<s>[INST] 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.
 [/INST] 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 $$\lim_{x \to 0}\frac{2^{x} - 1}{x} = \log 2$$
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