Complete and elementary proof that $(a^x - 1)/x $ converges as x goes to 0 Anybody who has taken a calculus course knows that
$$\lim_{x \to 0} \frac{a^x - 1}{x}$$
exists for any positive real number $a$, simply because the limit is by definition the derivative of the function $a^x$ at $x = 0$.  However, for this argument to be non-circular one must have an independent technique for proving that $a^x$ is differentiable.  The standard approach involves the following two steps:
1) Calculate the derivative of $\log_a x$ by reducing it to the calculation of
$$\lim_{h \to 0} (1 + h)^{\frac{1}{h}}$$
2) Apply the inverse function theorem.
I find this unsatisfying for two reasons.  First, the inverse function theorem is not entirely trivial, even in one variable.  Second, the limit in step 1 is quite difficult; in books it is often calculated along the sequence $h = \frac{1}{n}$ where $n$ runs over the positive integers, but doing the full calculation seems to be quite a bit more difficult (if one hopes to avoid circular reasoning).
So I would like a different argument which uses only the elementary theory of limits and whatever algebra is needed.  For instance, I would like to avoid logarithms if their use involves an appeal to the inverse function theorem.  Is this possible?

From the identity
$$ \frac{a^{x} - 1}{x} = \frac{a^{x/n} - 1}{x/n} \cdot \frac{1}{n} \cdot \left(1 + a^{x/n} + a^{2x/n} + \cdots + a^{(n-1)x/n} \right)$$
it shouldn't be terribly hard to show that the limit exists (although it might be a bit tedious to work through all of the detail). I can't say how illuminating this approach would be, though.
