First and second differentiability of the piecewise function $x^4sin(frac{1}{x})$ if $x neq 0$ and $0$ if...












2














I have the following function $f(x)$ defined as $x^4sin(frac{1}{x})$ if $x neq 0$ and $0$ if $x=0$. And I'm asked if the function is:

a) differentiable

b) two times differentiable

c) two times continuously differentiable





This function is of course quite similar to the function $x^2sin(frac{1}{x})$, which can be found on the internet for the same type of exercises. But I still want to be sure that I proceeded in the correct way, because I have no solution to this exercise. Thanks for your feedback.

a) At $x=0$, we have $$lim_{hto 0}frac{f(0+h)-f(0)}{h}=lim_{h to 0}frac{h^4sin(frac{1}{h})}{h}=lim_{h to 0}h^3sin(frac{1}{h})=0$$
Everywhere else, we have simply $4x^3 sin(frac{1}{x})-x^2cos(frac{1}{x})$.


Thus the function is differentiable everywhere, even at $0$ where its value is $0$.


b) Again, if $xneq 0$, we have simply $f^{prime prime}(x)=12x^2sin(frac{1}{x})-sin(frac{1}{x})-6xcos(frac{1}{x})$. At $x=0$, we have $$lim_{h to 0}frac{f'(0+h)-f'(0)}{h}=lim_{h to 0}frac{4h^3sin(frac{1}{h})-h^2cos(frac{1}{h})}{h}=lim_{h to 0}4h^2sin(frac{1}{h})-hcos(frac{1}{h})=0$$

So the function is also two times differentiable at $0$, where the derivative is $0$.

c) The function is however not two times continuously differentiable, because as said, the second derivative at $0$ is $0$, but if we take $lim_{x to 0_{pm}} 12x^2sin(frac{1}{x})-sin(frac{1}{x})-6xcos(frac{1}{x})$, we don't always reach $0$ because $sin(frac{1}{x})$ oscillates between $-1$ and $1$. So the function is not two times continuously differentiable, even if the second derivative is defined everywhere, even at $0$





Are my results correct ?
Thanks for your help !




real-analysis calculus limits derivatives piecewise-continuity







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    I have the following function $f(x)$ defined as $x^4sin(frac{1}{x})$ if $x neq 0$ and $0$ if $x=0$. And I'm asked if the function is:

    a) differentiable

    b) two times differentiable

    c) two times continuously differentiable





    This function is of course quite similar to the function $x^2sin(frac{1}{x})$, which can be found on the internet for the same type of exercises. But I still want to be sure that I proceeded in the correct way, because I have no solution to this exercise. Thanks for your feedback.

    a) At $x=0$, we have $$lim_{hto 0}frac{f(0+h)-f(0)}{h}=lim_{h to 0}frac{h^4sin(frac{1}{h})}{h}=lim_{h to 0}h^3sin(frac{1}{h})=0$$
    Everywhere else, we have simply $4x^3 sin(frac{1}{x})-x^2cos(frac{1}{x})$.


    Thus the function is differentiable everywhere, even at $0$ where its value is $0$.


    b) Again, if $xneq 0$, we have simply $f^{prime prime}(x)=12x^2sin(frac{1}{x})-sin(frac{1}{x})-6xcos(frac{1}{x})$. At $x=0$, we have $$lim_{h to 0}frac{f'(0+h)-f'(0)}{h}=lim_{h to 0}frac{4h^3sin(frac{1}{h})-h^2cos(frac{1}{h})}{h}=lim_{h to 0}4h^2sin(frac{1}{h})-hcos(frac{1}{h})=0$$

    So the function is also two times differentiable at $0$, where the derivative is $0$.

    c) The function is however not two times continuously differentiable, because as said, the second derivative at $0$ is $0$, but if we take $lim_{x to 0_{pm}} 12x^2sin(frac{1}{x})-sin(frac{1}{x})-6xcos(frac{1}{x})$, we don't always reach $0$ because $sin(frac{1}{x})$ oscillates between $-1$ and $1$. So the function is not two times continuously differentiable, even if the second derivative is defined everywhere, even at $0$





    Are my results correct ?
    Thanks for your help !




    real-analysis calculus limits derivatives piecewise-continuity







    share|cite|improve this question



























      2












      2








      2







      I have the following function $f(x)$ defined as $x^4sin(frac{1}{x})$ if $x neq 0$ and $0$ if $x=0$. And I'm asked if the function is:

      a) differentiable

      b) two times differentiable

      c) two times continuously differentiable





      This function is of course quite similar to the function $x^2sin(frac{1}{x})$, which can be found on the internet for the same type of exercises. But I still want to be sure that I proceeded in the correct way, because I have no solution to this exercise. Thanks for your feedback.

      a) At $x=0$, we have $$lim_{hto 0}frac{f(0+h)-f(0)}{h}=lim_{h to 0}frac{h^4sin(frac{1}{h})}{h}=lim_{h to 0}h^3sin(frac{1}{h})=0$$
      Everywhere else, we have simply $4x^3 sin(frac{1}{x})-x^2cos(frac{1}{x})$.


