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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?

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3  
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 Tamaroff Apr 25 '12 at 0:03
    
@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 0:07
2  
@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 Tamaroff Apr 25 '12 at 0:24

1 Answer 1

up vote 3 down vote accepted

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.

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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 0:58
1  
Reduction to previously proven cases is always useful. :-) –  robjohn Apr 27 '12 at 2:33

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