My Math Forum Proof of e^x, lnx and e as limit without circular dependency?

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 July 10th, 2019, 07:28 AM #1 Senior Member   Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8 People prove the derivative of $\displaystyle e^x$, $\displaystyle \ln(x)$ using either the formula of $\displaystyle e = \lim_{x \to 0} (1+x) ^ \frac{1}{x}$ or if they know one of the derivatives $\displaystyle e^x$, $\displaystyle \ln(x)$ they use it to prove the other. My textbook first gave me without proof that $\displaystyle \frac{d}{dx}e^x = e^x$ Then it found the limit of $\displaystyle \ln(x)$ using the chain rule: $\displaystyle f(x) = \ln(x) \Leftrightarrow e^{f(x)} = e^{\ln(x)} \Leftrightarrow e^{f(x)} = x \Leftrightarrow \frac{d}{dx}e^{f(x)} = \frac{d}{dx}x \Leftrightarrow e^{f(x)} \frac{df(x)}{dx} = 1 \Leftrightarrow \frac{df(x)}{dx} = \frac{1}{e^{f(x)} } \Leftrightarrow \frac{d}{dx}\ln x = \frac{1}{e^{\ln x} } \Leftrightarrow \frac{d}{dx}\ln x = \frac{1}{x}$ And then it uses the limit of ln(x) at x=1 as calculated above with the definition of the derivative in order to show that $\displaystyle e = \lim_{x \to 0} (1+x) ^ \frac{1}{x}$ And then others (on the internet) are using $\displaystyle e = \lim_{x \to 0} (1+x) ^ \frac{1}{x}$ to show that the derivative of $\displaystyle e^x$ is $\displaystyle e^x$. Well the circular problem is solved if you can some how find the derivative of $\displaystyle e^x$ without using any of the above. Then all the other derivatives and e as a limit can be proved by using $\displaystyle e^x$. Can you show me the right order? Well the book might have a good reason that it hasn't shown to me yet; maybe it's really difficult to calculate the limit of $\displaystyle e^x$. Thank you. Last edited by skipjack; July 10th, 2019 at 11:06 AM.
 July 10th, 2019, 08:32 AM #2 Senior Member   Joined: Oct 2009 Posts: 850 Thanks: 327 If you want to avoid circular reasonings, you'd need to start by adressing what is the DEFINITION of e, e^x and log(x).
July 10th, 2019, 08:37 AM   #3
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 Originally Posted by Micrm@ss If you want to avoid circular reasonings, you'd need to start by adressing what is the DEFINITION of e, e^x and log(x).
You mean the infinity series that give e and e^x?

July 10th, 2019, 08:56 AM   #4
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 Originally Posted by babaliaris You mean the infinity series that give e and e^x?
I usually see it as

$\ln(x) = \displaystyle \int_1^x~\dfrac{dt}{t}$

July 10th, 2019, 09:06 AM   #5
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 Originally Posted by romsek I usually see it as $\ln(x) = \displaystyle \int_1^x~\dfrac{dt}{t}$
hmm. But still the reason I know how to solve this integral is because I ask my self "What do i need to derive from to get 1/t?" so how does that integral make sense as a definition?

July 10th, 2019, 09:30 AM   #6
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 Originally Posted by babaliaris hmm. But still the reason I know how to solve this integral is because I ask my self "What do i need to derive from to get 1/t?" so how does that integral make sense as a definition?
A bit of reading on the web will provide you with all the answers to this question that exist.

July 10th, 2019, 11:38 AM   #7
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 Originally Posted by babaliaris . . . how does that integral make sense as a definition?
One needs to start somewhere. For $x > 0$, it defines a function of $x$ that can be called "$\ln(x)$" and has the properties you'd like for a logarithm.

One can now define $e$ as the value of $x$ such that $\ln(x) = 1$.

 Tags circular, dependency, limit, lnx, proof

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