Tricks for estimating $ \lim_{x\rightarrow 0} \frac{d}{dx} \bigl(-\frac{1}{x} \ln\bigl(1 + \frac{(e^{-xu}-1) (e^{-xv}-1)}{e^{-x}-1} \bigr) \bigr)$ I'm trying to find a Taylor approximation of $ f(x) =-\frac{1}{x} \ln\left(1 + \frac{(e^{-xu}-1) (e^{-xv}-1)}{e^{-x}-1} \right) $ at $x = 0$. 
For the derivation part Wolfram returns a quite a tedious expression to the problem, taking the limit of which does result in the answer I'm looking for. 
As I found this problem in a textbook which is not focused on honing derivation/approximation skills, I'm curious, whether there are some known tricks that would help easying up the calculation, or is this just a technical exercise?
Apologies if this is a dumb question.
 A: $\begin{array}\\
f(x) 
&=-\frac{1}{x} \ln\left(1 + \dfrac{(e^{-ax}-1) (e^{-bx}-1)}{e^{-x}-1} \right)\\
&=-\frac{1}{x} \ln\left(1 + \dfrac{(-ax+O(x^2)) (-bx+O(x^2))}{-x+O(x^2)} \right)\\
&=-\frac{1}{x} \ln\left(1 + \dfrac{abx^2+O(x^3)}{-x+O(x^2)} \right)\\
&=-\frac{1}{x} \ln\left(1 - abx+O(x^2) \right)\\
&=\frac{1}{x}(abx+O(x^2))\\
&=ab+O(x)\\
\end{array}
$
Take more terms
(e.g.,
$e^{-x} = 1-x+x^2/2+O(x^3)
$)
to get more precision.
Here is a try to get
one more term.
$\begin{array}\\
f(x) 
&=-\frac{1}{x} \ln\left(1 + \dfrac{(e^{-ax}-1) (e^{-bx}-1)}{e^{-x}-1} \right)\\
&=-\frac{1}{x} \ln\left(1 + \dfrac{(-ax+a^2x^2+O(x^3)) (-bx+b^2x^2/2+O(x^3))}{-x+x^2/2+O(x^3)} \right)\\
&=-\frac{1}{x} \ln\left(1 + x^2\dfrac{(a-a^2x+O(x^2)) (b-b^2x/2+O(x^2))}{-x+x^2/2+O(x^3)} \right)\\
&=-\frac{1}{x} \ln\left(1 + x\dfrac{(a-a^2x+O(x^2)) (b-b^2x/2+O(x^2))}{-1+x/2+O(x^2)} \right)\\
&=-\frac{1}{x} \ln\left(1 - x\dfrac{ab-(a^2b+ab^2)x+O(x^2)}{1-x/2+O(x^2)} \right)\\
&=-\frac{1}{x} \ln\left(1 - x(ab-(a^2b+ab^2)x+O(x^2))(1+x/2+O(x^2)) \right)\\
&=-\frac{1}{x} \ln\left(1 - abx(1-(a+b)x+O(x^2))(1+x/2+O(x^2)) \right)\\
&=-\frac{1}{x} \ln\left(1 - abx(1-(a+b+1/2)x+O(x^2)) \right)\\
&=-\frac{1}{x} \left( - abx(1-(a+b+1/2)x+O(x^2)) +((a+b+1/2)x+O(x^2))^2/2\right)\\
&=-\frac{1}{x} \left( - abx+ab(a+b+1/2)x^2+O(x^3)) +x^2((a+b+1/2)+O(x))^2/2\right)\\
&=-\frac{1}{x} \left( - abx+ab(a+b+1/2)x^2+O(x^3)) +x^2((a+b+1/2)^2+O(x))/2\right)\\
&=-\frac{1}{x} \left( -x\left(ab+ab(a+b+1/2)x+O(x^2) +x((a+b+1/2)^2+O(x))/2\right)\right)\\
&=-\frac{1}{x} \left( -x\left( ab +x(ab(a+b+1/2)+(a+b+1/2)^2\right)/2+O(x^2)\right)\\
&= ab +x(ab(a+b+1/2)+(a+b+1/2)^2)/2+O(x^2)\\
&= ab +x(ab(a+b+1/2))(1+(a+b+1/2))/2+O(x^2)\\
&= ab +xab(a+b+1/2)(a+b+3/2)/2+O(x^2)\\
\end{array}
$
You better check my algebra,
because
Prob(error) > 1/e.
A: Well, considering the Taylor expansion of $$ln(1+x)=x-\frac{x^2}{2}+o(x^2)\,\,\,\,\,\text{  for }x\to0$$
since
$$\frac{(e^{-ux}-1)(e^{-vx}-1)}{e^{-x}-1}\to 0\,\,\,\,\,\text{for }x\to 0$$
(it can be seen by taylor expand the exponential up to the first order)
then you get:
$$f(x)=-\frac{1}{x}\left[\left(\frac{(e^{-ux}-1)(e^{-vx}-1)}{e^{-x}-1}\right)-\frac{1}{2}\left(\frac{(e^{-ux}-1)(e^{-vx}-1)}{e^{-x}-1}\right)^2+...\right]$$
Then you taylor expand the exponential:
$$e^x=1+x+o(x)\,\,\,\text{ for }x\to 0$$
and get:
$$f(x)=-\frac{1}{x}\left[\left(\frac{(-ux)(-vx)}{-x}\right)-\frac{1}{2}\left(\frac{(-ux)(-vx}{-x}\right)^2+...\right]=-\frac{1}{x}\left[-uvx-\frac{1}{2}u^2v^2x^2+...\right]$$
You can clearly expand the exponential up to second order to get more terms.
