$$ and $$\frac nx$$ both tend to infinity so that

$$\lim_{n\to\infty}\left(1+\dfrac1n\right)^n=\lim_{n\to\infty}\left(1+\dfrac xn\right)^{n/x}=e.$$

Then by continuity, you can raise to the $$x^{th}$$ power inside the limit,

$$\lim_{n\to\infty}\left(1+\dfrac1n\right)^{nx}=\lim_{n\to\infty}\left(1+\dfrac xn\right)^n=e^x.$$

For negative $$x$$, you must modify the reasoning as the argument goes to minus infinity.

$$\lim_{n\to\infty}\left(1+\dfrac xn\right)^{n/x}=\lim_{n\to\infty}\left(1-\dfrac1n\right)^{-n}=\lim_{n\to\infty}\left(1-\dfrac1{n+1}\right)^{-1-n}=\lim_{n\to\infty}\left(1+\dfrac1n\right)^{n+1}\\=\lim_{n\to\infty}\left(1+\dfrac1n\right)^n\left(1+\dfrac1n\right)=\lim_{n\to\infty}\left(1+\dfrac1n\right)^n=e.$$

For a more intuitive approach, notice that for small $$\epsilon$$ you can linearize

$$(1+\epsilon)^x\approx 1+\epsilon x$$ so that for large $$n$$,

$$\left(1+\frac1n\right)^x\approx1+\frac xn$$ and

$$\left(1+\frac1n\right)^{nx}\approx\left(1+\frac xn\right)^n.$$

For larger and larger $$n$$, the approximation improves.

E.g.

$$1.001^3=1.003003001\approx 1.003$$ and $$1.001^{3000}=20.05545\approx 1.003^{1000}=19.99553\approx e^3=20.08553$$

Next, $$1.0001^3=1.000300030001\approx 1.0003$$ and $$1.0001^{30000}=20.08252\approx 1.0003^{10000}=20.07650\approx e^3=20.08553$$ $$\cdots$$

I will say my first thought.

If fix $$x$$, one can take $$\exp (x)$$ for "constant" intuitively. And it is known that for a real number $$x'$$ there are infinite sequences convergent to it, I mean there are many many sequences we do not construct (or notice) with the same limit $$x'$$, so let $$a_n = (1+\frac{x}{n})^n$$, $$b_n = (1+\frac{1}{n})^{xn}$$, sequences $$\{ a_n \} , \{ b_n \}$$ are just two cases( we are familiar with ) among them.

Maybe there is nothing special comparing to the rest.

• This doesn't say anything about the question. Why do $a_n$ and $b_n$ have the same limit? – David C. Ullrich Jul 11 at 21:08