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How did the Euler and Bernoulli know that $\lim_{x \to \infty} (1+\frac1x)^x$ exists?

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    $\begingroup$ What is the definition of $e$? $\endgroup$ May 2, 2013 at 1:53
  • $\begingroup$ I know my limit is the definition of $e$, I'm just asking how did the people who discover this limit (both Euler of Bernoulli had something to do with it) prove that the limit is equal to a constant. $\endgroup$
    – Ovi
    May 2, 2013 at 1:56
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    $\begingroup$ You start your question saying "Prove that the [limit] ... is actually equal to $e$." Sometimes that limit (whose existence needs to justified of course) is used as the definition of $e$, sometimes not. If you are assuming that, could you please adjust the first sentence accordingly? If you want to show that this limit corresponds to $e$ by some other definition, could you please add that information? $\endgroup$ May 2, 2013 at 1:58
  • $\begingroup$ Sorry for the confusion, I hope it's better now. $\endgroup$
    – Ovi
    May 2, 2013 at 2:00
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    $\begingroup$ The question hasn't become any clearer through the edit. It's still not clear on the basis of which definition you want to see a proof that the limit is $\mathrm e$. Also, it makes no sense to ask whether the limit is equal to a constant. The only question that can be asked without providing some other definition of $\mathrm e$ is whether the limit exists; if it exists, it is necessarily a constant, since the only variable in it is the dummy variable $x$. $\endgroup$
    – joriki
    May 2, 2013 at 2:15

4 Answers 4

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Let $A$ be a constant such that the derivative of $f(t) = A^t$ at $t = 0$ is less than one. Then for $t$ sufficiently small we have $1 + t > A^t$. Likewise if the derivative of $B^t$ is greater than one at $t=0$ then for $t$ small we have $1+t < B^t$. Setting $t = \frac{1}{x}$ shows that $A < (1+\frac{1}{x})^x < B$ for $x$ sufficiently large. So the limit is equal to the unique constant $e$ such that the derivative of $e^t$ at $t = 0$ is actually equal to one (my favorite definition of $e$).

I don't know how Euler proved it but if you want to find out you should read his book "Introduction to Analysis of the Infinite".

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Your edited question is simply, how do we know that $$\lim_{x\rightarrow\infty} (1+1/x)^x$$ exits? In modern terminology, we might define $f(x)=(1+1/x)^x$, show that $f$ is increasing and bounded above, and deduce that the limit is the supremum of the set of function values for positive $x$, i.e. $$\lim_{x\rightarrow\infty} f(x) = \sup \{f(x):x>0\}.$$

But, honestly, I think that Euler and his contemporaries largely took this kind of thing for granted, feeling that no proof was necessary.

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Expand both sides in a Taylor series to prove that they are equal.

This is the same idea as used to show $e^{ix} = \cos(x)+i\sin(x)$. If sine is expanded as a Taylor series, multiply it by $i$, and then added to the Taylor series for cosine, it equal the Taylor series for $e^{ix}$.

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It would be tedious before computers, but you can find that limit simply by plugging in some large numbers. If n = 100 you get 2.704. If n = 1000 you get 2.716. If n = 10000 you get 2.718. Plug in 69! (Wolfram Alpha won't go much higher) and you have 20 or so decimal places that won't change. The digits that already exist won't change no matter how high n gets. It follows that if n is infinite, then the resulting number has reached a point where the decimal places aren't changing at all.

The result takes the form of $\frac{(x+1)^x}{x^x}$, which may be somewhat easier to expand. I can offer something of a heuristic proof for a limit with an example in this — notice how as the numerator gets larger, the exponent of the denominator gets significantly larger with it.

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  • $\begingroup$ Is there a way however to prove that the limit exists? I know that with some limits if you plug in numbers it looks like it's going to give you a value but past a point it can quickly jump to another value. $\endgroup$
    – Ovi
    May 2, 2013 at 2:46
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    $\begingroup$ This does not answer the question. $\endgroup$ May 2, 2013 at 3:17

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