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How does one show that $\lim_{x\rightarrow\infty}\left(1+\frac{1}{x}\right)^x$ converges? I have a book that uses this to define $e$ but they don't show that it converges. I suppose we could give it an upper and lower bound but I don't know how to do this. I've seen this question before on here but I haven't seen an answer I like yet.

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closed as off-topic by Did, Paramanand Singh, M. Winter, Moishe Kohan, John B Oct 31 '17 at 21:24

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  • $\begingroup$ As witnessed unvoluntarily by the two answers instantly posted below, to address your question requires to know the tools available to you. IOW, what do you know about the logarithm, say? $\endgroup$ – Did Oct 31 '17 at 11:56
  • $\begingroup$ The properties of the natural log have not been defined yet. $\endgroup$ – David Oct 31 '17 at 12:00
  • $\begingroup$ "I haven't seen an answer I like yet." Wow. I admire your chutzpah. I would have thought that any correct and clear answer (and there are plenty of those) would suffice. I'm not sure I'm prepared to delve into the realm of personal aesthetics, and I certainly am not if you don't tell me the criteria in advance. This becomes something other than a math problem. $\endgroup$ – John Hughes Oct 31 '17 at 12:00
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    $\begingroup$ For example, a low-tech approach is to consider $$x_n=\left(1+\frac1n\right)^n=\sum_{k=0}^n\frac1{k!}\left(1-\frac1n\right)\cdots\left(1-\frac{k}n\right)$$ and to show that $(x_n)$ is increasing and bounded above by $$\sum_{k=0}^\infty\frac1{k!}$$ Both proofs are elementary. Then remains to extend this to every nonnegative real $x$, which is also elementary. $\endgroup$ – Did Oct 31 '17 at 12:03
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    $\begingroup$ "The properties of the natural log have not been defined yet" Which is the reason why you accept an answer based on these properties, I guess? Well done. $\endgroup$ – Did Oct 31 '17 at 13:15
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Set $h =\frac1x\to 0~~as ~~x\to\infty$ Then, $$\lim_{x\rightarrow\infty}\left(1+\frac{1}{x}\right)^x =\lim_{x\rightarrow\infty}\exp\left(\frac{\ln\left(1+\frac{1}{x}\right)}{\frac1x}\right) =\lim_{h\rightarrow 0}\exp\left(\frac{\ln\left(1+h\right)}{h}\right) = e$$

Given that $$\lim_{h\rightarrow 0}\frac{\ln\left(1+h\right)}{h} = 1$$

Edit: For OP doubt. I think he to know how can want guess the existence of $e?$ without appealling $\ln$ and $\exp$.

In fact Euler Predicted the existence of $e$ before the creation of the functions $\ln$ and $\exp$: it is explained here: https://www.youtube.com/watch?v=AuA2EAgAegE

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    $\begingroup$ I think you mean $1,$ not $0$. $\endgroup$ – Cameron Buie Oct 31 '17 at 11:54
  • $\begingroup$ Don't we have to know that the limit converges to some number $e$ before we use the log of $e$? As I said, this limit is used to define $e$. What if it diverges? Then it makes no sense to use the natural log in that case. $\endgroup$ – David Oct 31 '17 at 11:58
  • $\begingroup$ There are many definitions of $\ln$; one is that $\ln (x) = \int_1^x \frac{1}{t} dt$, which is known to exist (for $x > 0$) by the fundamental theorem of calculus, and even is known to be differentiable with derivative $\frac{1}{x}$ (by that same theorem), hence is a strictly increasing function. It's not even tough to prove it's a surjection from $R^{+}$ to $R$, hence invertible. Then one can define $e$ to be $ln^{-1}(1)$, and prove that it equals the limit you're wondering about. $\endgroup$ – John Hughes Oct 31 '17 at 12:04
  • $\begingroup$ @David It will be helful you let us konw how you want to define $\ln$ and $\exp$. $\endgroup$ – Guy Fsone Oct 31 '17 at 12:08
  • $\begingroup$ I define $e$ as that limit. The natural log has not yet been defined in my book. I just though that perhaps it could be shown that the limit exists without using $e$ and ln but I guess that's not the case. I'll define ln first and see what I can do. $\endgroup$ – David Oct 31 '17 at 12:14
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Consider $\ln b = \int\limits_1^b\frac{dx}{x}$ with $b = 1+\frac{1}{n}$ then

$\ln(1+\frac{1}{n}) = \int\limits_1^{1+\frac{1}{n}}\frac{dx}{x}$

The area represented by this integral lies between the area of two rectangles of height $\frac{n}{n+1}$ and $1$ and base $\frac{1}{n}$. These rectangles have areas $\frac{1}{1+n}$ and $\frac{1}{n}$, therefore:

$\frac{1}{n+1}\leq\ln\left(1+\frac{1}{n}\right)\leq\frac{1}{n}$

Multiplying through by $n$ and using $n\ln a = \ln a^n$:

$\frac{n}{n+1}\leq\ln\left((1+\frac{1}{n})^n\right)\leq1$

Since $\lim\limits_{n\to\infty}\frac{n}{n+1} = 1$ the middle quantity must approach $1$ by the squeeze theorem:

$\lim\limits_{n\to\infty}\ln\left((1+\frac{1}{n})^n\right)=1$

Now apply the continuous function $e^x$ to obtain the desired result:

$e^1 = e^{\lim\limits_{n\to\infty}\ln\left((1+\frac{1}{n})^n\right)} = \lim\limits_{n\to\infty}e^{\ln\left((1+\frac{1}{n})^n\right)} = \lim\limits_{n\to\infty}\left(1+\frac{1}{n}\right)^n$

Check out Rogawski Calculus section 7.5 pp 374-375.

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Consider $$A=\left(1+\frac{1}{x}\right)^x\implies \log(A)=x\log\left(1+\frac{1}{x}\right)$$ and use equivalents $$\log\left(1+\frac{1}{x}\right)\sim \frac{1}{x}$$ making $$\log(A)\sim 1\implies A \sim e$$

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