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Prove that $\lim_\limits{x\to 0}{\frac{e^x-1}{x}}=1$.

I currently know only one approach (using L'Hopital 's Rule and derivatives) as follows:

$$\lim_\limits{x\to 0}{\frac{e^x-1}{x}}=\lim_\limits{x\to 0}{\frac{\left(e^x-1\right)'}{x'}}=\lim_\limits{x\to 0}{\left(e^x\right)}=e^0=1$$

Here I ask for other proofs than those, preferably neither using derivatives in any way nor using Taylor, etc.

Note: An approach for $\lim_\limits{x\to 0^+}{(x\ln x)}$ without using derivatives can be found here.

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    $\begingroup$ We need to know which hands to tie behind our backs? What definition of $e^x$ is permissible? $\endgroup$ – Rob Arthan Sep 16 '15 at 17:13
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    $\begingroup$ @RobArthan The following definition(s) is(are) (all) allowed: $$e^x=\lim_\limits{n\to +\infty}{\left( 1+\frac{x}{n}\right)^n}=\lim_\limits{n\to +\infty}{\left[ \left( 1+\frac{1}{n}\right)^{n\cdot x}\right] }=\left[ \lim_\limits{n\to +\infty}{\left( 1+\frac{1}{n}\right)^n}\right] ^x$$ I would prefer to avoid the following definition (Taylor): $$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$ $\endgroup$ – Jason Sep 16 '15 at 17:21
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    $\begingroup$ We are splitting hairs since $\sum_{n\geq 0}\frac{x^n}{n!}$ and $\lim_{n\to +\infty}\left(1+\frac{x}{n}\right)^n$ are two equivalent definitions of $e^x$. You may prove it by noticing that both terms are continuous solutions of the functional equation $f(x)f(y)=f(x+y)$ with $f(0)=1$. $\endgroup$ – Jack D'Aurizio Sep 16 '15 at 17:26
  • $\begingroup$ @JackD'Aurizio: I thought that the last definition (the sum one) was proved using Taylor expansion, wasn't it? $\endgroup$ – Jason Sep 16 '15 at 17:32
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    $\begingroup$ A definition is a definition, there is nothing to prove about it. I am just saying that, equivalently, we may take one or the other as the definition of $e^x$, so to avoid one or the other is of little interest. $\endgroup$ – Jack D'Aurizio Sep 16 '15 at 17:35
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We can assume the "Important limit" $\lim_{x\rightarrow 0}(1+x)^{1/x}=e$. Then \begin{align*} \lim_{x\rightarrow 0} \frac{e^x-1}{x} &= \lim_{y\rightarrow 0}\frac{y}{\ln (1+y)}\\ &= \lim_{y\rightarrow 0}\frac{1}{\ln \left((1+y)^{1/y}\right)}\\ &= 1. \end{align*}

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Using $$e^x=\lim_{n\to\infty}(1+\frac{x}{n})^n=1+x+O(x^2)$$ Which is not directly using the Taylor Series, just the binomial expansion.

You can procede like in the proof of using the Taylor series.

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Using that

$$\exp(x)=\sum_{n=0}^\infty \frac{x^n}{n!}$$ and that the series $$\frac{e^x-1}{x}=\sum_{n\ge0}\frac{x^n}{(n+1)!}$$ is uniformly convergent on every interval containing $0$ so interchanging limit and $\sum$ is allowed and the result follows.

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  • $\begingroup$ This uses the taylor expansion, which was supposed to be avoided. $\endgroup$ – David Hill Sep 16 '15 at 17:13
  • $\begingroup$ The OP said "preferrably" avoiding Taylor. $\endgroup$ – Rob Arthan Sep 16 '15 at 17:14
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    $\begingroup$ -1: op asked not to use derivatives, taylor and related $\endgroup$ – Blex Sep 16 '15 at 17:14
  • $\begingroup$ @Blex: please see my comment below the main question. (+1) back. $\endgroup$ – Jack D'Aurizio Sep 16 '15 at 17:41
  • $\begingroup$ @JackD'Aurizio: If $exp(x) = \sum \frac{x^n}{n!}$ by definition then I agree, but I'm denied un-(-1) the answer. $\endgroup$ – Blex Sep 18 '15 at 19:30
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If you are allowed to use the fact that $e^z$ is analytic and hence you can take $z=iy\rightarrow 0$ and obtain the same limit (assuming it exists) the you get $\frac{\sin y}{y}$ whose limit can be obtained from geometrically.

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  • $\begingroup$ I will delete my post as it's redundant now!! $\endgroup$ – Chinny84 Sep 16 '15 at 17:34
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    $\begingroup$ I don't think it's redundant despite the similarity. $\endgroup$ – Shahar Even-Dar Mandel Sep 16 '15 at 17:37
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Using the definition of e, the limit becomes the same as:

$$\lim_{x\to 0}\lim_{n\to \infty} \frac{ \left(1+\frac{1}{n}\right)^{nx}-1}{x}$$

We can evaluate this along the path $n=\frac{1}{x}$. So, the limit becomes: $$\lim_{x\to 0} \frac{ \left(1+x\right)^{x/x}-1}{x}$$

This gives us:

$$ \lim_{x\to 0} \frac{ \left(1+x\right)-1}{x} = 1$$

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maybe this satisfies. $$ \frac{e^x-1}{x} = e^{x/2}\frac{e^{x/2}-e^{-x/2}}{x} \to e^{it}\frac{e^{it}-e^{-it}}{i2t} = \frac{\sin t}{t}e^{it} $$ Then taking limits to zero.

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  • $\begingroup$ Can you explain to me how you proceed with the $\to$? Where did $2it$ appear? Sorry to bother you... $\endgroup$ – Jason Sep 16 '15 at 17:46

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