Is the function is differentiable at 0?

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1
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Is the function given by



$ f(x) = begincases frac1x log2 - frac12^x-1,textif xneq 0 ,\frac 12 text if x=0 endcases$.



is differentiable at zero ?



My Attempts:



$f'(0) = fracf(x) - f(0)x-0 =fracfrac1x log2 - frac12^x-1- frac12 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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  • 1




    A Limit is missing!
    – Dr. Sonnhard Graubner
    Aug 27 at 15:40










  • @Dr.SonnhardGraubner.....ya i misses that
    – stupid
    Aug 27 at 15:41










  • 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 at 15:42














up vote
1
down vote

favorite












Is the function given by



$ f(x) = begincases frac1x log2 - frac12^x-1,textif xneq 0 ,\frac 12 text if x=0 endcases$.



is differentiable at zero ?



My Attempts:



$f'(0) = fracf(x) - f(0)x-0 =fracfrac1x log2 - frac12^x-1- frac12 x $



after that i can not able to proceed further



Pliz help me



Any hints/solution will be appreciated



thanks in advance







share|cite|improve this question
















  • 1




    A Limit is missing!
    – Dr. Sonnhard Graubner
    Aug 27 at 15:40










  • @Dr.SonnhardGraubner.....ya i misses that
    – stupid
    Aug 27 at 15:41










  • 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 at 15:42












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Is the function given by



$ f(x) = begincases frac1x log2 - frac12^x-1,textif xneq 0 ,\frac 12 text if x=0 endcases$.



is differentiable at zero ?



My Attempts:



$f'(0) = fracf(x) - f(0)x-0 =fracfrac1x log2 - frac12^x-1- frac12 x $



after that i can not able to proceed further



Pliz help me



Any hints/solution will be appreciated



thanks in advance







share|cite|improve this question












Is the function given by



$ f(x) = begincases frac1x log2 - frac12^x-1,textif xneq 0 ,\frac 12 text if x=0 endcases$.



is differentiable at zero ?



My Attempts:



$f'(0) = fracf(x) - f(0)x-0 =fracfrac1x log2 - frac12^x-1- frac12 x $



after that i can not able to proceed further



Pliz help me



Any hints/solution will be appreciated



thanks in advance









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 27 at 15:37









stupid

668111




668111







  • 1




    A Limit is missing!
    – Dr. Sonnhard Graubner
    Aug 27 at 15:40










  • @Dr.SonnhardGraubner.....ya i misses that
    – stupid
    Aug 27 at 15:41










  • 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 at 15:42












  • 1




    A Limit is missing!
    – Dr. Sonnhard Graubner
    Aug 27 at 15:40










  • @Dr.SonnhardGraubner.....ya i misses that
    – stupid
    Aug 27 at 15:41










  • 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 at 15:42







1




1




A Limit is missing!
– Dr. Sonnhard Graubner
Aug 27 at 15:40




A Limit is missing!
– Dr. Sonnhard Graubner
Aug 27 at 15:40












@Dr.SonnhardGraubner.....ya i misses that
– stupid
Aug 27 at 15:41




@Dr.SonnhardGraubner.....ya i misses that
– stupid
Aug 27 at 15:41












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 at 15:42




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 at 15:42










1 Answer
1






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up vote
3
down vote



accepted










$$frac12^x-1=frac1exp(xln2)-1=frac1xln 2+x^2(ln 2)^2/2+O(x^3)
=frac1xln 2left(1-fracln 22+O(x)right)$$
etc.






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    3
    down vote



    accepted










    $$frac12^x-1=frac1exp(xln2)-1=frac1xln 2+x^2(ln 2)^2/2+O(x^3)
    =frac1xln 2left(1-fracln 22+O(x)right)$$
    etc.






    share|cite|improve this answer
























      up vote
      3
      down vote



      accepted










      $$frac12^x-1=frac1exp(xln2)-1=frac1xln 2+x^2(ln 2)^2/2+O(x^3)
      =frac1xln 2left(1-fracln 22+O(x)right)$$
      etc.






      share|cite|improve this answer






















        up vote
        3
        down vote



        accepted







        up vote
        3
        down vote



        accepted






        $$frac12^x-1=frac1exp(xln2)-1=frac1xln 2+x^2(ln 2)^2/2+O(x^3)
        =frac1xln 2left(1-fracln 22+O(x)right)$$
        etc.






        share|cite|improve this answer












        $$frac12^x-1=frac1exp(xln2)-1=frac1xln 2+x^2(ln 2)^2/2+O(x^3)
        =frac1xln 2left(1-fracln 22+O(x)right)$$
        etc.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 27 at 15:39









        Lord Shark the Unknown

        88.4k955115




        88.4k955115



























             

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