differentiability of function at $0$

Question

I have to check differentiability of the following function at 0
$$f(x)=\begin{cases} \frac{1}{x\ln2}-\frac{1}{2^x-1} & \text{for } x\neq 0 \\ 1/2 & \text{for }x=0 \end{cases} $$ I tried using L'hospital rule but did not get the solution. Any Help will be appreciated

From Hint:
$lim_{x \to 0}\frac {\frac{1}{xln2}-\frac{1}{2^x-1}-1/2}{x}$=$lim_{x \to 0}\frac {\frac{1}{xln2}(\frac{1}{2 - x\ln2}-1/2)}{x}$=$\frac{-\ln 2}{4} $ but answer was $\frac{-\ln 2}{12} $ where is my mistake happen? Please help me out

Answer 1

HINT : $$2^x-1=e^{x\ln(2)}-1\underset{(0)}{=}x\ln\left(2\right)+x^2\frac{(ln(2))^2}{2}+o\left(x^2\right) $$

Answer 2

Using $2^h-1=h\log(2)+\frac12(h\log(2))^2+\frac16(\log(2)h)^3+O(h^4)$, we can write

$$\begin{align} \lim_{h\to 0}\frac{f(h)-f(0)}{h}&=\lim_{h\to 0}\frac{\frac{1}{h\log(2)}-\frac{1}{2^h-1}-\frac12}{h}\\\\ &\lim_{h\to 0}\frac{(2-\log(2)h)(2^h-1)-2\log(2)h}{2h^2(2^h-1)\log(2)}\\\\ &=\lim_{h\to 0}\frac{(2-\log(2)h)(h\log(2)+\frac12(h\log(2))^2+\frac16(\log(2)h)^3+O(h^4))-2\log(2)h}{2\log(2)h^2((\log(2)h)+O(h^2))}\\\\ &=\lim_{h\to 0}\frac{-\frac16(\log(2)h)^3+O(h^4)}{2\log^2(2)h^3+O(h^4)}\\\\ &=-\frac{\log(2)}{12} \end{align}$$

as was to be shown!