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

You approach is correct, but the substitution is wrong! You should have proceeded in this manner:-
$$f'\left( 0 \right) = \lim_{x \rightarrow 0}\dfrac{f \left( x \right) - f \left( 0 \right)}{x - 0}$$
$$= \lim_{x \rightarrow 0}{\dfrac{ \left( \dfrac{1}{x \log 2} - \dfrac{1}{2^{x} - 1} \right) - \left( \dfrac{1}{2} \right)}{x}}$$
$$= \lim_{x \rightarrow 0} \dfrac{2 \left( 2^{x} - 1 - x \log 2 \right) - \left( 2^x - 1 \right)x \log 2}{2 x^2 \log 2 \left( 2^x - 1 \right)}$$
Use L'Hospital's Rule to get the answer!
