Derivative at $x=0$ for $\frac{1}{x\ln(2)} - \frac{1}{2^x - 1}$

Question

A function $f$ is defined on $\Bbb R$ as $\frac{1}{x\ln 2 } - \frac{1}{2^x -1}$ for $x \neq 0$ and $\frac{1}{2}$ for $x = 0$. Show that $f$ is continuous and find $f'(0)$

From my knowledge

$$f'(0) = \lim_{x\to 0}\frac{f(x) - f(0)}{x}\\ =\frac{\frac{1}{x\ln(2)} - \frac{1}{2^x-1} - \frac{1}{2}}{x}\\ = \frac{\frac{x \ln(2)}{2(2^x-1)} - \frac{1}{2}}{x}\\ = \frac{x \ln(2) - (2^x -1)}{2x(2^x - 1)}\\ = \frac{x^2 \ln^2(2)}{4x(2^x - 1)}$$

Which on using mclaurin series of $2^x$ gives $\frac{-\ln 2}{4}$ but the correct answer is $\frac{-\ln 2}{12}$

Answer 1

We have $$\frac{1}{x\ln(2)}-\frac{1}{2^x-1}-\frac{1}{2}=\frac{(2^x-1)2-2x\ln(2)-(2^x-1)x\ln(2)}{x\ln(2)(2^x-1)\cdot 2}$$

Answer 2

You did not include all terms in the series. In intermediate steps, there is also a term $\frac{x^3\ln^3(2)}{3!}$ which is missed.

$$\frac{\frac{1}{x\ln(2)} - \frac{1}{2^x-1} - \frac{1}{2}}{x} = \frac{2(2^{x}-1)-2(x \ln(2))-x\ln(2)(2^{x}-1)}{2x^2 \ln(2)(2^{x}-1)} \\ = \frac{2(x\ln(2)+\tfrac{x^2 \ln^2(2)}{2}+\tfrac{x^3 \ln^3(2)}{6})-2x\ln 2 - x \ln(2)(x\ln(2)+\tfrac{x^2 \ln^2(2)}{2})}{2x^2 \ln(2)(2^{x}-1)}\\ = \frac{-\tfrac{x^3 \ln^3(2)}{6}}{2x^2 \ln(2)(2^{x}-1)}$$

which will give $ \frac{-\ln(2)}{12}$ as the limit.