Taking as our definition of exponentiation repeated multiplication (extended to real exponents by continuity), can we show that the limit
$$\lim_{h\to 0}\dfrac{a^h-1}{h}$$
exists, without l'Hôpital, $e$, or even natural logarithm? Sure, l'Hôpital will work, but that's circular if we're developing calculus of transcendentals from first principles. There is a good answer to this question already by user Neal, but he uses the exponential function with base $e$ (it's been answered many times: see also here, here, and here).
But using the special properties of $e$ strikes me as circular too; not literally logically circular, in the sense that we are invoking results we're trying to prove, since there are definitions of $e^x$ which make it trivial to verify the derivative. But perhaps pedagogically circular for a complete novice, the special properties of $e$ appear unmotivated because they cannot be justified without reference to the very derivative we are trying to compute (or else a detour through logarithms, but let's not).
Can we find nice squeeze theorem bounds like Neal has, but for the function $a^x$ instead of $e^x$, with the additional handicap that we can't just write $a^x=e^{x\log a}$? I thought to substitute a series expansion for $\log a$, but didn't come up with any bounds that were nicely polynomial in both $x$ and $a$.
I wonder whether the geometric proof of $\lim (\sin x)/x$ (see for example, robjohn's answer here) could be adapted.
Obviously without a reference to natural logarithm, we cannot compute the value of the limit. But I just want to show it exists (via squeeze theorem or monotone convergence). Once we know this limit exists, we can show it behaves like a logarithm, whence there is a unique base for which the limit is 1, which we call $e$. The rest of the development of calculus of exponentials and logs follows easily. This seems like the approach that would appear the most accessible yet motivated to a novice calculus student.
An analogous limit to $\lim\dfrac{a^h-1}{h}$ for understanding to differentiating exponential functions, are the limits $\lim\limits_{n\to\infty} (1+\frac{1}{n})^n$ and $\lim\limits_{n\to\infty} (1+\frac{1}{n})^{n+1}$ for differentiating the logarithm, if you prefer to start with that as your primitive concept. Both limits are shown to exist using Bernoulli's inequality (see WimC's answer here for the first limit, and see David Mitra's answer here for the second limit). I tried without success to use Bernoulli's inequality to show my sequence was monotone. This limit can also be analyzed using the AM-GM inequality as seen in user94270's answer to this question. So that inequality may help here.
I would also accept an explanation of why the limit cannot be computed without transcendental techniques, or an opinion why this is not a pedagogically sound approach to introducing the calculus of exponentials and logarithms.
Edit: This question has a nice solution by Paramanand Singh to a closely related problem.
Answer
Let $a>1.$ I assume $a^x$ is continuous, and that the basic exponent law $a^{x+y}=a^xa^y$ holds.
Claim: $a^x$ is convex on $[0,\infty).$ Proof: Because $a^x$ is continuous, it suffices to show $a^x$ is midpoint convex. Suppose $x,y\in [0,\infty).$ Using $(uv)^{1/2} \le (u+v)/2$ for nonnegative $u,v,$ we get $a^{(x+y)/2} = (a^{x} a^{y})^{1/2} \le (a^x+a^y)/2.$
Now if $f$ is convex on $[0,\infty),$ then $(f(x)-f(0))/x$ is an increasing function of $x$ for $x\in(0,\infty).$ This is a simple and easily proved property of convex functions.
Claim: $\lim_{x\to 0^+}(a^x-1)/x$ exists. Proof: All of these quotients are bounded below by $0.$ As $x$ decreases to $0,(a^x-1)/x$ decreases by the above. Because of the lower bound of $0,$ the limit exists.
It follows that $\lim_{x\to 0}(a^x-1)/x$ exists. This follows from the above and the fact that if $x>0,$ then $a^{-x} = 1/a^{x}.$ To handle $0
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