Proving $e$ exist

Asked Feb 12 '23 at 10:06

Active Feb 12 '23 at 10:17

Viewed 46 times

I saw a cool definition for $e$. Thats cool because they showed how to arrive to it.

Most of the time the definition I see is $\lim_{n\to\infty}(1+1/n)^n$ but no one explains how they got it.

The new definition is $e$ is the number such that if you place it in $x$ $\lim_{h\to 0} \frac{x^h - 1}{h} = 1$

The question I have is how can I show that such number exist.

Don't really know how to tackle this and what information can I get from $\lim_{h\to 0} \frac{x^h - 1}{h} = 1$ to solve this.

Thanks to anyone that can help.

edited Feb 12 '23 at 10:17

QC_QAOA

asked Feb 12 '23 at 10:06

Okosh

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos Feb 12 '23 at 10:11
It is well known that the derivate of $e^x$ is again $e^x$ (In general , $a^x$ has derivate $\ln(a)\cdot a^x$ for $a>0$). This is used in this "new definition". To calculate $e$ , the equality $$e=\sum_{j=0}^{\infty} \frac{1}{j!}$$ is by the way much more useful , we could consider this to be the "deifnition of $e$" as well. – Peter Feb 12 '23 at 10:15
1

it should. i check it right away. and thx – Okosh Feb 12 '23 at 10:31

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