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I want to prove that if $$\lim_{h \to 0} \frac{a^h-1}{h}=1$$ then $a$ must equal Euler's constant, denoted as "$e$." However, I have some specific constraints for this proof:

1.$e$ is defined only with the limit definition ($\left(1+\frac{1}{x}\right)^x$ as x tends to infinity).

  1. I only want to start the proof by manipulating $\lim_{h \to 0} \frac{a^h-1}{h}=1$ and I don't immediately want to start with the limit definition of e. I want to solve it as if I just discovered the limit and did not know that $a=e$
uggupuggu
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    Bruv your question is already answered here ( albeit slightly different question but you can manipulate the solution to prove your statement) https://math.stackexchange.com/questions/152605/proving-that-lim-limits-x-to-0-fracex-1x-1 – The Grim Reaper Oct 11 '23 at 16:13
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    can you use derivatives? if so, you can notice that the limit means that the derivative of $a^x$ must be 1 if $x=0$. – finch Oct 11 '23 at 16:14
  • See this answer specifically: https://math.stackexchange.com/a/177837/513093 – Dennis Oct 11 '23 at 16:15
  • I said I want to start from scratch, I don't want to immediately jump to the conclusion that it's e. – uggupuggu Oct 11 '23 at 16:27
  • If you really want to start from scratch you need to work step by step. First define $a^x$ for $a>0$ and real $x$ and then using this definition show that $\lim_{h\to 0}(a^h-1)/h$ exists for every $a>0$ and thus defines a function of $a$, say $f(a) $. Next prove that $f$ is continuous and strictly increasing and its range is whole of $\mathbb{R} $. Then there exists a unique number $a>1$ such that $f(a) =1$ and then prove that $a=\lim_{x\to\infty} (1+(1/x))^x$. – Paramanand Singh Oct 12 '23 at 11:01
  • An approach as outlined in my previous comment is available in my blog post: https://paramanands.blogspot.com/2014/05/theories-of-exponential-and-logarithmic-functions-part-3.html?m=0 – Paramanand Singh Oct 12 '23 at 11:04

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