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?

The most common definition of $e$ is $$e:=\lim_{x\to0}\left(1+x\right)^{1/x}$$ although you often see it with $n=1/x$ and as $n\to\infty$.
Now $$\begin{aligned}\lim_{x\to0}\left(\frac{e^x-1}{x}\right)&=\lim_{x\to0}\left(\frac{\left(\lim_{y\to0}\left(1+y\right)^{1/y}\right)^x-1}{x}\right)\\
&=\lim_{x\to0}\left(\frac{\lim_{y\to0}\left(1+y\right)^{x/y}-1}{x}\right)\\
&=\lim_{x\to0}\lim_{y\to0}\left(\frac{\left(1+y\right)^{x/y}-1}{x}\right)
\end{aligned}$$ 
Now are you willing to believe that $$\lim_{(x,y)\to(0,0)}\left(\frac{\left(1+y\right)^{x/y}-1}{x}\right)
$$ exists? If so it equals the last line above, and it also equals  $$\lim_{x\to0}\left(\frac{\left(1+x\right)^{x/x}-1}{x}\right)
$$ by tracking along the line $y=x$. This last quantity is clearly $1$. From here, you can establish that the derivative of $e^x$ is $e^x$, from which the Chain Rule gives that the derivative of $a^x$ is $a^x\ln(a)$, from which you can get that $\lim_{x\to0}\left(\frac{a^x-1}{x}\right)=\ln(a)$.
See if you can find justification for the existence of that limit as $(x,y)\to(0,0)$.
