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.

  • $\begingroup$ It is easy to show that the derivative of $e^x$ is itself using nothing more than the fact that $e^x$ is the nth root of 1 + x/n as n goes to infinity, together with the limit definition of derivative. Since $b^x = exp(xln(b))$ the result follows. $\endgroup$ – Vik78 Apr 9 '16 at 0:23
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    $\begingroup$ The problem is that in finding derivatives in using L'Hopital's rule, you'd already have to know the result you're trying to prove here. $\qquad$ $\endgroup$ – Michael Hardy Apr 9 '16 at 1:10

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!

  • $\begingroup$ perfect! I'll have to secure some of this stuff because it's a bit new to me, but I'm glad I have a concrete reason why this is true. Thanks a lot! $\endgroup$ – cameron Apr 9 '16 at 0:36
  • $\begingroup$ You're welcome. My pleasure. -Mark $\endgroup$ – Mark Viola Apr 9 '16 at 1:09
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    $\begingroup$ Note that if $0 < c < \dfrac12$, then $\dfrac1{1-c} < 1+2c$. Therefore, if $h \log(b) < \dfrac12$, then $\frac{\log(b)}{1-h\log(b)} < \log(b)(1+2h\log(b))=\log(b)+2h\log^2(b)$ which makes squeezing easy. $\endgroup$ – marty cohen Apr 9 '16 at 1:26
  • $\begingroup$ @martycohen Nice addendum! +1 $\endgroup$ – Mark Viola Apr 9 '16 at 1:42
  • $\begingroup$ Hey again, sorry to bother you... I've proved Bernoulli's Inequality on paper and algebraically gone from your line (2) to (3). I'm just having a little trouble setting up the bounded limit from your (2) and (3). Do you think you could explain in short how you arrived it? Again, sorry I'm just learning this stuff and don't have a lot of experience under my belt quite yet. $\endgroup$ – cameron Apr 9 '16 at 3:59

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}

  • $\begingroup$ Hey, thanks for the other approach to this question. I really like my current teacher-- he asked us to prove everything he taught us this year except this one expression and wanted us to try it ourselves if we wanted to, which is exactly what I did. This approach to teaching is exactly what needs to be done. Other classes had poorer understanding because they were just told to assume such and such was true. $\endgroup$ – cameron Apr 9 '16 at 23:40
  • $\begingroup$ ''and from this inequality it is possible to show that $g′(x)=1/x$ for all $x>0$.'' Just this line I cant figure out how you did it. Do you mind going over this a bit? $\endgroup$ – cameron Apr 9 '16 at 23:51
  • $\begingroup$ @Cameron: see updated answer. $\endgroup$ – Paramanand Singh Apr 10 '16 at 3:51
  • $\begingroup$ Man, the feeling when something finally makes sense. Thanks very much mister $\endgroup$ – cameron Apr 10 '16 at 4:19
  • $\begingroup$ @Cameron: Glad to know that I was of some help. I think every textbook on calculus must develop a theory of logarithmic, exponential (and circular functions) with full rigor (the material may be kept out of exams, but definitely should be included in books for the interested students). If this is done properly like in Hardy's "A Course of Pure Mathematics" then everything in calculus will "finally make sense". $\endgroup$ – Paramanand Singh Apr 10 '16 at 6:11

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.


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 $e$

$$e = 1 + \frac{n}{n}+\frac{n(n-1)}{2n^2}+\frac{n(n-1)(n-2)}{3!n^3}...\\e^h = 1 + h + \frac{h^2}{2!}+\frac{h^3}{3!}+\cdots$$

$$(e^x) \lim_\limits{h\to 0} \dfrac{e^{h} - 1}{h} = (e^x)\lim_\limits{h\to 0}\dfrac{h + \frac{h^2}{2!}+\frac{h^3}{3!}+\cdots}{h}$$

  • $\begingroup$ You're exchanging limits, which is a pretty nontrivial thing. $\endgroup$ – YoTengoUnLCD Apr 9 '16 at 3:23

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