Finding derivative by using L'Hospital's rule $$ f(x)= \begin{cases} \frac{1}{x\ln 2} - \frac{1}{2^x -1} \text { , if } x \neq 0 \\ \\ \frac{1}{2} \text{ , if }x=0 \end{cases}$$
I'm supposed to find derivative at point 0. I thought I should start by finding limits of both directions, showing that they equal by applying L'Hospital as many times as needed.

Forgive me, but I am going to illustrate how to find the derivative at $x=0$ using a Taylor expansion of the form $f(0)+f'(0) x+\cdots$.  We do this by finding the coefficient of $x$ in this expansion as follows:
$$\begin{align}\frac{1}{x \log{2}} - \frac{1}{2^x-1} &= \frac{1}{x \log{2}} - \frac{1}{e^{x \log{2}}-1}\\ &= \frac{1}{x \log{2}} -\frac{1}{1+x \log{2}+\frac12 x^2 \log^2{2}+\frac16 x^3 \log^3{2}\cdots-1}\\ &=\frac{1}{x \log{2}} \left ( 1-\frac{1}{1+\frac12 x \log{2} + \frac16 x^2 \log^2{2}+\cdots}\right )\\ &=\frac{1}{x \log{2}} \left [1-\left (1-\left [\frac12 x \log{2} + \frac16 x^2 \log^2{2}+\cdots \right ]+ \\\left [\frac12 x \log{2} +\cdots \right ]^2 +\cdots \right ) \right ] \\ &= \frac12 -\frac{\log{2}}{12} x + \cdots\end{align}$$
Therefore, the sought-after derivative at $x=0$ is $-\log{2}/12$.
