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I would like a proof as to why as the limit of $n$ approaches infinity, $(1+\frac{1}{n})^n$ means the same thing as the derivative of $e^x$ = $e^x$.

I have tried proving this by converting $(e^x)'$ into the limit definition of e, and vice versa but keep ending up at a dead end.

I am a Precal BC student, so I do not have much Calculus experience and would love an explanation that involves as little Calculus as possible.

By the way, here is the work that I have done so far:

$(e^x)'$ = $e^x$ $\Rightarrow$ derivative formula = $\lim_{h \to 0} \frac{f(x+h)- f(x)}{h}$ $\Rightarrow$ plug it in $\Rightarrow$ $\lim_{h \to 0} \frac{e^{(x+h)} - e^x}{h}$ = $e^x \Rightarrow \lim_{h \to 0} \frac{e^x(e^h - 1)}{h}$ = $e^x$ $\Rightarrow$ h = $\frac{1}{n}$ $\Rightarrow$ $n \to \infty$ $\Rightarrow$ $\lim_{n \to \infty}$ $\frac{e^x(e^{(1/n)} - 1)}{\frac{1}{n}}$ = $e^x$ $\Rightarrow$ $\lim_{n \to \infty} (e^{\frac{1}{n}} - 1)n = 1$

This is the part where I get stuck. I am not supposed to transfer the limit over to the other side so I don't know what to do. Please help and thanks in advance!

user21820
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Freedom Eagle
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    The limit is simply $e$ (without any derivative or exponent). You should maybe say what your definition of $e$ is since some people would define it exactly that way. – Carlos Esparza Nov 22 '18 at 22:38
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    A lot of people use this limit as the definition of $e$. We can prove that it's equivalent to other definitions and find the decimal approximation to the number with it – Yuriy S Nov 22 '18 at 22:39
  • https://math.stackexchange.com/questions/2067849/why-does-lim-n-to-infty1-frac1nn-e-instead-of-1?rq=1 – amWhy Nov 22 '18 at 22:51
  • You should clarify what exactly you are looking for. Are you aware that $(1+\frac{1}{n})^n\to e$? Whay are you referring to $e^x$? – user Nov 22 '18 at 22:59
  • @gimusi That sort of clarification ought to be sought, and obtained, before jumping the gun to answer. – amWhy Nov 22 '18 at 23:02
  • @amWhy For my experience, I see that often with new contributors give an answer can solicit their involving in a discussion more than comments or downvoting. – user Nov 22 '18 at 23:06
  • I am trying to figure out how I can prove that (e^x)' = e^x and e = lim as n --> infinity of (1+1/n)^n mean the same thing by converting (e^x)' into lim as n --> infinity of (1+1/n)^n or vice versa – Freedom Eagle Nov 22 '18 at 23:17
  • By the way, I don't really know how to use this site, and it is not showing me some of the comments that it was showing me before, so sorry if I don't reply. I really appreciate the help! – Freedom Eagle Nov 22 '18 at 23:18
  • how do I make this question not on hold? i just edited it – Freedom Eagle Nov 24 '18 at 02:50
  • You have to wait for people to vote to reopen it. Anyway, there is no simple correct explanation that does not involve a bit of real analysis (which is rigorous calculus). The main problem is that you haven't even defined what $e^x$ means, and once you do that and prove its properties, correctly, you would have done enough to make the equivalence you ask about trivial. – user21820 Nov 24 '18 at 10:05
  • Once you construct the real exponential function $\exp$ and prove that $\exp' = \exp$ and $\exp$ is strictly increasing, define $\ln$ as its inverse and $a^b = \exp(b·\ln(a))$ for reals $a,b$ such that $a > 0$. Then $e^x = \exp(x)$ for any real $x$. (Note that it is wrong to write "$(e^x)'$" since "$f'$" only makes sense for a function $f$.) And familiar properties can be derived, such as $a^{b+c}$ $= \exp((b+c)·\ln(a))$ $= \exp(b·\ln(a)+c·\ln(a))$ $= \exp(b·\ln(a))+\exp(c·\ln(a))$ $= a^b·a^c$ for any reals $a,b,c$ such that $a>0$. – user21820 Nov 24 '18 at 10:09
  • And for any real $x$, as natural $n \to \infty$ we have $(1+\frac{x}{n})^n$ $= \exp(n·\ln(1+\frac{x}{n}))$ $\in \exp(n·(\frac{x}{n}+O((\frac{x}{n})^2)))$ $⊆ \exp(x)·\exp(O(\frac{x^2}{n}))$ $⊆ \exp(x)·(1+O(\frac{x^2}{n}))$ $\to \exp(x)$ where we have used the asymptotic expansions of $\exp$ and $\ln$ since $\frac{x}{n} \to 0$ and $\frac{x^2}{n} \to 0$. For more on asymptotic analysis see here. – user21820 Nov 24 '18 at 10:16
  • As you can see, my last comment above is essentially a 1-line proof of a much more general fact than you're asking for, but of course it depends on all the hard work that went into defining $\exp$ and proving its properties, which translates to properties of real exponentiation. (If you have never questioned the 'laws' of exponents, now's not too late to start!) There is no need to prove 'the other direction' of the desired equivalence; if we define $E = \lim_{n\in\mathbb{N}\to\infty} (1+\frac1n)^n$ after proving the limit exists, we know that $E = \exp(1)$ so immediately $E^x = \exp(x)$. – user21820 Nov 24 '18 at 10:24

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That limit expression gives the number $e$ itself: for $e^x$ it's

$$\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^n .$$

if you differentiate this, you get a $1/n$ from inside - the coefficient of $x$; and a $n$ from the exponent, which cancel out. You are left with $n-1$ as the exponent ... but this makes no difference in the limit as $n\rightarrow\infty$ ... infact you could take that limit formula & add any finite real number to either occurence of $n$ in it,

$$\lim_{n\rightarrow\infty}\left(1+\frac{x}{n+\mu}\right)^{n+\nu} ,$$

& it would make no essential difference to it, because of $n$ tending to $\infty$.

AmbretteOrrisey
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  • How would the expression of $e^x$ be lim n→∞ $(1+/frac{x}{n})^n$ rather than it being $(1+/frac{1}{n})^xn$? And, by the way, how did you get your limit sign to look like that? – Freedom Eagle Nov 25 '18 at 00:31
  • Really sorry for the formatting issues. What I meant to ask was how the expression of $e^x$ be lim n→∞ $(1+\frac{x}{n})^n$ rather than it being $(1+\frac{1}{n})^{xn}$? – Freedom Eagle Nov 25 '18 at 00:52
  • It pans out equivalent - precisely because of the limiting process. Observe how it's equivalent to $$\lim_{\frac{n}{x}\rightarrow\infty}\left(1+\frac{x}{n}\right)^{x\frac{n}{x}}$$ – AmbretteOrrisey Nov 25 '18 at 10:34
  • The letting $n\rightarrow\infty$ allows various strangelooking 'liberties' to be taken ... like that substitution of $\mu$ & $\nu$ that I did. And you'll see I've answered your other question! – AmbretteOrrisey Nov 25 '18 at 10:38