<s>[INST] Is the function is differentiable at $0$? Is the function  given by
$\displaystyle  f(x) =  \begin{cases} \dfrac{1}{x\log(2)} - \dfrac{1}{2^x -1}, \quad &x \neq 0, \\ \dfrac{1}{2}  , &x = 0 \end{cases}$
differentiable at $0$ ?
My attempt : $f'(0) =  \frac {f(x)  -f(0)}{x-0}$
$$\lim _{x\to 0}f'(0) = \lim _{x\to 0}\frac {\frac{1}{x\log(2)} - \frac{1}{2^x -1} - \frac{1}{0\log(2)} - \frac{1}{2^0 -1}}{x- 0}$$
$$\lim _{x\to 0}\frac {\frac{1}{x\log(2)} - \frac{1}{2^x -1} - \frac{1}{0} - \frac{1}{0}}{x- 0}$$
Now I am not able  to proceed further.
Any hints /solution   will  be appreciated.
Thanks in advance
 [/INST] Yes, it is.
Let us start with the function$$\begin{array}{rccc}g\colon&\mathbb{R}&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}\frac1{\log2}-\frac x{2^x-1}&\text{ if }x\neq0\\0&\text{ otherwise.}\end{cases}\end{array}$$Since the Taylor series of $\frac{2^x-1}x$ centered at $0$ is$$\log(2)+\frac{\log^2(2)}{2!}x+\frac{\log^3(2)}{3!}x^2+\cdots,$$the Taylor series of $\frac x{2^x-1}$ at $0$ begins with$$\frac1{\log 2}-\frac x2+\frac{\log 2}{12}x^2+\cdots$$Therefore, the Taylor series of $g$ centered at $0$ begins with$$\frac x2-\frac{\log 2}{12}x^2+\cdots$$and your function is $\frac{g(x)}x$.
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