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?

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