# Existence of the limit $\lim_{h\to0} \frac{b^h-1}h$ without knowing $b^x$ is differentiable

When trying to derive, from first principles, the fact that exponential functions $$a^x$$ (where $$a>1$$ is real) are differentiable, we easily see that $$\lim_{h\to0} \frac{a^{x+h}-a^x}h = a^x \lim_{h\to0} \frac{a^h-1}h,$$ provided the latter limit exists. It's even pretty easy to see that $$\lim_{h\to0} \frac{a^h-1}h = ( \log_b a ) \lim_{h\to0} \frac{b^h-1}h$$ for any other real $$b>1$$, provided the latter limit exists. (And then one can define $$e$$ to be the number such that $$\lim_{h\to0} \frac{e^h-1}h = 1$$ and continue.)

So my question, which doesn't seem to have an answer on this site (though I'd be happy to be proved wrong) nor in the textbooks I've consulted: how can one justify the existence of any limit of the form $$\lim_{h\to0} \frac{b^h-1}h$$ $$(b>1)$$, without using the as-yet-underived fact that $$b^x$$ is differentiable? (Edited to add: I also want to avoid infinite series.)

• This question seems like a paradox to me. $b^x$ is differentiable because the mentioned limit exists. That’s the definition of differentiability. Commented Sep 29, 2020 at 0:05
• How do you define an exponential function? Do you use a series? Commented Sep 29, 2020 at 0:10
• @RaiyanChowdhury I think OP is asking how to show that the limit exists. Commented Sep 29, 2020 at 0:12
• @RaiyanChowdhury The exact derivative of $b^x$ for all $x\in\Bbb R$, for every $b>1$, can be established if one can show that a single limit of the form $\lim_{h\to0} \frac{b^h-1}h$ exists at all. The latter task seems weaker in three different ways (just one $x$, just one $b$, just existence and not value). It's the latter task for which I'm seeking a proof. Commented Sep 29, 2020 at 0:16
• @paulinho I want to avoid infinite series as well (I'll add that to the OP). So $a^x$ would be defined for integers in the usual recursive way, for rational numbers by inference (a la Cauchy's functional equation), and for real numbers as the supremum of $a^r$ over all $r<x$. Commented Sep 29, 2020 at 0:17

This is just to address some comments by Greg Martin. I place it here for it is long for the comment section.

• Convexity alone will imply differentiability except on a countable exceptional set.

It is easy to check that convexity of a function $$\phi$$ is equivalent to any of the inequalities \begin{align} \frac{\varphi(u)-\varphi(x)}{u-x}\leq\frac{\varphi(y)-\varphi(x)}{y-x}\leq \frac{\varphi(y)-\varphi(u)}{y-u}\tag{1}\label{convex-equiv} \end{align} For fixed $$a, inequalities $$\eqref{convex-equiv}$$ show that the map $$u\mapsto \tfrac{\varphi(u)-\varphi(x)}{u-x}$$ decreases as $$u\searrow x$$ and increases as $$u\nearrow x$$. Consequently,
the maps \begin{align} \alpha(x):=\sup_{a satisfy \begin{align} \alpha(x)\leq\beta(x)\leq\alpha(y),\quad a

Lemma: The functions $$\alpha$$ and $$\beta$$ are monotone increasing and left continuous and right continuous respectively. Furthermore, $$\alpha(x+)=\beta(x)$$ and $$\alpha(x)=\beta(x-)$$.

Proof: Let $$x\in(a,b)$$ be fixed, and consider the sequence $$x_n=x+\tfrac{1}{n}$$. From $$\eqref{leftrightderivative}$$, it follows that $$\beta(x)\leq\alpha(x+\tfrac1n)\leq \beta(x+\tfrac1n)\leq n(\varphi(x+\tfrac2n)-\varphi(x+\tfrac1n))$$. Letting $$n\nearrow\infty$$, we obtain $$\beta(x)\leq\alpha(x+)\leq\beta(x+)\leq\beta(x)$$. The corresponding statement for left limits follows by using $$x_n=x-\tfrac1n$$ instead.

