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

Question

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?

Answer 1

As to your comment:

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.

Answer 2

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 $\lim_{y \rightarrow 0}{\dfrac{1}{\ln{(y+1)}^\frac{1}{y}}} = \dfrac{1}{\ln{(\lim_{y \rightarrow 0}{(y+1)^\frac{1}{y}})}} = \dfrac{1}{\ln{e}} = \dfrac{1}{1} = 1.$

Answer 3

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, $$ \begin{align} \lim_{x\to0}\frac{e^x-1}{x} &=\lim_{x\to0}\frac{e^x-e^0}{x-0}\\ &=\left.\frac{\mathrm{d}}{\mathrm{d}x}e^x\right|_{x=0}\\ &=\left.e^x\right|_{x=0}\\ &=1\tag{4} \end{align} $$

Answer 4

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.

Answer 5

In THIS ANSWER I used the limit definition of the exponential function along with Bernoulli's Inequality to show that the exponential function satifies the estimates for $x<1$

$$ 1+x\le e^x\le \frac1{1-x}\tag1$$

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Applying $(1)$ for $0<x<1$ we find that

$$1 \le \frac{e^{x}-1}{x}\le \frac{1}{1-x}\tag2$$

while for $x<0$ we find

$$\frac1{1-x} \le \left(\frac{e^{x}-1}{x}\right)\le 1\tag3$$

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Finally, applying the squeeze theorem to $(2)$ and $(3)$ we arrive at the coveted limit

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

as was to be shown!

TOOLS USED: The limit definition of the exponential function, Bernoulli's Inequality, Basic Algebra, and The Squeeze Theorem

Answer 6

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 $$\,\,e^x=y\iff^{\textrm{def.}} \log y=x$$ and assuming we know the usual about the logarithmic function we get $$y=e^x\implies \log y=x\implies\frac{1}{y}\,dy=dx\implies \frac{dy}{dx}=y=e^x\implies (e^x)'=e^x$$ and now using the definition of derivative, for any real we have $$e^x=(e^x)'=\lim_{h\to 0}\frac{e^{x+h}-e^x}{h}=e^x\lim_{h\to 0}\frac{e^h-1}{h}\implies \lim_{h\to 0}\frac{e^h-1}{h}=1$$since, by this def. of the exponential function, we get using the basic properties of logarithm $$e^{x+h}=e^xe^h\iff \log(e^xe^h)=x+h ....$$$${}$$

(2) By one of the most usual, and elementary, of the definitions of the number $\,e\,$ , we get: $$e:=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\implies e^h=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{hn}=\lim_{k\to\infty}\left(1+\frac{h}{k}\right)^k$$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!

Answer 7

Starting from the foundamental limit

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

by $y=\frac1n \to 0^+$ we have

$$\left(1+\frac1n\right)^n=e^{\frac{\log(1+y)}y}\to e \implies \frac{\log(1+y)}y\to 1$$

and by $y=e^x-1 \to 0^+$ as $x\to 0^+$

$$\frac y{\log(1+y)}=\frac{e^x-1}{x} \to 1$$

Answer 8

I have an approach using non standard Analysis, The required limit can be rewritten as, $$ st\left(\frac{e^{dx}-1}{dx}\right) $$ According to Euler's treatise, "The analysis of infinite" chapter 7, $e^{dx}=1+dx$. $$ st\left(\frac{1+dx-1}{dx}\right)$$ Which gives us $1$. So, $$\lim\limits_{x \to 0}\frac{e^x-1}{x} = 1$$