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 $n \rightarrow \infty$?


  • 1
    $\begingroup$ 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

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.


$$1.001^3=1.003003001\approx 1.003$$ 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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.