Since the functions $$\alpha$$ and $$\beta$$ are nondecreasing, we conclude that, except for a countable set of common discontinuities where jumps are equal, $$\alpha=\beta$$ on $$(a,b)$$.

Theorem: If $$\varphi:(a,b)\rightarrow\mathbb{R}$$ convex, then $$\varphi$$ is continuous; moreover, $$\varphi$$ is differentiable everywhere, except on a countable set, and

\begin{aligned} \varphi(y)-\varphi(x)=\int^y_x\beta(t)\,dt=\int^y_x\alpha(t)\,dt \end{aligned} for all $$a.

Proof: Suppose $$a and let $$x=x_0<\ldots. Then $$\beta(x_{m-1})(x_m-x_{m-1})\leq\varphi(x_m)-\varphi(x_{m-1}) \leq \alpha(x_m)(x_m-x_{m-1})$$ Adding all terms gives $$\sum^n_{m=1}\beta(x_{m-1})(x_m-x_{m-1})\leq\varphi(y)-\varphi(x) \leq \sum^n_{m=1}\alpha(x_m)(x_m-x_{m-1}).$$ Consequently, $$\varphi(y)-\varphi(x)=\int^y_x\beta(t)\,dt=\int^y_x\alpha(s)\,ds$$; hence, $$\varphi$$ is continuous on any closed interval, and differentiable everywhere except in the countable set $$N$$ of discontinuities of $$\beta$$.

Comment 1: There is no need to appeal to integral calculus to show continuity of $$\phi$$. I am sure the OP knows many ways to achieve this.

Comment 2: Using the fact that the left and right derivatives $$\alpha$$ and $$\beta$$ are monotone along with the left-right continuity relations between them, one can conclude that $$\phi$$ is differentiable at every point with the exceptions of a countable set where $$\alpha$$ and $$\beta$$ have jump discontinuities. All this, I believe, makes the arguments suitable for a course of differential calculus prior the introduction of Riemann integration.

• For the exponential function, if convexity can be proven, then differentiability at every point will follow immediately:

Suppose $$\phi(x)=a^x$$ is differentiable at $$x_0$$ (such $$x_0$$ exists from the discussion above. From the existence of $$\lim_{h\rightarrow0}\frac{\phi(x_0+h)-\phi(x_0)}{h}=\lim_{h\rightarrow0}\phi(x_0)\frac{\phi(h)-1)}{h}$$, it follows the existence of $$\lim_{h\rightarrow0}\frac{\phi(h)-1}{h}$$. From this, the differentiable it’s at any point follows.

Alternative method:

I undust a couple of my old soviet textbooks (Kudriavtsev, L. D., Curso de Análisis Matemático, Vol 1, and Nilkosky, S. M., A Course of Mathematical Analysis, Vol. I) and this is more or less how the derivative of exponential functions are presented without the defining the log function as an integral:

1. Assuming that the exponential function $$\phi_a(x)=a^x$$ has been introduced and continuity and strict monotonic properties are established (starting from exponential at rational numbers, exteding to irrational, etc)
2. The existence of $$\lim_{h\rightarrow0}\big(1+h\big)^{1/h}=e$$ and $$2 is established (starting from $$\lim_{n\rightarrow\infty}\Big(1+\tfrac1n\Big)^n$$ and then to $$\lim_{h\rightarrow0}(1+h)^{1/h}$$ using standard tricks)

then, for $$a>1$$

1. the $$\log_a:(0,\infty)\rightarrow\mathbb{R}$$ function, being the inverse of a strictly monotone increasing and continuous function $$\phi_a$$, is itself continuous and strictly monotone increasing.

2. $$\lim_{x\rightarrow0}\frac{\log_a(x+1)}{x}=\lim_{x\rightarrow0}\log_a\Big(\big(1+x\big)^{1/x}\Big)=\log_ae$$.

