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
 A: 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!
A: 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 $\frac{g(x)}x$.
A: The idea is to use an expansion near $x=0$:
$$\frac{1}{2^x-1}=\frac{1}{e^{x \ln(2)}-1}=\frac{1}{1+x \ln(2) +\frac{x^2 \ln(2)^2}{2}+\frac{x^3 \ln(2)^3}{6}-1+o(x^3)}=\frac{1}{x \ln(2) \left(1+\frac{x \ln(2)}{2}+\frac{x^2 \ln(2)^2}{6}+o(x^2) \right)}$$
so near $x=0$:
\begin{align}
f(x)&=\frac{1}{x \ln(2)}-\frac{1}{x \ln(2) \left(1+\frac{x \ln(2)}{2}+\frac{x^2 \ln(2)^2}{6}+o(x^2) \right)}\\
&=\frac{1}{x \ln(2)}-\frac{1}{x \ln(2)} \left(1-\frac{x \ln(2)}{2}-\frac{x^2 \ln(2)^2}{6}+\left(\frac{x \ln(2)}{2}\right)^2+o(x^2) \right)\\
&=\frac{1}{2}+x \left(\frac{\ln(2)}{6}-\frac{\ln(2)}{4} \right)+o(x)
\end{align}
so, as $f(0)=\frac{1}{2}$:
$$\frac{f(x)-f(0)}{x-0}=\frac{\frac{1}{2}-x \frac{\ln(2)}{12}+o(x)}{x}=-\frac{\ln(2)}{12}+o(1)$$
so $f$ is differentiable at $x=0$ and:
$$f'(0)=-\frac{\ln(2)}{12}$$
