# 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 ?

# $f'(0) = \frac{f(x) - f(0)}{x-0} =\frac{\frac{1}{x \log2} - \frac{1}{2^x-1}- \frac{1}{2} }{x}$

after that i can not able to proceed further

Pliz help me

Any hints/solution will be appreciated

• Make those denominators of fractions involving $x$ over the line be the same. If you have learned the Maclaurin formula then you can applied it and asymptotically transform the numerator to polynomials. – xbh Aug 27 '18 at 15:42
$$\frac1{2^x-1}=\frac1{\exp(x\ln2)-1}=\frac1{x\ln 2+x^2(\ln 2)^2/2+O(x^3)} =\frac1{x\ln 2}\left(1-\frac{\ln 2}2+O(x)\right)$$ etc.