3. The punch line: To compute $$\lim_{h\rightarrow0}\frac{e^h-1}{h}$$, let $$t=e^h-1$$ so that $$h=\ln(t+1)$$, $$t>-1$$. Then $$h\rightarrow0$$ is equivalent to $$t\rightarrow0$$. From this, $$\lim_{h\rightarrow0}\frac{e^h-1}{h}=\lim_{t\rightarrow0}\frac{t}{\ln(1+t)}=1$$

As already mentioned,we define the exponential function on $$\Bbb R_{\ge 0}$$ as $$f(x)=\lim\limits_{n\to\infty}\left(1+\frac{x}n\right)^n\quad (*)$$.

For $$x>0$$ and $$n\in\Bbb N,$$ the following holds:

$$1\le\underbrace{\frac{\left(1+\frac{x}n\right)^n-1}x=\frac1n\sum_{k=0}^{n-1}\left(1+\frac{x}n\right)^{n-1-k}}_{\text{difference of powers}}\le\frac1n\cdot n\left(1+\frac{x}n\right)^{n-1}=\left(1+\frac{x}n\right)^{n-1}\le\left(1+\frac{x}n\right)^n\tag 1$$

By letting $$n\to+\infty$$ in $$(1)$$, we obtain: $$1\le\frac{f(x)-1}x\le f(x)\tag 2$$

For $$x<0$$,replacing $$x$$ by $$-x$$ in $$(2)$$ and dividing by $$f(-x)>0,$$ we obtain:

$$\frac1{f(-x)}\le\frac{\frac1{f(-x)}-1}x\le 1\tag 3$$

Now, from $$(2)$$ and $$(3)$$ and the definition $$(*)$$ of the exponential function,$$\forall x\ne 0$$, we have

$$0<\min\{1,e^x\}\le\frac{e^x-1}x\le\max\{1,e^x\}\tag 4$$

If $$|x|<1,$$ then $$e^{|x|} and $$\max\{1,e^x\} so it follows that $$0<|e^x-1|\le e|x|$$ and squeezing we get $$\lim\limits_{x\to 0}e^x=1$$. Now, let's write:

$$\min\{1,e^x\}=\frac{1+e^x-|1-e^x|}2\\\max\{1,e^x\}=\frac{1+e^x+|1-e^x|}2\tag 5$$ We see from $$(5)$$ that $$\lim\limits_{x\to 0}\min\{1,e^x\}=1$$ and $$\lim\limits_{x\to 0}\max\{1,e^x\}=1$$ and we apply the squeeze theorem to $$(4)$$.

And $$b^h$$ can be written as $$e^{h\ln(b)},$$ so we end up with the limit: $$\lim_{h\to 0}\frac{b^h-1}h=\lim_{h\to 0}\frac{e^{h\ln(b)}-1}{h\ln(b)}\ln(b)$$

If we allow ourselves to use the limit $$\lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n = e,$$ we can subtract the constant inside the limit and use the difference of powers formula to show that $$e^{1/n} - \left(1 + \frac{1}{n} \right) = o \left(\frac{1}{n}\right)$$:

\begin{align*} 0 &= \lim_{n \to \infty} \left[e - \left(1 + \frac{1}{n} \right)^n \right] \\ = \lim_{n \to \infty} \left( e^{1/n} - \left(1 + \frac{1}{n} \right) \right) & \left[ \left(1 + \frac{1}{n} \right)^{n-1}+e^{1/n}\left(1 + \frac{1}{n} \right)^{n-2} + ... + e^{(n-1)/n} \right], \ \end{align*}

