Differentiating $e^x$ from first principles using limits.

Question

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?

Answer 1

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: $$1=\lim_{x\to0}\frac{e^x-1}x$$

Answer 2

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