Differentiating $e^x$ from first principles using limits. I have been trying to differentiate the exponential function from first principles without the use of Taylor's series or the derivative of its inverse function ($\frac{d}{dx} (\ln x) = \frac{1}{x}$ and $\ln (e^x) = x$.
Let $f(x) = e^x$, then differentiating $f(x)$ from first principles,
$$f^\prime(x) = \lim_{\delta x \to 0} \frac{f(x+\delta x) - f(x)}{\delta x} = \lim_{\delta x \to 0} \frac{e^{x+\delta x} - e^x}{\delta x} = \lim_{\delta x \to 0}\frac{e^x(e^{\delta x} - 1)}{\delta x} = e^x \lim_{\delta x \to 0} \frac{e^{\delta x} - 1}{\delta x}$$
Therefore, in order to prove from first principles that $\frac{d}{dx}(e^x) = e^x$, I would need to first show that
$$\lim_{\delta x \to 0} \frac{e^{\delta x} - 1}{\delta x} = 1$$
However, I am not sure how to evaluate this limit and the use of L'Hôpital's rule requires preliminary knowledge on the derivative of $e^x$. 
Is it possible to prove the derivative of $e^x$ from first principles solely using limits, or is it impossible as the knowledge of its derivative is a prerequisite to its discovery by Bernoulli?
 A: As I've said fairly often in the last few days (for some reason), one of my favorite equations is:
$$e^x\ge x+1$$
The reason, partly, is that it uniquely defines $e$ without calculus. Hint for a proof: use this. (By the way, do equations need equals signs? Or is it equalities that need equals signs?)
Now, replacing $x$ by $-x$, we get $e^{-x}\ge1-x$, so:
$$e^x\le\frac1{1-x}$$
(The inequality gets reverse for $x>1$, as the right-hand side is negative there. But we only care about when $x$ is near zero.)
Thus:
\begin{align}
x+1\le{}&e^x\le\frac1{1-x}\\
x\le{}&e^x-1\le\frac x{1-x}\\
1\le^*{}&\frac{e^x-1}x\le^*\frac1{1-x}
\end{align}
*Since we just divided by $x$, the inequalities get reversed if $x$ is negative. It doesn't affect the argument.
Let $x$ tend to zero. By the squeeze theorem:
$$1=\lim_{x\to0}\frac{e^x-1}x$$
A: Sadly I don't have enough reputation to comment, so I guess I will post an answer instead. I hope its not inappropriate.
As indicated by the comments, the answer varies on your definition of $e^x$. In fact one definition is implicitly by:
$$ \lim_{h \to 0} \frac{e^h - 1}{h} = 1$$
Defining $e^x$ this way, you are finished. If you define it as 
$$ e^x = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^n$$
you are left with a bit more work.
Regardless, you will probably find all the answers you're looking for in the following thread: Proof of derivative of $e^x$ is $e^x$ without using chain rule
