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In many places I have read that $$\lim_{h\to 0}\frac{b^h - 1}{h}$$ is by definition $\ln(b)$. Does that mean that this is unsolvable without using that fact or a related/derived one?

I can of course solve it with L'Hospitals rule, but that uses the derivative of the exponential function which is exactly what I want to derive by solving this limit.

Since the derivative of the logarithm can be derived from the derivative of the exponential, using the fact that they are inverses, means that deferring this limit to something that can be solved using the derivative of log seams also cheating.

Others have asked that before, but all the "non-L'Hospital"-solutions seem to defer it to some other limit that they claim obvious. For example two solutions from Proving that $\lim_{h\to 0 } \frac{b^{h}-1}{h} = \ln{b}$ use $$\lim_{x\to 0}\frac{\log_a(1+x)}{x}=\log_a e$$ and $$\lim_{x\to 0}\frac{e^x-1}x=1$$ none of which is more obvious (to me) as the original.

On that same page, the 1st of the above 2 is "derived" using a Taylor expansion (if I am not mistaken) which (if I remember correctly) is based on derivatives (in this case of the logarithm), which is related to the derivative of exp as I meantioned above. So this seems to be circular reasoning too. (a very large circle though)

So is this limit not solvable at all without using smth that is based on something this meant to prove? Can this only be defined to equal $\ln(b)$; and numerically determined to some precision?

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    $\begingroup$ Given a definition of $x^y$, one can prove the result. When one does analysis rigorously, a precise definition is given. There are several approaches. $\endgroup$ – André Nicolas Apr 27 '13 at 6:30
  • $\begingroup$ Have you tried using the limit-wise definition of $e$ to see the first limit is being held? $\endgroup$ – mrs Apr 27 '13 at 6:33
  • $\begingroup$ @AndréNicolas: Maybe here's a different way to phrase the question: Define $f$ as such: $f(1)=2$ and $f(x+y)=f(x)\cdot(y)$ for all $x,y\in\mathbb{Q}$. This yields a unique function $f:\mathbb{Q}\rightarrow\mathbb{Q}$, and this function is continuous. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, there is at most one continuous extension $\hat f:\mathbb{R}\rightarrow\mathbb{R}$. Let's assume that such a function $\hat f$ exists. Without explicitly constructing $\hat f$, can we show that $\hat f$ is differentiable at $0$? $\endgroup$ – Managu Apr 28 '13 at 3:46
  • $\begingroup$ @Managu: That is one of the possible definitions, for the not particularly special case $x=2$. It is even a natural definition, unpopular mainly because getting to the main results takes longer, and we tend to put excessive value on slick. Mostly it is done the other way, make a clever definition and then later verify that it agrees with the earlier intuition about $x^{p/q}$. Your suggestion is I think pedagogically preferable. $\endgroup$ – André Nicolas Apr 28 '13 at 4:52
  • $\begingroup$ math.stackexchange.com/questions/1828962/… $\endgroup$ – user301988 Jul 24 '16 at 0:53
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This depends on how you introduce the functions. If you start by defining

$$\log x=\int_1^x \frac{1}{t}\,dt$$

(for $x>0$) then you know from the fundamental theorem that the derivative is $\log' x=1/x$. Thus the function is increasing; it's easy to show the main property $\log(ab)=\log a+\log b$:

$$\int_1^{ab}\frac{1}{t}\,dt= \int_1^a\frac{1}{t}\,dt + \int_a^{ab}\frac{1}{t}\,dt $$

and the substitution $t=au$ in the second integral will give the result. Since $\log2>0$, we get $\log2^n=n\log2$ and this implies that $\lim_{x\to\infty}\log x=\infty$. Since $\log(1/x)=-\log x$ because $\log1=0$, we also get $\lim_{x\to0}\log x=-\infty$. Thus $\log$ takes on all real values and its inverse function $x\mapsto\exp x$ is defined and differentiable on the whole real line, with $\exp' x=\exp x$.

This allows to define $b^x = \exp(b\log x)$ (for $b>0$) and it's easy to show that, for all $x$ and $y$, $b^{(x+y)}=b^xb^y$, which means in particular that

$$b^n=\underbrace{b\cdot b\cdot\dots\cdot b}_{\text{$n$ times}}$$

(easy induction) so the notation $b^x$ is not ambiguous. It's also clear that the derivative of $g(x)=b^x$ is $g'(x)=b^x\log b$, so your limit

$$\lim_{h\to0}\frac{b^h-1}{h}=g'(0)=b^0\log b=\log b$$

without any circular reasoning.

