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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 ?

My Attempts:

$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

thanks in advance

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    $\begingroup$ A Limit is missing! $\endgroup$ – Dr. Sonnhard Graubner Aug 27 '18 at 15:40
  • $\begingroup$ @Dr.SonnhardGraubner.....ya i misses that $\endgroup$ – jasmine Aug 27 '18 at 15:41
  • $\begingroup$ 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. $\endgroup$ – xbh Aug 27 '18 at 15:42
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$$\frac1{2^x-1}=\frac1{\exp(x\ln2)-1}=\frac1{x\ln 2+x^2(\ln 2)^2/2+O(x^3)} =\frac1{x\ln 2}\left(1-\frac{\ln 2}2+O(x)\right)$$ etc.

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