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I'm currently messing around with the limit of the Euler's constant. These two in particular: $$ \lim_{x\to 0} (1+x)^\frac{1}{x}=e $$ $$ \lim_{x\to \infty} (1+\frac{1}{x})^x=e $$ I really want to find an interpretation or a proof for them that doesn't use l'Hopital's rule, but I am quite lost to say the least. My current idea is to try finding a closed form for a finite sum $$ \sum_{x=0} ^{n} \frac{1}{x!} $$ and perhaps in the case where n tends to infinity it will match one of the limits. But I'm not too familiar with finding closed forms of series with factorials. I tried forming a reccurence relation and solving it, but it just loops back to a sum of factorials. Forming a differential equation doesn't seem to work either, because I have to differentiate a factorial or get zero, because n isn't in the formula for the general term and I just differentiate a constant. I'd really appriciate any help with this! And terribly sorry if this is actually simple and I'm just overlooking something.

DifferentiateWithoutRespectToX
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  • Check this: https://math.stackexchange.com/q/39170/42969, or this: https://math.stackexchange.com/q/433442/42969, or this: https://math.stackexchange.com/q/69806/42969 – Martin R Jul 19 '22 at 07:39
  • I know a way using the area under $y=1/x$, that $e$ and $ln$ are inverse, and the squeeze theorem for limits. – calc ll Jul 19 '22 at 07:44
  • @MartinR linked a post with the exact same question that has already been answered, so if anyone else is interested, I recommend you check the questions he linked. Thank you very much! This helps :) – DifferentiateWithoutRespectToX Jul 19 '22 at 07:59
  • For what it's worth, the phrase Euler's constant is almost always used for another number. Although some internet sites use "Euler's number" or other phrases for $e,$ pretty much for the last 200 years or more just about everyone has called it e -- verbally say and/or write the letter 'e', in the same way that people refer to $\pi$ by its Greek letter name (verbally or in written form). – Dave L. Renfro Jul 19 '22 at 09:34

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Take a look at a Riemann sum of size one under $1/x$.

Get the double inequality $$1/(n+1)\le\int_n^{n+1}1/x\rm dx \le1/n$$

Which leads to $$1/(n+1)\le \ln((n+1)/n)\le1/n$$

Now take $e$ $$e^{1/(n+1)}\le 1+1/n\le e^{1/n}$$.

Raise to the $n$, and you can take limits

$$e^{n/(n+1)}\le (1+1/n)^n\le e$$.

A little adjustment will get you $e^x=\lim_{n\to\infty}(1+x/n)^n$.

calc ll
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  • Thanks a lot for this, a very clear and elegant method :) I wanted to clarify the very first line though - how do we know the area under 1/x is between 1/n and 1/(n+1)? I plugged this into desmos and played around with different values of n - no idea how, but it is true – DifferentiateWithoutRespectToX Jul 19 '22 at 08:21
  • Sorry, nevermind, I think using the fact that the function is convex and comparing the integral to the area of a trapezium between x=n and x=n+1 works Thanks a lot again, I really appreciate this :) – DifferentiateWithoutRespectToX Jul 19 '22 at 08:30
  • It's easy to see if you draw a picture: the area under the graph is between the areas under the rectangles of base $1$ and heights $1/n$ and $1/(n+1)$. You remember upper and lower Riemann sums. Well these only have one term. – calc ll Jul 19 '22 at 09:07