I was told to assume that
$$\ln b=\lim_{h\to 0} \frac{\left(b^h-1\right)}{h}$$
where b is a positive, real, base.
Unfortunately, being told to assume something isn't good enough.
When using L'Hopital's with a base of $e$, it can be shown that the limit approaches $e^0$, which of course equals 1, or, $\ln e$. However, I was hung up on proving that for any base, the limit will approach the natural log of the base, without using the direct proof that
$$\frac{d}{dx}b^x=b^x(\ln b)$$
which is what's trying to be proved in the first place.
Is L'Hopital's even the right route to go?
Thanks in advance.
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$\begingroup$Note that we have $$\frac{b^h-1}{h}=\frac{e^{h\log(b)}-1}{h} \tag 1$$
Now, in THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequality
$$e^x\ge 1+x \tag 2$$
From $(2)$ (along with the property $e^xe^{-x}=1$) it is easy to see that for $x<1$
$$e^x\le \frac{1}{1-x} \tag 3$$
Using $(2)$ and $(3)$, we can bound $(1)$ as
$$\log(b) \le \frac{e^{h\log(b)}-1}{h}\le \frac{\log(b)}{1-h\log(b)}$$
whereupon application of the squeeze theorem yields the coveted limit
$$\lim_{h\to 0}\frac{b^h-1}{h}=\log(b)$$
And we are done!
$\endgroup$ 11 $\begingroup$I like your sentence "Unfortunately, being told to assume something isn't good enough". But these days rarely do students have this attitude. Even worse are the books which ask students to assume anything which requires even slightest effort to prove.
In this answer I have dealt with the limit of $(a^{h} - 1)/h$ as $h \to 0$ and I wish to add some further remarks here with notation specific to your post. From your post it appears that your goal is to find the derivative of the function $f(x) = b^{x}$. The tough part is to define $b^{x}$ for any real number $x$. When $x$ is rational the symbol $b^{x}$ can be defined using algebra, but when $x$ is irrational then things are bit complicated and there are multiple approaches to define $b^{x}$ (all of the approaches are hard for a beginner and discussed in my blog posts).
Once you have a definition of $f(x) = b^{x}$ for $b > 0$ it is easy to see that $$f'(x) = b^{x}\lim_{h \to 0}\frac{b^{h} - 1}{h} = b^{x}g(b)$$ where $g(b)$ denotes the limit of $(b^{h} - 1)/h$ as $h \to 0$. I have proved in my linked answer that the limit $g(b)$ exists for all $b > 0$. It can be further proved easily that function $g$ satisfies the following relations $$g(ab) = g(a) + g(b),\, g(a/b) = g(a) - g(b),\, g(1) = 0$$ Further we have inequalities $$\frac{x - 1}{x} \leq g(x) \leq x - 1$$ for $x \geq 1$ and from this inequality it is possible to show that $g'(x) = 1/x$ for all $x > 0$. This function $g(x)$ is traditionally denoted by symbol $\log x$ (or $\ln x$ which I don't prefer) and hence we have $$f'(x) = (b^{x})' = b^{x}g(b) = b^{x}\log b$$
On request of OP (via comment) I am giving the derivation of $g'(x) = 1/x$ here (this is available with more details in my blog post linked earlier). Let $x > 1$ and then by dividing the inequality related to $g(x)$ by $(x - 1)$ we get $$\frac{1}{x} \leq \frac{g(x)}{x - 1} \leq 1$$ and using Squeeze theorem when $x \to 1^{+}$ we get $$\lim_{x \to 1^{+}}\frac{g(x)}{x - 1} = 1$$ The same result holds when $x \to 1^{-}$. If $x \to 1^{-}$ we put $x = 1/y$ so that $y \to 1^{+}$ as $x \to 1^{-}$ and then $$\lim_{x \to 1^{-}}\frac{g(x)}{x - 1} = \lim_{y \to 1^{+}}y\frac{g(1/y)}{1 - y} = \lim_{y \to 1^{+}}\frac{g(1/y)}{1 - y} = \lim_{y \to 1^{+}}\frac{g(y)}{y - 1} = 1$$ because from $g(a/b) = g(a) - g(b), g(1) = 0$ we easily get $g(1/y) = -g(y)$.
