Show $\lim_{h\to 0} \frac{(a^h-1)}{h}$ exists without l'Hôpital or even referencing $e$ or natural log

Solution 1:

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 $0<a<1,$ look at $[(1/a)^x-1]/x$ to see $\lim_{x\to 0}(a^x-1)/x$ exists.

Solution 2:

If $a = 1$ the limit is obviously $0$. Let $a > 1$ and $0 < b < 1$. Using simple algebra it is easy to show that $$\frac{a^{r} - 1}{r} > \frac{a^{s} - 1}{s},\,\,\frac{1 - b^{r}}{r} < \frac{1 - b^{s}}{s}\tag{1}$$ where $r, s$ are positive rationals with $r > s$. (see equation $(11)$ of this post). On OP's request I am providing the proof of the above inequality here itself.

First let's us assume that $r, s$ are positive integers. Clearly we know that $a^{i} < a^{r}$ for all $i = 0, 1, 2,\dots, r - 1$ and hence on adding these equations we get $$1 + a + a^{2} + \dots + a^{r - 1} < ra^{r}$$ and multiplying the above equation by $(a - 1) > 0$ we get $$a^{r} - 1 < ra^{r}(a - 1) = ra^{r + 1} - ra^{r}$$ or $$(r + 1)a^{r} - r - 1 < ra^{r + 1} - r$$ or $$(r + 1)(a^{r} - 1) < r(a^{r + 1} - 1)$$ so that we have finally $$\frac{a^{r} - 1}{r} < \frac{a^{r + 1} - 1}{r + 1}$$ It thus follows that the sequence $t_{n} = (a^{n} - 1)/n$ is strictly increasing and hence $t_{r} > t_{s}$ for positive integers $r,s$ with $r > s$. This proves the first inequality of $(1)$ with the restriction that $r, s$ are positive integers. The second inequality dealing with $0 < b < 1$ can be proved similarly starting with $b^{i} > b^{r}$ for all $i = 0, 1, 2, \dots, r - 1$.

Next we extend the inequality $(1)$ to the case when $r, s$ are positive rational numbers with $r = p/q, s = m/n$ where $p, q, m, n$ are positive integers and $r > s$ so that $np > mq$. Let $c = a^{1/nq}$ so that $c > 1$ and therefore via inequality $(1)$ (with restriction of positive integral indexes) we get $$\frac{c^{np} - 1}{np} > \frac{c^{mq} - 1}{mq}$$ Multiply the above equation by $nq > 0$ we get $$\frac{a^{r} - 1}{r} > \frac{a^{s} - 1}{s}$$ so that the inequality $(1)$ is proved for the case when $r, s$ are positive rationals.

If we extend the definition of $a^{x}$ to real exponents $x$ by any method (like one using Dedekind cuts suggested by OP), we see that the above inequalities hold true even if $r, s$ are positive reals with $r > s$. However any such procedure is effectively based on limiting processes and hence we can only obtain the weaker versions of the above inequalities in this manner. We are however lucky that we only need the weaker version here. Thus we have $$\frac{a^{r} - 1}{r} \geq \frac{a^{s} - 1}{s},\,\,\frac{1 - b^{r}}{r} \leq \frac{1 - b^{s}}{s}\tag{2}$$ where $r, s$ are real numbers with $r > s > 0$ and $a, b$ are real numbers with $a > 1 > b > 0$. It is now obvious that for $a > 1$ the function $f(x) = (a^{x} - 1)/x$ in increasing in $(0, \infty)$. Also $f(x) > 0$ for all $x$. It thus follows that $f(x) \to L$ as $x \to 0^{+}$ and moreover $L \geq 0$.

If $x < 0$ so that $x = -y$ and $y > 0$ then we can see that $$f(x) = f(-y) = \frac{a^{y} - 1}{ya^{y}} = \frac{a^{y} - 1}{y}\frac{1}{a^{y}} \to L$$ as $y \to 0^{+}$. It thus follows that $f(x) \to L$ as $x \to 0$ and $L \geq 0$.

If $0 < a < 1$ then we can use the ineuqalities related to $b$ above and show that $f(x) \to L$ and $L \leq 0$ as $x \to 0$. With slightly more effort we can show that $L = 0$ if and only if $a = 1$. The existence of this limit for all $a > 0$ defines a new function of $a$ which is normally called the logarithm of $a$. From here we can develop a theory of exponential and logarithmic function. See more details in this post.

Solution 3:

I'll sketch one way. Assume that $a > 1$. (The case $a < 1$ can be done similarly.)

1. Show that $f(x) = a^x$ is continuous (in fact continuous from the right is enough) and strictly increasing.
2. Every continuous monotone function is differentiable almost everywhere. (This is a well-known, but not entirely trivial theorem.)
3. Show that $f(x+y) = f(x)f(y)$. Thus, if $f$ is differentiable at some point $x=x_0$, then $f$ is also differentiable at $x=0$, which shows that your limit exists.

Solution 4:

This addresses showing that $\lim c_n=\lim n(a^{1/n}-1)$ exists. Let $u=a^{1/n}$ and apply the identity $u^n-1=(u-1)(1+u+\cdots + u^{n-1}).$ Then after multiplying numerator and denominator of $c_n$ by the second factor of that identity, one gets to $c_n=(a-1)/I_n,$ where $$I_n=\frac{1}{n} [ a^{0/n}+a^{1/n}+\cdots +a^{(n-1)/n}],$$ which is a left endpoint Riemann sum for the integral $\int_0^1 a^x \ dx,$ so in the limit converges to that integral, showing as desired the limit exists.

Typically at least, a calc 2 student would be familiar with left hand Riemann sums and their convergence to the associated integral.