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Thanks for reading!

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!

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

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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: $$\begin{align} \lim_{x\to \infty} \left( 1+\frac{a}{x} \right) ^{bx} &= \lim_{x\to \infty} \left[ \left( 1+\frac{a}{x} \right)^{x/a} \right]^{ab} \\ &= \lim_{n\to \infty} \left[ \left( 1+\frac{1}{n} \right)^n \right]^{ab} \quad (\textrm{take } n=x/a) \\ &= \left[ \lim_{n\to \infty} \left( 1+\frac{1}{n} \right)^n \right]^{ab} \\ &= e^{ab}. \end{align}$$

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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.

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

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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.

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

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