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Define $e := \lim_{x\to\infty} (1 + 1/x)^x$. Why is the following true? $$\lim_{x\to\infty} (1 - 1/x)^x = e^{-1}$$

If $\lim_{x\to - \infty} (1 + 1/x)^x = e$, then the above immediately follows, but I'm not sure how to prove this is true given the definition of $e$.

Toby Mak
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keepfrog
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    More generally: https://math.stackexchange.com/q/35486/42969 – Martin R Jul 28 '22 at 07:09
  • Perhaps consider $\lim\limits_{x\to\infty} (1 + 1/x)^x(1 - 1/x)^x = \lim\limits_{x\to\infty} (1 - 1/x^2)^x$ and show this is $1$ – Henry Jul 28 '22 at 07:12
  • @MartinR I guess my question comes down to: if we define $e = \lim_{x\to\infty} (1+1/x)^x$, then is it true that $e = \lim_{x\to-\infty} (1+1/x)^x$? – keepfrog Jul 28 '22 at 07:12

3 Answers3

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Note that $$ \begin{align*} (1-1/x)^x &= \left(\frac{x-1}{x}\right)^x \\ &= \left(\frac{1}{x/(x-1)}\right)^x \\ &= \left(\frac{1}{\frac{x-1+1}{x-1}}\right)^x \\ &= \left(\frac{1}{1+\frac{1}{x-1}}\right)^x \\ &= \frac{1}{\left(1+\frac{1}{x-1}\right)^x} \\ &= \frac{1}{\left(1+\frac{1}{x-1}\right)^{x-1}}\cdot \frac{1}{1+\frac{1}{x-1}} \to \frac{1}{e} \cdot 1 = e^{-1} \end{align*} $$

Ignacio Henríquez
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Clearly $e=\displaystyle \lim_{n \to+\infty}(1+\dfrac{1}{n})^{n}$.

Now consider $(1+\dfrac{x}{n})^{n}$. This can be written as $(1+\dfrac{1}{\frac{n}{x}})^{n}$ which is equal to

$[(1+\dfrac{1}{\frac{n}{x}})^\frac{n}{x}]^{x}$. Clearly the limit of the brackets is $e$, hence the limit is $e^{x}$. Setting $x=-1$ we get the required equality! i.e.

$\displaystyle \lim_{ x\to +\infty}(1-\dfrac{1}{x})^{x}=e^{-1}$.

As an answer to a comment I give a very simple proof that $\displaystyle \lim_{ n\to+\infty}(1+\dfrac{x}{n})^{n}=e^{x}$ for ALL $x$ positive, negative or zero! Let $x>0$. Then, (sorry for not typing)

enter image description here

  • When you set $x=-1$ and take the limit of $n\to\infty$, basically it is computing $\lim_{x\to -\infty} (1 + 1/x)^x$. My question is, by defining $e = \lim_{x\to \infty} (1 + 1/x)^x$, how can I prove that $e=\lim_{x\to -\infty} (1 + 1/x)^x$? – keepfrog Jul 28 '22 at 07:46
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    Read my answer carefully! I prove that assuming $e=\displaystyle \lim_{x \to+\infty}(1+\dfrac{1}{x})^{x}$ we get $\displaystyle \lim_{n \to+\infty}(1+\dfrac{x}{n})^{n}=e^{x} $. Then setting $x=-1$ we get the result! I just use n instead of $x$. The limit (when $x\to +\infty$) is the same for all sequences tending to $+\infty$ and hence for $x_{n}=n$. –  Jul 28 '22 at 12:46
  • But, when $x<0$, "Clearly the limit of the brackets is $e$" is not obvious, because $e = \lim_{n\to-\infty} (1 + 1/n)^n$ is not proven yet. If $x>0$, then it immediately follows from the definition, but if $x<0$, that requires proving $e = \lim_{n\to-\infty} (1 + 1/n)^n$ – keepfrog Jul 29 '22 at 21:46
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    Read my picture edit to see what happens. It has nothing to do with $x$ being positive,negative or zero!! –  Jul 30 '22 at 00:10
  • Thanks, this is the proof that I was looking for. – keepfrog Jul 30 '22 at 21:19
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I'm assuming you're wondering, "how can we use the provided definition of $e$ to our advantage to prove the said statement," right? We can start our proof by first writing that definition and working with it by taking the reciprocal on both sides and letting $ u = -x$ like this:

$$\eqalign{e &= \lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x \cr e^{-1} &= \left(\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x\right)^{-1} \cr &= \lim_{x\to\infty}\left(\left(1+\frac{1}{x}\right)^x\right)^{-1} \cr &= \lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{-x} \cr &= \lim_{u\to-\infty}\left(1+\frac{1}{-u}\right)^{-(-u)} \cr &= \lim_{u\to-\infty}\left(1-\frac{1}{u}\right)^{u}. \cr }$$

Does that answer your question?

Accelerator
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  • Yes, I was wondering how to derive $e^{-1}$ from the definition of $e$ (first line). This is helpful, but is it obvious that $\lim_{u \to -\infty} (1 - 1/u)^u = \lim_{u \to \infty}(1 - 1/u)^u$? I'm struggling to understand this, just as I don't understand why $\lim_{x \to \infty} (1+1/x)^x = \lim_{x \to -\infty} (1+1/x)^x$ (the left hand is $e$ by definition). – keepfrog Jul 28 '22 at 08:10
  • @user789100 That last equality is true but I don't see how it's relevant. What I did was let $u = -x$, then I did $\lim_{x\to\infty} u = \lim_{x\to\infty} -x = -\infty$, which means that $u$ goes to $-\infty$. This is more theory-based, but you can read this link here: https://math.stackexchange.com/questions/167926/formal-basis-for-variable-substitution-in-limits – Accelerator Jul 28 '22 at 08:19
  • My original question was why is it $\lim_{u\to+\infty}(1 -1/u)^u = e^{-1}$, while your answer shows $\lim_{u\to-\infty}(1 -1/u)^u = e^{-1}$. I'm just wondering why taking the limit to $+\infty$ is the same as taking the limit to $-\infty$. Also, I thing the last equation is relevant, as $\lim_{x\to\infty} (1-1/x)^x = \lim_{x\to\infty} ((1+1/(-x))^{-x})^{-1} = \lim_{u\to -\infty} (1+1/u)^u)^{-1} = e^{-1}$, which I'm guessing is basically the same argument as yours. – keepfrog Jul 28 '22 at 08:25
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    I think I see what you're asking. To be honest, I don't know a way to prove that $\lim_{u \to \infty}(1-1/u)^u = e^{-1}$ by using the definition. My best guess is to prove separately that $\lim_{u \to \infty}(1-1/u)^u = e^{-1}$ and that $\lim_{u \to -\infty}(1-1/u)^u = e^{-1}$, but I don't know how to use the given definition to prove that first equality. – Accelerator Jul 28 '22 at 08:56