TITLE: Interchanging two limits in a proof. Is it legitimate?
QUESTION [8 upvotes]: 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:

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

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?

REPLY [3 votes]: 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.