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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.

For the purposes of this post, I define the exponential by any of the following limits: $$e^x =\lim_\limits{n\to +\infty}{\left( 1+\frac{x}{n}\right)^n}=\lim_\limits{n\to +\infty}{\left[ \left( 1+\frac{1}{n}\right)^{n\cdot x}\right] }=\left[ \lim_\limits{n\to +\infty}{\left( 1+\frac{1}{n}\right)^n}\right] ^x.$$

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

Xander Henderson
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    We need to know which hands to tie behind our backs? What definition of $e^x$ is permissible? – Rob Arthan Sep 16 '15 at 17:13
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    @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):

    $$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$

     –  Sep 16 '15 at 17:21
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    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 $f(0)=1$. – Jack D'Aurizio Sep 16 '15 at 17:26
  • @JackD'Aurizio: I thought that the last definition (the sum one) was proved using Taylor expansion, wasn't it? –  Sep 16 '15 at 17:32
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    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. – Jack D'Aurizio Sep 16 '15 at 17:35
  • See this answer http://math.stackexchange.com/a/541330/72031 which uses $e^{x} = \lim_{n \to \infty}(1 + (x/n))^{n}$. – Paramanand Singh Sep 18 '15 at 13:30
  • Posts about the same limit: Proving that $\lim\limits_{x \to 0}\frac{e^x-1}{x} = 1$ and other posts linked there. In particular this one seems closely related: Calculate $\lim_{x \to 0} (e^x-1)/x$ without using L'Hôpital's rule – Martin Sleziak Sep 19 '15 at 05:52

7 Answers7

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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*}

Gordon
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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.

MrYouMath
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Since $1 + x \le e^x$ for all $x$ and we have $e^x \le 1/(1 - x)$ for $x < 1$ so $$1 \le (e^x - 1)/x \le 1/(1 - x) \to 1$$ as $x\to 0$.

Ertxiem - reinstate Monica
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Keith A. Lewis
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  • While this is not in the question, the "admissible" definition(s) of the exponential is(are) given in comments. Your definition is not among those "admissible" definitions. Before proving the desired result, you should demonstrate that your definition is, indeed, equivalent to the "admissible" definition(s). – Xander Henderson Sep 25 '19 at 16:05
  • This definition implies the "admissible" definition. Note $1 - x \le e^{-x}$ so $e^x \le 1/(1 - x)$ for $x < 1$. Hence $(1 + x/n)^n \le e^x \le (1 - x/n)^{-n}$ Both sides converge to $\sum_{n\ge 0} x^n/n!$ – Keith A. Lewis Dec 05 '19 at 23:34
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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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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.

Shahar Even-Dar Mandel
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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:

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

Avatrin
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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.

Chinny84
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  • Can you explain to me how you proceed with the $\to$? Where did $2it$ appear? Sorry to bother you... –  Sep 16 '15 at 17:46