elementary way to prove strict monotonicity of $\frac{\mathrm d}{\mathrm d a} \big( a\mapsto a^x \big)$ at $x=0$

Question

is there a way using basic means (no derivatives, no using the fact that $\ln$ and $\exp$ are mutually inverse) to show that $$a \mapsto \lim_{h \to 0} \frac{a^h -1}h$$is strictly monotonously increasing?
I would like to use something along the lines of $(a+\delta)^h \overset{(1)}{>} a^h$ for $h,\delta>0$ but that only leads to $$\lim_{h \to 0} \frac{(a+\delta)^h -1}h \geq \lim_{h \to 0} \frac{a^h -1}h$$ in the limit, of course. There should be a better estimate for $(1)$, I think, but I cannot come up with something.
Any hint is highly aprreciated. Thanks a lot!

Answer 1

We will show this for only for $a>1$ and with limits from the right, but the methods will be easily modified to give a correct general proof for the left hand limit and the case $a<1$ as well. The only assumptions will be that exponentiation has already been defined as a continuous extension of rational exponentiation, as done here, for example.

First, note that once you believe the limit exists for each $a$, you can establish this pretty easily. That is, you can observe that for each $n\in\mathbb N$, and each $b>1$, we have

$$\frac{b^{\frac{1}{n}}-1}{\frac{1}{n}} = \frac{\left(\frac{b-1}{b^{\frac{n-1}{n}}+b^{\frac{n-2}{n}}+\dots+1}\right)}{\frac{1}{n}} \geq \frac{\left(\frac{b-1}{nb}\right)}{\frac{1}{n}}= \frac{b-1}{b}>0,$$ so if the limit exists, we have $$\lim_{h\to 0^+} \frac{b^h-1}{h} \geq \frac{b-1}{b}>0.$$

Then for $a_1<a_2$, we let $b=\frac{a_2}{a_1}$, and obtain

$$\lim_{h\to 0^+}\frac{a_2^h-1}{h} - \lim_{h\to 0^+}\frac{a_1^h-1}{h} = \lim_{h\to 0^+}\frac{a_2^h-a_1^h}{h} = \lim_{h\to 0^+}a_1^h\left(\frac{b^h-1}{h}\right) >0.$$

It remains to show the limit exists. This will follow immediately from the greatest lower bound property, provided we can show that the map $h\mapsto \frac{a^h-1}{h}$ is increasing.

To prove that the aforementioned map is increasing, we observe that by density of the rationals and continuity of exponentiation in $h$, we must only show this for $h$ rational, which in turn implies it suffices to show that for $p,q\in \mathbb N^+$ we have

$$\frac{a^{\frac{p+1}{q}}-1}{\frac{p+1}{q}}>\frac{a^{\frac{p}{q}}-1}{\frac{p}{q}},$$ or equivalently, $$\frac{a^{\frac{p+1}{q}}-1}{a^{\frac{p}{q}}-1}>\frac{\frac{p+1}{q}}{\frac{p}{q}}=1+\frac{1}{p}.$$ To see this, we simply let $b=a^{\frac{1}{q}}$ and expand the left side

$$\frac{b^{p+1}-1}{b^p-1} = \frac{b^p+b^{p-1}+\dots+1}{b^{p-1}+\dots+1} = 1+ \frac{b^p}{b^{p-1}+\dots+1} > 1+ \frac{b^p}{pb^p} = 1+\frac{1}{p}.$$

Update

While we can easily mimic the arguments above for the left hand limit, which I have left as an exercise, I should have mentioned that it still remains to establish that the left hand limit and right hand limit then coincide. To see this, we need merely observe that \begin{align*} \lim_{h\to 0^+}\frac{a^h-1}{h}-\lim_{h\to 0^-}\frac{a^{h}-1}{h} &= \lim_{h\to 0^+}\left(\frac{a^h-1}{h}-\frac{a^{-h}-1}{-h}\right) = \lim_{h\to 0^+}\frac{a^h+a^{-h}-2}{h}\\ &= \lim_{h\to 0^+}\left(a^{-h}\cdot \frac{(a^{h}-1)}{h}\cdot (a^h-1)\right)=0, \end{align*} with the last equality following from the fact we have already shown the middle factor converges to a finite number.