Limit using L'Hopital's Rule I am trying to solve the limit 
$$ \lim_{x\to 0}\left(\frac{\frac{1}{x \ln2} - \frac{1}{2^x-1} - \frac{1}2}x\right)$$
using L'Hopital's Rule. However, it seems that I am doing something incorrectly, as after using the rule a couple of times, it does not get me anywhere but only complicates the existing limit. 
What am I missing?

We can proceed as follows
\begin{align}
L &= \lim_{x \to 0}\dfrac{\dfrac{1}{x\log 2} - \dfrac{1}{2^{x} - 1} - \dfrac{1}{2}}{x}\notag\\
&= \frac{1}{2\log 2}\lim_{x \to 0}\frac{2(2^{x} - 1) - 2x\log 2 - x\log 2(2^{x} - 1)}{x^{2}(2^{x} - 1)}\notag\\
&= \frac{1}{2\log 2}\lim_{x \to 0}\dfrac{2(2^{x} - 1) - 2x\log 2 - x\log 2(2^{x} - 1)}{x^{3}\cdot\dfrac{(2^{x} - 1)}{x}}\notag\\
&= \frac{1}{2(\log 2)^{2}}\lim_{x \to 0}\frac{2(2^{x} - 1) - 2x\log 2 - x\log 2(2^{x} - 1)}{x^{3}}\notag\\
&= \frac{1}{2(\log 2)^{2}}\lim_{x \to 0}\frac{2^{x}\log 2 - \log 2 - (\log 2)^{2}x\cdot 2^{x}}{3x^{2}}\text{ (via L'Hospital's Rule)}\notag\\
&= \frac{1}{2(\log 2)^{2}}\lim_{x \to 0}\frac{ - (\log 2)^{3}x\cdot 2^{x}}{6x}\text{ (via L'Hospital's Rule)}\notag\\
&= -\frac{\log 2}{12}\notag
\end{align}
Alternatively it is much easier to use Taylor series and get
\begin{align}
\frac{1}{2^{x} - 1} &= \frac{1}{e^{kx} - 1}\text{ where }k = \log 2\notag\\
&= \dfrac{1}{kx + \dfrac{k^{2}x^{2}}{2} + \dfrac{k^{3}x^{3}}{6} + o(x^{3})}\notag\\
&= \frac{1}{kx}\left(1 + \frac{kx}{2} + \frac{k^{2}x^{2}}{6} + o(x^{2})\right)^{-1}\notag\\
&= \frac{1}{kx}\left(1 - \frac{kx}{2} + \frac{k^{2}x^{2}}{12} + o(x^{2})\right)\notag\\
&= \frac{1}{kx} - \frac{1}{2} + \frac{kx}{12} + o(x)\notag\\
&= \frac{1}{x\log 2} - \frac{1}{2} + \frac{x\log 2}{12} + o(x)\notag
\end{align}
and therefore $$\frac{1}{x\log 2} - \frac{1}{2^{x} - 1} - \frac{1}{2} = -\frac{x\log 2}{12} + o(x)$$ and upon dividing the above equation by $x$ the desired limit is clearly seen to be $-(\log 2)/12$.
