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In calculus, $e$ is defined as the number such that

$$\lim_{h\rightarrow0}\left(\frac{e^h -1}h\right)=1$$

The main motivation of this limit that is equal to $1$ is to simplify the differentiation formula of exponential functions.

But how do we prove, or how was it found, that the value of $e$ that will satisfy the above equation is equal to the value discovered by Jacob Bernoulli while he was working on a problem in compound interest, which is the following:

$$e=\lim_{n\rightarrow\infty}\left(1+\frac1n\right)^n = 2.71828... $$

JLC
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    Well, the first definition is odd, since you'd first have to define $e^h$. I would have said the more common definition was $e=\sum \frac 1{n!}$. Anyway, perhaps this duplicate gives you want you want. – lulu Jun 24 '25 at 18:17
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    There is some difficulty in defining $e^h$ for irrational $h$. You can find more detail in this thread. This issue is often swept under the rug in introductory calculus courses. But to conduct a rigorous proof of a statement about $e^x$, you can't sweep it under the rug. – Joe Jun 24 '25 at 18:20
  • I would have said that this question was a better source of information (but it uses the power series notation, $\exp (x)=\sum \frac {x^n}{n!}$. – lulu Jun 24 '25 at 18:20
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    Although the first definition you give of $e$ can be made to make sense, you have to (i) first define $a^x$ for $a>0$ and $x\in\mathbb R$, and then (ii) show that there exists a unique number $a>0$ such that $\lim_{h\rightarrow0}\left(\frac{a^h -1}h\right)=1$. Both (i) and (ii) are nontrivial tasks. – Joe Jun 24 '25 at 18:22
  • Quick beginner guide for asking a well-received question + please avoid "no clue" questions. – Anne Bauval Jun 24 '25 at 19:06
  • One way to prove $y_n=(1+1/n)^n$ converges to a specific number as $n\rightarrow\infty$ is to consider $\log(y_n)=n\log(1+1/n)$ and use properties of the derivative of the natural logarithm. This means we first need to define "natural log," then use properties of this natural log to find that $y_n$ converges to the number such that the natural log of that number is 1. We can call that number $e$. – Michael Jun 24 '25 at 19:59

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