Thus we have proved that $g(x)/(x - 1) \to 1$ as $x \to 1$. This means that $g(1 + x)/x \to 1$ as $x \to 0$. We have \begin{align} g'(x) &= \lim_{h \to 0}\frac{g(x + h) - g(x)}{h}\notag\\ &= \lim_{h \to 0}\frac{g((x + h)/x)}{h}\notag\\ &= \lim_{h \to 0}\frac{g(1 + (h/x))}{(h/x)}\cdot\frac{1}{x}\notag\\ &= \frac{1}{x}\lim_{t \to 0}\frac{g(1 + t)}{t}\text{ (putting }t = h/x)\notag\\ &= \frac{1}{x}\notag \end{align}
$\endgroup$ 5 $\begingroup$First the easy part: \begin{align} \frac d {dx} b^x & = \lim_{h\to 0} \frac{b^{x+h} - b^x} h \\[10pt] & = \lim_{h\to0} \left( b^x\ \frac{b^h - 1} h \right) & & (\text{just algebra}) \\[10pt] & = b^x \lim_{h\to0} \frac{b^h - 1} h & & (\text{because $b^x$ does not change as $h$ changes}) \\[10pt] & = (b^x\cdot\text{constant}). & & (\text{The limit is a “constant'' because it} \\ & & & \phantom{({}} \text{ does not change as $x$ changes.}) \end{align}
Next we need this fact: $$ \text{If } b=e\approx 2.71828\ldots \text{then the “constant'' is $1$; otherwise it is some other number.} $$ How do we know that? More on that below.
Hence we have $\dfrac d{dx} e^x = e^x\cdot 1$.
What then is the “constant” when $b\ne e$? Here one can use the chain rule: \begin{align} \frac d {dx} b^x = \frac d {dx} e^{x\log_e b} = e^{x\log_e b} \cdot \frac d{dx} (x\log_e b) = b^x \frac d{dx} (x\log_e b) = \cdots. \end{align} And $$ \frac d{dx} (x\log_e b) = \log_e b $$ for the same reason that $\dfrac d {dx} (x\cdot5) = 5$, i.e. $\log_e b$ is a constant, meaning it does not change as $x$ changes.
The proof of the chain rule involves a slight subtlety that does not come up in most of the basic proofs involving derivatives. I've posted on that here before.
One can show that this "natural" base, $e$, must be more than $2$ by observing that $\dfrac{2^1 - 2^0}{1-0} = 1$, so $2^x$ must be changing more slowly than $1$ when $x=0$, and that the natural base must be less than $4$ because $\dfrac{4^0 - 4^{-1/2}}{0 - (-1/2)} =1$, so $4^x$ must be changing at a rate faster than $1$ when $x=0$. So we've narrowed it down to somewhere between $2$ and $4$. Narrowing it down to $2.71828\ldots$ by this method is inefficient; but other more efficient methods exist.
$\endgroup$ $\begingroup$How have you defined $e$?
I first learned it as:
$$e = \lim_\limits{n\to \infty} (1+\frac{1}{n})^n$$
$$\frac{d}{dx} e^x = \lim_\limits{h\to 0} \dfrac{e^{x+h} - e^x}{h}\\ (e^x) \lim_\limits{h\to 0} \dfrac{e^{h} - 1}{h}$$
binomial expansion from the definition of $e$
$$e = 1 + \frac{n}{n}+\frac{n(n-1)}{2n^2}+\frac{n(n-1)(n-2)}{3!n^3}...\\e^h = 1 + h + \frac{h^2}{2!}+\frac{h^3}{3!}+\cdots$$
$$(e^x) \lim_\limits{h\to 0} \dfrac{e^{h} - 1}{h} = (e^x)\lim_\limits{h\to 0}\dfrac{h + \frac{h^2}{2!}+\frac{h^3}{3!}+\cdots}{h}$$
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