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Landau defines

$$\log x = \lim\limits_{k \to 0} {x^k-1 \over k}$$

I wanted to prove the elemental properties of the logaritm with this, namely:

  1. $\log xy = \log x +\log y $
  2. $\log x^a = a\log x $
  3. $1-\dfrac 1 x\leq\log x \leq x-1 $
  4. $\lim\limits_{x\to 0}\dfrac{\log(1+x)}{x}=1 $
  5. $\dfrac{d}{dx}\log x = \dfrac 1 x$

I proved them all, however, in the last case I did this

$$\eqalign{ & \frac{d}{dx}\log x = \lim \limits_{h \to 0} \frac{\log \left( x + h \right) - \log x}{h} \cr & = \lim\limits_{h \to 0} \lim \limits_{k \to 0} \frac{\left( x + h \right)^k - x^k}{kh} \cr & = \lim \limits_{k \to 0} \frac{1}{k}\lim \limits_{h \to 0} \frac{\left( x + h \right)^k - x^k}{h} \cr & = \lim \limits_{k \to 0} \frac{1}{k}\lim \limits_{h \to 0} kx^{k - 1} = \lim \limits_{k \to 0} \lim \limits_{h \to 0} x^{k - 1} = x^{ - 1} } $$

Since I'm not familiar with multivariable calculus, I don't know how to justify this. What could work here?

Michael Hardy
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Pedro
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    I'd reduce 5. to 4. and 1. by rewriting the fraction as $$\frac{\log{(x+h)}-\log{x}}{h} = \frac{1}{x} \cdot \frac{\log{(1+h/x)}}{h/x},$$ then pulling the $\frac{1}{x}$ out of the limit and substituting $k=h/x$, noting that $h \to0$ if and only if $k \to 0$. – t.b. Apr 24 '12 at 23:48
  • @Peter, if i'm understanding well your question, is about to justify the change of the order of the limits ? – Abdelmajid Khadari Apr 24 '12 at 23:51
  • @AbdelmajidKhadari Yes. However, I think t.b.'s aproach is best. – Pedro Apr 25 '12 at 00:03
  • @PeterTamaroff : Can you tell me where you learned your TeX coding technique? I keep seeing strange things in TeX code in this forum that I see nowhere else. Things like {\frac{{({x+h})^{k}}}{{h}}} where \frac{(x+h)^k}{h} would serve. Is there some manual that tells you to write things that way in this forum? – Michael Hardy Apr 25 '12 at 00:07
  • @Michael I use MathType when I have to write a lot of $\TeX$ codes, to make things faster. The program automatically places ${$ and $}$ in trivial ways which might not always be necesary. Basically, I write my equations in MT and it codes its templates into TeX or other similar codes, or changes into an OLE graphic compatible with Office. It's a great program. – Pedro Apr 25 '12 at 00:08
  • @Peter : A remark that might not make you happy ; when you expand the $\log$s as limits, you put the same index for both $\log$s (i.e. $k \to 0$), but it would be more accurate to say that one $\log$ is $\lim_{k \to 0}$ and the other $\log$ is $\lim_{m \to 0}$. Perhpas that doesn't help but at least makes you rigorous. – Patrick Da Silva Apr 25 '12 at 00:10
  • @PatrickDaSilva Why would that be more accurate? – Pedro Apr 25 '12 at 00:11
  • I agree with t.b. In fact, proving 5 involves essentially the same manipulations as 4. So, I don't understand how you could have done 4 with no problem exchanging the limit and yet had trouble with doing the same thing in 5. – George Lowther Apr 25 '12 at 00:20
  • Well because your two limits that define the two logs are not related one another, so there's no reason to index the limits with the same letter, except if you don't make them interact along your proof. But anyway, I think I agree with t.b.'s argument and that this is going nowhere. – Patrick Da Silva Apr 25 '12 at 00:21
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    @George For $4.$ I simply used the squeeze theorem with $$\frac{1}{{1 + x}} \leqslant \frac{{\log \left( {1 + x} \right)}}{x} \leqslant 1$$ which stems from $3.$ I really like tb's approach because it follows my systematic regression from the case in $n$ to the case in $n-1$. – Pedro Apr 25 '12 at 00:24
  • @PatrickDaSilva Indeed, I'm sticking to that MO. I don't think it is necessary to overcomplicate things. – Pedro Apr 25 '12 at 00:26

1 Answers1

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On Peter's request I'm posting my comment as an answer:

Note that 1. with $y=1+ \frac{x}{h}$ gives $$ \log{(x+h)}-\log{x} = \log{(1+h/x)}, $$ so, $$\frac{d}{dx}\log{x} = \lim_{h\to0}\frac{\log{(x+h)}-\log{x}}{h} = \frac{1}{x} \cdot \lim_{h\to0} \frac{\log{(1+h/x)}}{h/x} = \frac{1}{x} \cdot \lim_{k\to0} \frac{\log{(1+k)}}{k} \stackrel{4.}{=} \frac{1}{x}, $$ as desired.

t.b.
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  • If you want, I might add something on interchanging limits tomorrow but it's getting too late here to write something coherent about that right now. Meanwhile, this text looks quite decent. – t.b. Apr 25 '12 at 00:58
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    Reduction to previously proven cases is always useful. :-) – robjohn Apr 27 '12 at 02:33