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I am a high school student (sophomore) and have come across something I would like explained.

I was watching 3blue1brown for an explanation of calculus, when he used the formula: lim a->0 (d/dx(n^x)=(n^x)((n^a)-1)/a)). (7:05 of https://www.youtube.com/watch?v=m2MIpDrF7Es&list=PL0-GT3co4r2wlh6UHTUeQsrf3mlS2lk6x&index=5)

He was putting in different numbers for n, lamenting in the fact that ((n^a)-1)/a) approached some constant, which was ln(n). However, this sparked a question of what value n has to be in order for ((n^a)-1)/a) to equal e.

So, like any normal person, I went straight to Desmos and graphed ((n^x)-1)/x) and y=e, looking for what number I had to put in n to approximate the point at which they equal at around zero (which I know is undefined, but considering that we are approaching 0, it shouldn't matter).

Now, I got to a number approximately 15.15426, which I then looked up and got a paper by Walter Gottschalk called "The Hyperpower Function #66 of Gottschalk's Gestalts", which said that a very close number to this was, unbelievably, e^e, which I would say is most likely not a coincidence.

I put this number in as n and, wouldn't you know it, it approximated exactly e (I understand this juxtaposition, but just bear with me).

I have tried putting this function to equal e in algebra solvers and got told there was no solutions for n. I tried looking up this problem or any uses for e^e and have found nothing except the exceptional uses for vitamin e in the human body.

I am no where near smart enough or have the right tools necessary to solve this problem so I'm hoping someone on here will read of my plight and help, which may be you, if you are reading this

Thank you.

P.S. I hope this problem isn't super easy and I look like and idiot.

Andrew Thorson
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  • 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 Jun 18 '24 at 07:13
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    So you wish to verify that $$ \lim_{a \to 0} \frac{(e^e)^a-1}{a} = e $$ This is a generalization of the same fact that was noted in the video: $$ \lim_{a \to 0} \frac{n^a-1}{a} = \ln n $$ (Notice that if $n=e^e$, then $\ln(e^e) = e$.) The above limit is arguably a definition of the logarithm, one of many out there. But let's suppose it's not; how can we verify the claim?

    It can be taken as a corollary of the more standard claim for $n=e$: $$ \lim_{a \to 0} \frac{e^a-1}{a} = 1 $$ [cont.]

     – PrincessEev Jun 18 '24 at 07:29
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    Then$$ \lim_{a \to 0} \frac{n^a-1}{a} = \lim_{a \to 0} \frac{e^{a \ln n}-1}{a} = \ln n \lim_{a \to 0} \frac{e^{a \ln n}-1}{a \ln n} = \ln n \lim_{b = a \ln n \to 0} \frac{e^b - 1}{b} = \ln n $$ The standard result is discussed in more detail here. (Maybe it's discussed elsewhere on MSE without appealing to this result as well and someone can link that instead.) – PrincessEev Jun 18 '24 at 07:29

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