# Proving that $\lim\limits_{x \to 0}\frac{e^x-1}{x} = 1$

I was messing around with the definition of the derivative, trying to work out the formulas for the common functions using limits. I hit a roadblock, however, while trying to find the derivative of $e^x$. The process went something like this:

$$\begin{align} (e^x)' &= \lim_{h \to 0} \frac{e^{x+h}-e^x}{h} \\ &= \lim_{h \to 0} \frac{e^xe^h-e^x}{h} \\ &= \lim_{h \to 0} e^x\frac{e^{h}-1}{h} \\ &= e^x \lim_{h \to 0}\frac{e^h-1}{h} \end{align}$$

I can show that $\lim_{h\to 0} \frac{e^h-1}{h} = 1$ using L'Hôpital's, but it kind of defeats the purpose of working out the derivative, so I want to prove it in some other way. I've been trying, but I can't work anything out. Could someone give a hint?

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use the power series representation of $e^x$. – user20266 Jun 1 '12 at 20:20

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What is the definition of $\exp(x)$ you are working with? – user17762 Jun 1 '12 at 20:22

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What is your definition of $\exp x$? Often its the solution to $y'=y,y(0)=1$, in which case the derivative of $e^x$ comes for free. Otherwise the series representation works. – anon Jun 1 '12 at 20:22
@Thomas: Isn't using the power series representation of $e^x$ pretty much equivalent to saying that $(e^x)' = e^x$? – Javier Badia Jun 1 '12 at 20:22

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I cannot stress enough how incredibly useful the estimates $1+x \leq e^x \leq 1/(1-x)$ are, where $x < 1$ for the upper bound. – WimC Jun 1 '12 at 20:23
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## 5 Answers

Consider the differential equation

$$y - \left( {1 + \frac{x}{n}} \right)y' = 0$$

It's solution is clearly $$y_n={\left( {1 + \frac{x}{n}} \right)^n}$$

If we let $n \to \infty$ "in the equation" one gets

$$y - y' = 0$$

One should expect that the solution to this is precisely

$$\lim_{n \to \infty} y_n =y=\lim_{n \to \infty} \left(1+\frac x n \right)^n := e^x$$

Also note $$\mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{x}{n}} \right)^n} = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{x}{{xn}}} \right)^{xn}} = \mathop {\lim }\limits_{n \to \infty } {\left[ {{{\left( {1 + \frac{1}{n}} \right)}^n}} \right]^x}$$

My approach is the following:

I have as a definition of $\log x$ the following:

$$\log x :=\lim_{k \to 0} \frac{x^k-1}{k}$$

Another one would be

$$\log x = \int_1^x \frac{dt}t$$ Any ways, the importance here is that one can define $e$ to be the unique number such that

$$\log e =1$$

so that by definition

$$\log e =\lim_{k \to 0} \frac{e^k-1}{k}=1$$

From another path, we can define $e^x$ as the inverse of the logarithm. Since

$$(\log x)'=\frac 1 x$$

the inverse derivative theorem tells us

$$(e^x)'=\frac{1}{(\log y)'}$$

where $y=e^x$

$$(e^x)'=\frac{1}{(1/y)}$$

$$(e^x)'=y=e^x$$

The looking at the difference quotient, one sees that by definition one needs

$$\mathop {\lim }\limits_{h \to 0} \frac{{{e^{x + h}} - {e^x}}}{h} = {e^x}\mathop {\lim }\limits_{h \to 0} \frac{{{e^h} - 1}}{h} = {e^x}$$

so that the limit of the expression is $1$. One can also retrieve from the definition of the logarithm that

$$\eqalign{ & \frac{x}{{x + 1}} <\log \left( {1 + x} \right) < x \cr & \frac{1}{{x + 1}} < \frac{{\log \left( {1 + x} \right)}}{x} <1 \cr}$$

Thus

$$\mathop {\lim }\limits_{x \to 0} \frac{{\log \left( {1 + x} \right)}}{x} = 1$$

a change of variables $e^h-1=x$ gives the result you state. In general, we have to go back to the definition of $e^x$. If one defines $${e^x} = 1 + x + \frac{{{x^2}}}{2} + \cdots$$

Then

$$\frac{{{e^x} - 1}}{x} = 1 + \frac{x}{2} + \cdots$$

$$\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \left( {1 + \frac{x}{2} + \cdots } \right) = 1$$

from the defintion we just chose.

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I'm seeing that I haven't really thought this through. I would say that the only definition of $e^x$ that makes this problem an interesting problem is the compound interest one, since the ones you suggest are related to the fact that $(e^x)' = e^x$. But then it seems that the problem reduces to showing that the derivative of $\lim_{n \to \infty}(1+\frac{x}{n})^n$ is equal to itself. – Javier Badia Jun 1 '12 at 20:37
I guess I was looking for answers that use mostly the definition of a limit together with some of $e^x$'s properties, but I'm not sure that's practical. – Javier Badia Jun 1 '12 at 20:40
Solve $$y - \left( {1 + \frac{x}{n}} \right)y' = 0$$ Then let $n\to \infty$. – Peter Tamaroff Jun 1 '12 at 20:51
That works, then. Don't get me wrong, the rest of your answer is great too, it's just not precisely what I had in mind when I thought of the question. Thanks! – Javier Badia Jun 1 '12 at 21:15

If (as is fairly commonly done in calculus courses) one defines $\ln x$ by $$\ln x=\int_1^x \frac{dt}{t},$$ then it is easy to show that the derivative of $\ln x$ is $\frac{1}{x}$. One can either appeal to the Fundamental Theorem of Calculus, or operate directly via a squeezing argument.