The Napier-Euler number $e$ can be defined by $\log e=1$ (there is a unique solution of this equation because $\log$ is increasing) and by definition we get $\exp x=e^x$. Also the fundamental limit

$$\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x=e$$

can be computed in the same vein; just take logarithms:

$$\lim_{x\to\infty}x\log\left(1+\frac{1}{x}\right)= \lim_{t\to0}\frac{\log(1+t)}{t}= \lim_{t\to0}\frac{\log(1+t)-\log1}{t}= \log'1=1$$

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    $\begingroup$ I personally think that egreg's approach is the most elegant one and in most college calculus books (Like Stewarts) that approach is included. That's how I learned it and that's how I teach my students. $\endgroup$ – imranfat Apr 28 '13 at 1:27
  • $\begingroup$ The definition of $\log x$ as an integral is sort of very easy to work with (i.e. properties of $\log x$ can be established easily), but is difficult in principle as we need to show the existence of integral which itself requires deep result that continuous functions are integrable. A simpler definition is $\log x = \lim\limits_{n \to \infty}n(x^{1/n} - 1)$ because we can establish the existence of this limit easily and with this definition it is slightly harder to establish properties of $\log x$. $\endgroup$ – Paramanand Singh Jul 15 '13 at 15:12
  • $\begingroup$ @ParamanandSingh One can present the logarithm as an area, appealing to geometric intuition and it's not really difficult to show the main property by using an affine transformation that scales the $x$-axis by $1/a$ and the $y$-axis by $a$, so that areas of rectangles don't change and the hyperbola $y=1/x$ is fixed. Later on this can be made rigorous with integrals. $\endgroup$ – egreg Jul 15 '13 at 15:17
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There are two standard approaches to the logarithm and exponential.

  1. Define the exponential, perhaps via Taylor series or via $\frac{d}{dx} e^x =e^x$. Then, using this, derive properties of its inverse $\ln x$.

  2. Define the logarithm, perhaps via Taylor series or via $\frac{d}{dx} \ln x = \frac 1x$ or via the limit in the OP. Then, using this, derive properties of its inverse $e^x$.

Similarly, there are multiple approaches to generalizing the logarithm and exponential to different bases, e.g. $b$.

You seem to object to each approach because of the existence of the other approach. The ability to prove both $A\rightarrow B$ and $B\rightarrow A$ does not make either proof circular. As @Andre wrote, you need to choose a definition to get started, and then things won't be circular.

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  • $\begingroup$ Perhaps not circular, but is it possible to derive the value of $\frac{\mathrm{d}}{\mathrm{d}x}b^x|_{x=0}$ based only upon the usual "precalculus" definition of exponentials and logarithms? $\endgroup$ – Managu Apr 28 '13 at 3:49
  • $\begingroup$ @Managu: In precalculus we normally get properties, not definitions. I don't know how to define $b^x$ without using limits, derivatives, series, or integrals. $\endgroup$ – vadim123 Apr 28 '13 at 3:51
  • $\begingroup$ Ok, then, so can the derivative be calculated from the usual properties of exponentials? The properties $b^1=b$; $b^{x+y}=b^x\cdot b^y$; and $f(x)=b^x$ is continuous uniquely determine $b^x$ for all $x\in \mathbb{R}$. $\endgroup$ – Managu Apr 28 '13 at 3:58
  • $\begingroup$ To talk about continuity and derivatives you need limits, which do not exist in precalculus. Once you add those topics, you're back to the situation I describe. $\endgroup$ – vadim123 Apr 28 '13 at 4:00
  • $\begingroup$ Put another way, one can (for example) rigorously derive the value of $\frac{\mathrm{d}}{\mathrm{d}x} \sqrt{x}|_{x=1}$ without giving a formal definition of the square root function; algebra and limit laws suffice. Can something similar be done with the exponential function? $\endgroup$ – Managu Apr 28 '13 at 4:07
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If we accept

$$\lim_{x\to 0}(1+x)^{1/x}=\text{e}$$

then

$$\lim_{x\to 0}\frac{\log_a(1+x)}{x}=\lim_{x\to 0}\frac{1}{x}\frac{\log_a(1+x)}{x}=\lim_{x\to 0}[\log_a(1+x)]^{1/x}=\lim_{x\to 0}\log_a(1+x)^{1/x}=\log_a\mathrm{e}$$

by applying some known theorems.

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    $\begingroup$ Hope to see you before I head to bed (not anytime very soon, as of this posting of my comment! ;-) Miss you! $\endgroup$ – Namaste May 1 '13 at 0:46

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