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?
 A: Prove that 
$$\lim_n \left( 1+\frac{1}{n} \right)^n = \lim_n\left( 1+\frac{1}{n} \right)^{n+1}$$
exists, and call this limit $e$. The reason why these two limit exists is because it can be proven with the Bernoulli inequality that $\left( 1+\frac{1}{n} \right)^n$ is increasing and $\left( 1+\frac{1}{n} \right)^{n+1}$ is decreasing.
It follows that they are both convergent, and their ratio converges to $1$, thus the limit exists.
From here, we also get immediately that
Now, for each $x \in (0, \infty)$ we have $\lfloor x \rfloor \leq x <  \lfloor x \rfloor  +1$. For simplicity I will denote $n:=\lfloor x \rfloor$. Then, we get
$$\left( 1+\frac{1}{n+1} \right)^n \leq \left( 1+\frac{1}{x} \right)^x \leq \left( 1+\frac{1}{n} \right)^{n+1}$$
As $n \to \infty$ when $x \to \infty$, by a squeeze type argument we get
$$\lim_{x \to \infty} \left( 1+\frac{1}{x} \right)^x=e$$
Using $1-\frac{1}{x}=\frac{x-1}{x}=\frac{1}{1+\frac{1}{x-1}}$ we also get
$$\lim_{x \to \infty} \left( 1-\frac{1}{x} \right)^x=e^{-1}$$
and then
$$\lim_{y \to 0} \left( 1+y \right)^\frac{1}{y} =e$$
Let $y=a^x-1$ [Note: no logarithms are used here, we just use the fact that if the limit exists for all $y \to 0$, it also exists for this articular choice of $y$.]
Then, we get
$$\lim_{x \to 0} \left( a^x \right)^\frac{1}{a^x-1} =e$$
Thus, we proved that 
$$\lim_{x \to 0} a^\frac{x}{a^x-1} =e \,,$$
exists. At this point logarithms would solve the problem, but you can probably finish the argument without using logarithms. For example, you can prove that $a^y$ is strictly increasing/decreasing (which reduces to $x >y \Rightarrow a^{x-y} >1$) and prove the following lemma:
Lemma If $f$ is strictly monotonic and continuous,  and for some $c$ the limit 
$$\lim_{x \to c} f(g(x))$$
exists, then $\lim_{x \to c} g(x)$ exists.
A: 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 $(x,y)\to(0,0)$.
A: 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.
A: This question is ultimately related to conceptual foundation of exponentiation and logarithm. The limit of $(a^{x} - 1)/x$ as $ x \to 0$ requires that we first define the symbol $a^{x}$. This itself is a tough problem (for a beginner in calculus) and there are many approaches possible. Based on the question given here, one possible approach goes like this.
1) Let $f(x) = \lim_{n \to \infty}n(x^{1/n} - 1)$. Then it can be shown (using various theorems on limits and nothing else) that this limit exists for all $x > 0$ and hence defines a function $f(x)$ which has following properties:


*

*$f(1) = 0$

*$f(xy) = f(x) + f(y)$

*$f(x)$ is strictly increasing
function of $x$ and maps the domain $(0, \infty)$ into range
$(-\infty, \infty)$


2) Next we can use the inverse function theorem to define a new function $g(x)$ such that $y = g(x)$ and $x = f(y)$ are equivalent. Let $e = g(1)$ so that $f(e) = 1$ and then we define function $a^{x} = g(xf(a))$.
3) It can then be proved that $\lim_{x \to 0}(a^{x} - 1)/x = f(a)$ and $\lim_{h \to 0}(1 + h)^{1/h} = e = g(1)$
In traditional notation $f(x)$ is represented by $\log x$ and $g(x)$ is represented by $e^{x}$ or $\exp(x)$. The above presentation does not involve any circular reasoning, but does require the deep conceptual framework of real numbers and limits. Even at the beginning of this presentation we need to know the meaning of $x^{1/n}$ which can not be given without the completeness property of real numbers.
UPDATE: After looking at OP's comments on my answer I think what is needed here is to show that the limit $\lim_{x \to 0}(a^{x} - 1)/x$ exists under the following definition of $a^{x}$:
If $x$ is rational then $a^{x}$ can be defined to some extent via algebra as is done in courses of elementary algebra.
If $x$ is irrational then let $x_{n}$ be a sequence of rationals tending to $x$ as $ n \to \infty$ and then we define $a^{x} = \lim_{n \to \infty}a^{x_{n}}$.
This definition is intuitive but has technical challenges namely a) to show that $\lim_{n \to \infty}a^{x_{n}}$ exists and also show that b) if $x_{n}, y_{n}$ are rational and $x_{n} \to x, y_{n} \to x$ then $\lim_{n \to \infty}a^{x_{n}} = \lim_{n \to \infty}a^{y_{n}}$
Using this definition we can proceed as follows:
1) Show that if $x_{n}$ is a sequence of reals tending to $x$ then $a^{x} = \lim_{n \to \infty} a^{x_{n}}$
2) Show that $\lim_{x \to 0} a^{x} = 1$
3) Using 2) above show that it is sufficient to consider $x \to 0+$ when calculating $\lim (a^{x} - 1)/x$
4) Establish that $\lim_{n \to \infty}n(a^{1/n} - 1) = L$ exists. Use this to find bounds for $a^{1/n}$ like $a^{1/n} = 1 + \{(L + \phi(n))/n\}$ where $\phi(n)$ to zero as $n \to \infty$
5) Using 4) above estimate $a^{x}$ for small $x$ i.e. for $1/(n + 1) < x < 1/n$ and then show that $(a^{x} - 1)/x$ also tends to $L$ as $x \to 0+$.
All this can be done rigorously with epsilon-delta type arguments. Proving 4) will require the theorem that bounded and monotone sequences are convergent. Thus we can provide a rigorous proof that $\lim_{x \to 0}(a^{x} - 1)/x$ exists for all $a > 0$ and use this limit as definition of $\log a$ if one so desires.
A: If we know,
from the properties of exponents,
 that
$f(x) = a^x$
satisfies
$f(x+y) = f(x)f(y)$,
then
$f(x+h)-f(x)
=a^{x+h}-a^x
=a^xa^h-a^x
=a^x(a^h-1)
$
so
$\frac{f(x+h)-f(x)}{h}
=a^x\frac{a^h-1}{h}
=a^x\frac{f(h)-f(0)}{h}
$.
All we need,
therefore,
is that
$a^x$ is differentiable at $0$.
Then,
as has been said in other solutions,
we have to look at
some foundation stuff.
