limits - Proving that $limlimits_{x to 0}frac{e^x-1}{x} = 1$

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?

Answer

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.