and clearly $$\left( \left(1 + \frac{1}{n} \right)^{n-1}+e^{1/n}\left(1 + \frac{1}{n} \right)^{n-2} + ... + e^{(n-1)/n} \right) \geq n,$$ which implies $$e^{1/n} - \left(1 + \frac{1}{n} \right) = o \left(\frac{1}{n}\right)$$, or rearranging, $$e^{1/n} - 1 = \frac{1}{n} + o \left(\frac{1}{n} \right),$$ giving $$\lim_{n \to \infty} n(e^{1/n} - 1) = 1,$$ and changing variable to $$h = 1/n$$ we get $$\lim_{h \to 0^+} \frac{e^h - 1}{h} = 1.$$ We can use this right-hand limit to prove that $$\lim_{h \to 0^+} \frac{b^h - 1}{h} = \ln(b),$$ and in particular, taking $$b = 1/e$$ shows $$\lim_{h \to 0^+} \frac{e^{-h} - 1}{h} = -1,$$ which by taking $$k = -h$$ gives us the left-hand limit $$\lim_{k \to 0^-} \frac{e^{k} - 1}{-k} = -1,$$ or $$\lim_{k \to 0^-} \frac{e^{k} - 1}{k} = 1,$$ and we are done.

• I guess technically we need something more in deducing the full limit $h\to0+$ from the subsequence limit $\frac1n\to0$. But your main point, I think, is that some sort of squeeze theorem argument should work here. Commented Sep 29, 2020 at 18:29
• @greg You could do it with $x$ instead of $n$ and the argument works out basically the same way, although the difference of powers piece is a little messier when $n$ isn’t an integer. Commented Sep 29, 2020 at 18:54

You could prove it using Riemann sums: Let's attempt to evaluate $$\int_0^1 b^x dx$$ as a Riemann sum. The curve is continuous and bounded, so there is a well-defined area under it which we can find using Riemann sums.

Split the interval up into subintervals of size $$h$$, and let $$N=\lfloor \frac1h\rfloor$$, i.e. we divide $$[0,1]$$ into $$[0,h), [2h,3h), ... [N h, 1]$$. The left Riemann sum is given by $$\sum_{n=0}^{N-1} b^{nh} h + b^{Nh} (1-Nh)$$ so we have that $$\lim_{h\rightarrow 0} \sum_{n=0}^{N-1} b^{nh} h + b^{Nh} (1-Nh)$$ is convergent. The sum is a geometric series, which we can simplify as $$\sum_{n=0}^{N-1} b^{nh} h = h\sum_{n=0}^{N-1}(b^h)^n = h\left(\frac{b^{hN} - 1}{b^{h}-1}\right)$$ Since $$\lim_{h\rightarrow 0} hN = \lim_{h\rightarrow 0}h\lfloor \frac1h\rfloor$$ converges to $$1$$, we can conclude that $$\lim_{h\rightarrow 0} \sum_{n=0}^{N-1} b^{nh} h + b^{Nh} (1-Nh) = \lim_{h\rightarrow 0} h\left(\frac{b^{hN} - 1}{b^{h}-1}\right) + 0 = (b-1)\lim_{h\rightarrow0} \frac{h}{b^h - 1}$$ converges. Thus either $$\lim_{h\rightarrow0} \frac{b^h-1}h$$ converges, or it diverges to infinity, but diverging to infinity would mean $$\int_0^1 b^x dx = 0$$, which is false because $$b^x > 0$$ for all $$x\in [0,1]$$, so the integral must be strictly positive.

• Fair enough, but if we wanted to allow integration techniques, then we might as well take the smoothest route and define $\ln x = \int_1^x \frac 1x\,dx$ right out of the gate; all the other properties of exponentials and logarithms follow from that. Commented Sep 29, 2020 at 18:27
• It's true. But for what I have written, you don't need any integral properties other than the fact that they exist, so it is actually more elementary than the log definition. Commented Sep 29, 2020 at 18:58
• Well, buried in the logic somewhere is the very broad fact that Riemann sums converge to areas in the first place.... Commented Sep 29, 2020 at 20:16
• Good point. I'll see if I can get a non-integral argument here; I think the geometric series is a promising approach though. Commented Sep 30, 2020 at 16:20