If we now define $e^x$ as the inverse function of $\ln x$, then the fact that the derivative of $e^x$ is $e^x$ follows from the basic theorem about the derivative of an inverse function.

There are many ways to define the exponential function. The proof of the result you are after depends heavily on the approach we choose to take.

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Define $$f_n(x)=\left(1+\frac{x}{n}\right)^n\tag{1}$$ Note that $$f_n^{\,\prime}(x)=\left(1+\frac{x}{n}\right)^{n-1}\tag{2}$$ On compact subsets of $\mathbb{R}$, both $(1)$ and $(2)$ converge uniformly to $e^x$. This means that $$\frac{\mathrm{d}}{\mathrm{d}x}e^x=e^x\tag{3}$$ Therefore, $$\begin{align} \lim_{x\to0}\frac{e^x-1}{x} &=\lim_{x\to0}\frac{e^x-e^0}{x-0}\\ &=\left.\frac{\mathrm{d}}{\mathrm{d}x}e^x\right|_{x=0}\\ &=\left.e^x\right|_{x=0}\\ &=1\tag{4} \end{align}$$

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Let say $y=e^h -1$, then $\lim_{h \rightarrow 0} \dfrac{e^h -1}{h} = \lim_{y \rightarrow 0}{\dfrac{y}{\ln{(y+1)}}} = \lim_{y \rightarrow 0} {\dfrac{1}{\dfrac{\ln{(y+1)}}{y}}} = \lim_{y \rightarrow 0}{\dfrac{1}{\ln{(y+1)}^\frac{1}{y}}}$. It is easy to prove that $\lim_{y \rightarrow 0}{(y+1)}^\frac{1}{y} = e$. Then using Limits of Composite Functions $\lim_{y \rightarrow 0}{\dfrac{1}{\ln{(y+1)}^\frac{1}{y}}} = \dfrac{1}{\ln{(\lim_{y \rightarrow 0}{(y+1)^\frac{1}{y}})}} = \dfrac{1}{\ln{e}} = \dfrac{1}{1} = 1.$

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1) Using Apostol's definition in "Calculus I", 6.12, or in Finney's " Calculus and Analytic Geometry". 6.3 , or in Swokowski's "Calculus and Analytic Geometry", def. 7.8, we have that $$\,\,e^x=y\Longleftrightarrow^{def.} \log y=x$$ and assuming we know the usual about the logarithmic function we get $$y=e^x\Longrightarrow \log y=x\Longrightarrow\frac{1}{y}\,dy=dx\Longrightarrow \frac{dy}{dx}=y=e^x\Longrightarrow (e^x)'=e^x$$ and now using the definition of derivative, for any real we have $$e^x=(e^x)'=\lim_{h\to 0}\frac{e^{x+h}-e^x}{h}=e^x\lim_{h\to 0}\frac{e^h-1}{h}\Longrightarrow \lim_{h\to 0}\frac{e^h-1}{h}=1$$since, by this def. of the exponential function, we get using the basic properties of logarithm $$e^{x+h}=e^xe^h\Longleftrightarrow \log(e^xe^h)=x+h ....$$$${}$$

(2) By one of the most usual, and elementary, of the definitions of the number $\,e\,$ , we get: $$e:=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\Longrightarrow e^h=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{hn}=\lim_{k\to\infty}\left(1+\frac{h}{k}\right)^k$$putting $\,\,k:=hn\,\,$, so$$\lim_{h\to 0}\frac{e^h-1}{h}=\lim_{h\to 0}\frac{\lim_{n\to\infty}\left(1+\frac{h}{n}\right)^n-1}{h}=\lim_{h\to 0}\frac{\lim_{n\to\infty}\sum_{k=0}^n\binom{n}{k}\left(\frac{h}{n}\right)^k-1}{h}=$$$$=\lim_{h\to 0}\lim_{n\to\infty}\frac{\sum_{k=1}^n\binom{n}{k}\left(\frac{h}{n}\right)^k}{h}=1+\lim_{h\to 0}\lim_{n\to\infty}\rlap{/}{h}\frac{\sum_{k=2}^n\binom{n}{k}\frac{h^{k-1}}{n^k}}{\rlap{/}{h}}=1$$as every single summand in what's left in that sum is multiplied by a positive power of $\,h$...

As far as I know, most authors go with the first approach, I couldn't find even one that goes with the second approach, and some even go with the definition of the exponential function as a power series...now you choose!

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