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.
 A: 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 $-\log{2}/12$.
A: Apply the approximation 
$$
2^{x}-1\sim x\ln2+x^{2}\frac{\left(\ln2\right)^{2}}{2}+x^{3}\frac{\left(\ln2\right)^{3}}{6}+\cdots
$$
then
\begin{eqnarray*}
 &  & \lim_{x\rightarrow0}\frac{f\left(x\right)-f\left(0\right)}{x}\\
 & = & \lim_{x\rightarrow0}\frac{\frac{1}{x\ln2}-\frac{1}{2^{x}-1}-\frac{1}{2}}{x}\\
 & = & \lim_{x\rightarrow0}\frac{2\left(2^{x}-1\right)-2\ln2\cdot x-\ln2\cdot x\left(2^{x}-1\right)}{2\ln2\cdot x^{2}\left(2^{x}-1\right)}\\
 & = & \lim_{x\rightarrow0}\frac{2\left(x\ln2+x^{2}\frac{\left(\ln2\right)^{2}}{2}+x^{3}\frac{\left(\ln2\right)^{3}}{6}\right)-2\ln2\cdot x-\ln2\cdot x\left(x\ln2+x^{2}\frac{\left(\ln2\right)^{2}}{2}\right)}{2\ln2\cdot x^{2}\left(x\ln2\right)}\\
 & = & \lim_{x\rightarrow0}\frac{\frac{1}{3}\left(\ln2\right)^{3}x^{3}-\frac{1}{2}\left(\ln2\right)^{3}x^{3}}{2\left(\ln2\right)^{2}x^{3}}\\
 & = & -\frac{1}{12}\ln2
\end{eqnarray*}
A: I give it a try without any powerful theorems of calculus, because I think the question belongs to an introductory course on calculus.
We have
$\displaystyle \begin{aligned}f'(0) &= \lim_{x \to 0}\frac{f(x) - f(0)}{x}\\
&= \lim_{x \to 0}\dfrac{\dfrac{1}{x\log 2} - \dfrac{1}{2^{x} - 1} - \dfrac{1}{2}}{x}\\
&= \lim_{x \to 0}\dfrac{2(2^{x} - 1) - 2x\log 2 - x(2^{x} - 1)\log 2}{2x^{2}(2^{x} - 1)\log 2}\\
&= \lim_{x \to 0}\dfrac{2(2^{x} - 1) - 2x\log 2 - x(2^{x} - 1)\log 2}{2x^{3}\log 2}\dfrac{x}{2^{x} - 1}\\
&= \lim_{x \to 0}\dfrac{2^{x + 1} - x\log 2 - 2^{x}x\log 2 - 2}{2x^{3}(\log 2)^{2}}\\
&(\text{apply L'Hospital rule})\\
&= \lim_{x \to 0}\dfrac{2^{x + 1}\log 2 - \log 2 - (2^{x} + 2^{x}x\log 2)\log 2}{6x^{2}(\log 2)^{2}}\\
&= \lim_{x \to 0}\dfrac{2^{x}\log 2 - \log 2 - 2^{x}x(\log 2)^{2}}{6x^{2}(\log 2)^{2}}\\
&(\text{apply L'Hospital rule})\\
&= \lim_{x \to 0}\dfrac{2^{x}(\log 2)^{2} - (2^{x} + 2^{x}x\log 2)(\log 2)^{2}}{12x(\log 2)^{2}}\\
&= \lim_{x \to 0}\dfrac{ -2^{x}x(\log 2)^{3}}{12x(\log 2)^{2}}\\
&= -\lim_{x \to 0}2^{x}\cdot\frac{\log 2}{12}\\
&= -\frac{\log 2}{12}\end{aligned}$
I have used the fact that $$\lim_{x \to 0}\frac{2^{x} - 1}{x} = \log 2$$
