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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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    @PatrickDaSilva Indeed, I'm sticking to that MO. I don't think it is necessary to overcomplicate things.2012-04-25

1 Answers 1

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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.

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    Reduction to previously proven cases is always useful. :-)2012-04-27