TITLE: show $\lim_{x\to 0}\frac{e^x-1}{x}=1$ without L'Hopital
QUESTION [13 upvotes]: how would you show that $$\lim_{x\to 0}\frac{e^x-1}{x}=1$$ without using derivatives or l'hopital but using basic ideas that are generally introduced just before derivatives in a typical introductory calculus course. my attempt.. $$\lim_{x\to 0}\frac{e^x-1}{x}=\lim_{x\to 0}\lim_{n\to \infty}\frac{\left(1+\frac{x}{n}\right)^n-1}{x}$$ then.. this is 
$$ =\lim_{x\to 0}\lim_{n\to \infty}\frac{1+{n \choose 1}\frac{x}{n} + {n \choose 2}\left(\frac{x}{n}\right)^2 +\cdots+ {n \choose n}\left(\frac{x}{n}\right)^n -1}{x}$$ 
$$ =\lim_{x\to 0}\lim_{n\to \infty} \left[ 1+ \frac{   {n \choose 2}\left(\frac{x}{n}\right)^2 +\cdots+ {n \choose n}\left(\frac{x}{n}\right)^n }{x}\right]$$ 
then we could switch the limit order but the justification for this seems to be not very basic and I would like to provide a basic explanation, as basic as possible. 
by the way I am doing this to provide a pre-derivatives proof of $$\lim_{x\to c}\frac{e^x-e^c}{x-c}=e^c$$
thank you.

REPLY [2 votes]: Here's a possible method, which uses no power series.  Let's focus on the right-hand limit $x \to 0^+$.  Since
$$\lim_{x\to0^+}\frac{e^x - 1}{x} = \lim_{x\to0^+}\frac{1-e^{-x}}{xe^{-x}} = \lim_{x\to0^+} \frac{e^{-x}-1}{-x} = \lim_{x\to0^-} \frac{e^x-1}{x}$$
we lose no generality doing so, and it will let us work out inequalities more easily later.
We have that $e^x = \lim_{n\to\infty} \left(1+\frac{x}{n+1}\right)^{n+1} = \lim_{n\to\infty} \left(1+\frac{x}{n+1}\right)^n$.  Using this,
$$\begin{align*}
\lim_{x\to0^+} \frac{e^x-1}{x} &= \lim_{x\to0^+}\lim_{n\to\infty} \frac{\left(1+\frac{x}{n+1}\right)^{n+1} - 1^{n+1}}{x}\\
& = \lim_{x\to0^+}\lim_{n\to\infty} \frac{(1+\frac{x}{n+1} - 1)(1 + (1+\frac{x}{n+1}) + \cdots + (1 + \frac{x}{n+1})^n)}{x}\\
& = \lim_{x\to0^+}\lim_{n\to\infty} \frac{1}{n+1} \sum_{k=0}^n a_k \\
&= \lim_{x\to0^+}\lim_{n\to\infty} \frac{n}{n+1}\frac{1}{n} \sum_{k=0}^n a_k\end{align*}$$
Where we have used the formula for the difference of $n^\mbox{th}$ powers to get the second line, and $a_k = \left(1+\frac{x}{n+1}\right)^k$.  We have that 
$$1+\frac{k}{n}x \le a_k \le \left(1+\frac{x}{n+1}\right)^n$$
so,
$$\lim_{x\to0^+}\lim_{n\to\infty}\frac{1}{n} \sum_{k=0}^n 1+\frac{k}{n}x \le \lim_{x\to0^+}\frac{e^x-1}{x} \le \lim_{x\to0^+}\lim_{n\to\infty} \frac{1}{n} \sum_{k=0}^n\left(1+\frac{x}{n+1}\right)^n$$
$$\lim_{x\to0^+}1 + \frac{n(n-1)}{2n^2}x \le \lim_{x\to0^+}\frac{e^x-1}{x} \le \lim_{x\to0^+}\lim_{n\to\infty} \left(1+\frac{x}{n+1}\right)^n$$
(where we have removed the factors $\frac{n}{n+1} \to 1$ as $n\to\infty$ for brevity.)
Taking the limits as $n\to \infty$,
$$\lim_{x\to0^+}1 + \frac{1}{2}x \le \lim_{x\to0^+}\frac{e^x-1}{x} \le \lim_{x\to0^+} e^x$$
and so evaluating the two limits we find $\lim_{x\to0^+}\frac{e^x-1}{x} = 1$, as desired.