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From the definition of $e$ as the number such that $\lim_{h\rightarrow0}\frac{e^h-1}{h}=1$, is it possible to derive the formula $e=\lim_{n\rightarrow\infty}\left(1+\frac1n\right)^n$ withtout using any approximation or series expansion of the function $e^x$ ?

I'm looking for a way to introduce the number $e$ at the high school level, and defining it to be the value of $a$ such that $a^x$ has slope 1 at $x=0$ seems promising. I can then prove that the derivative of $e^x$ is itself; which is one of the main reasons for using $e$.

But, this doesn't really give a way to compute $e$ with arbitrary precision, while $e=\lim_{n\rightarrow\infty}\left(1+\frac1n\right)^n$ is such a way (just calculate $\left(1+\frac1n\right)^n$ with $n$ large). I can prove this formula from the above definition, but only by using Taylor expansion, which is not really taught at high school level. Hence my question.

Sephi
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    See here for a couple of proofs. – Dietrich Burde Aug 13 '24 at 16:56
  • For $\lim_{h\to 0}\frac {e^h-1}h=1$ to be a definition we have to have somehow proven there exist a unique real number, $w$, so that $\lim_{h\to 0}\frac{w^h-1}h=1$. And that unique number is defined to be $e$. If so then all we need to do is prove that $\lim_{h\to 0} \frac {[\lim_{n\to \infty}(1+\frac 1n)^n]^h-1}h =1$. By squeeze theorem $\lim_{h\to 0}f(h)=\lim_{n\to\infty}f(\frac 1n)$ and... – fleablood Aug 13 '24 at 18:06

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On the left hand side, $\ln(e)=1$.

On the right hand side, $$ \ln\left(\lim_{n\to\infty} \left( 1 + \frac{1}{n}\right)^n \right) = \lim_{n\to\infty} \ln\left( \left( 1 + \frac{1}{n}\right)^n \right) = \lim_{n\to\infty} n \ln \left( 1 + \frac{1}{n} \right) = \lim_{n\to\infty} \frac{ \ln \left( 1 + \frac{1}{n} \right) }{\left(\frac{1}{n}\right)} = \dots $$ where (1) the first equality follows from the fact that natural logarithm is continuous on its domain, (2) the second equality is an exponent property of natural logarithm.

By L'Hopitals rule (check that the conditions are satisfied), we take the derivative of numerator and denominator to continue the above equality: $$ \dots = \lim_{n\to\infty} \frac{-n^{-2} \, \left(1+\frac{1}{n}\right)^{-1}}{-n^{-2}} = \lim_{n\to\infty} \left(1+\frac{1}{n}\right)^{-1} = 1^{-1} = 1 $$

NazimJ
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