<s>[INST] Is the function is differentiable at 0? Is  the function given by
$ f(x)  = \begin{cases} \frac{1}{x \log2} - \frac{1}{2^x-1},\text{if} \  x\neq 0 ,\\\frac{ 1}{2} \ \text   {if} \  x=0 \end{cases}$.
is differentiable at zero ?
My Attempts:
$f'(0) =  \frac{f(x) -  f(0)}{x-0} =\frac{\frac{1}{x \log2} - \frac{1}{2^x-1}- \frac{1}{2} }{x} $
after that i can not able to proceed further
Pliz  help me
Any hints/solution will be appreciated
thanks in advance
 [/INST] $$\frac1{2^x-1}=\frac1{\exp(x\ln2)-1}=\frac1{x\ln 2+x^2(\ln 2)^2/2+O(x^3)}
=\frac1{x\ln 2}\left(1-\frac{\ln 2}2+O(x)\right)$$
etc.
</s>