Is the function is differentiable at 0?
Clash Royale CLAN TAG#URR8PPP
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
real-analysis
add a comment |Â
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
real-analysis
1
A Limit is missing!
â Dr. Sonnhard Graubner
Aug 27 at 15:40
@Dr.SonnhardGraubner.....ya i misses that
â 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.
â xbh
Aug 27 at 15:42
add a comment |Â
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
real-analysis
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
real-analysis
asked Aug 27 at 15:37
stupid
668111
668111
1
A Limit is missing!
â Dr. Sonnhard Graubner
Aug 27 at 15:40
@Dr.SonnhardGraubner.....ya i misses that
â 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.
â xbh
Aug 27 at 15:42
add a comment |Â
1
A Limit is missing!
â Dr. Sonnhard Graubner
Aug 27 at 15:40
@Dr.SonnhardGraubner.....ya i misses that
â 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.
â xbh
Aug 27 at 15:42
1
1
A Limit is missing!
â Dr. Sonnhard Graubner
Aug 27 at 15:40
A Limit is missing!
â Dr. Sonnhard Graubner
Aug 27 at 15:40
@Dr.SonnhardGraubner.....ya i misses that
â stupid
Aug 27 at 15:41
@Dr.SonnhardGraubner.....ya i misses that
â 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.
â 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.
â xbh
Aug 27 at 15:42
add a comment |Â
1 Answer
1
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.
add a comment |Â
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.
add a comment |Â
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.
add a comment |Â
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.
$$frac12^x-1=frac1exp(xln2)-1=frac1xln 2+x^2(ln 2)^2/2+O(x^3)
=frac1xln 2left(1-fracln 22+O(x)right)$$
etc.
answered Aug 27 at 15:39
Lord Shark the Unknown
88.4k955115
88.4k955115
add a comment |Â
add a comment |Â
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1
A Limit is missing!
â Dr. Sonnhard Graubner
Aug 27 at 15:40
@Dr.SonnhardGraubner.....ya i misses that
â 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.
â xbh
Aug 27 at 15:42