      Thus the function is differentiable everywhere, even at $0$ where its value is $0$.


      b) Again, if $xneq 0$, we have simply $f^{prime prime}(x)=12x^2sin(frac{1}{x})-sin(frac{1}{x})-6xcos(frac{1}{x})$. At $x=0$, we have $$lim_{h to 0}frac{f'(0+h)-f'(0)}{h}=lim_{h to 0}frac{4h^3sin(frac{1}{h})-h^2cos(frac{1}{h})}{h}=lim_{h to 0}4h^2sin(frac{1}{h})-hcos(frac{1}{h})=0$$

      So the function is also two times differentiable at $0$, where the derivative is $0$.

      c) The function is however not two times continuously differentiable, because as said, the second derivative at $0$ is $0$, but if we take $lim_{x to 0_{pm}} 12x^2sin(frac{1}{x})-sin(frac{1}{x})-6xcos(frac{1}{x})$, we don't always reach $0$ because $sin(frac{1}{x})$ oscillates between $-1$ and $1$. So the function is not two times continuously differentiable, even if the second derivative is defined everywhere, even at $0$





      Are my results correct ?
      Thanks for your help !




      real-analysis calculus limits derivatives piecewise-continuity







      share|cite|improve this question















      I have the following function $f(x)$ defined as $x^4sin(frac{1}{x})$ if $x neq 0$ and $0$ if $x=0$. And I'm asked if the function is:

      a) differentiable

      b) two times differentiable

      c) two times continuously differentiable





      This function is of course quite similar to the function $x^2sin(frac{1}{x})$, which can be found on the internet for the same type of exercises. But I still want to be sure that I proceeded in the correct way, because I have no solution to this exercise. Thanks for your feedback.

      a) At $x=0$, we have $$lim_{hto 0}frac{f(0+h)-f(0)}{h}=lim_{h to 0}frac{h^4sin(frac{1}{h})}{h}=lim_{h to 0}h^3sin(frac{1}{h})=0$$
      Everywhere else, we have simply $4x^3 sin(frac{1}{x})-x^2cos(frac{1}{x})$.


      Thus the function is differentiable everywhere, even at $0$ where its value is $0$.


      b) Again, if $xneq 0$, we have simply $f^{prime prime}(x)=12x^2sin(frac{1}{x})-sin(frac{1}{x})-6xcos(frac{1}{x})$. At $x=0$, we have $$lim_{h to 0}frac{f'(0+h)-f'(0)}{h}=lim_{h to 0}frac{4h^3sin(frac{1}{h})-h^2cos(frac{1}{h})}{h}=lim_{h to 0}4h^2sin(frac{1}{h})-hcos(frac{1}{h})=0$$

      So the function is also two times differentiable at $0$, where the derivative is $0$.

      c) The function is however not two times continuously differentiable, because as said, the second derivative at $0$ is $0$, but if we take $lim_{x to 0_{pm}} 12x^2sin(frac{1}{x})-sin(frac{1}{x})-6xcos(frac{1}{x})$, we don't always reach $0$ because $sin(frac{1}{x})$ oscillates between $-1$ and $1$. So the function is not two times continuously differentiable, even if the second derivative is defined everywhere, even at $0$





      Are my results correct ?
      Thanks for your help !




      real-analysis calculus limits derivatives piecewise-continuity




      real-analysis calculus limits derivatives piecewise-continuity






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      edited Jan 4 at 16:01









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      asked Jan 4 at 15:31









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          Your results are correct.



          However your explanation on the fact that $12x^2sin(frac{1}{x})-sin(frac{1}{x})-6xcos(frac{1}{x})$ doesn't have a limit at $0$ is a bit fuzzy. For example $g(x)=sin(frac{1}{x})-sin(frac{1}{x})=0$ do have a limit at $0$ while $sin(frac{1}{x})$ don't. The sum of two maps that don't have a limit at a point may have a limit!



          You should say that both $12x^2sin(frac{1}{x})$ and $6xcos(frac{1}{x})$ converges towards $0$ at $0$ while $sin(frac{1}{x})$ doesn't have a limit. Hence the sum of the three terms can't have a limit at zero.






          share|cite|improve this answer





















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            1 Answer
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            active

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            active

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            active

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            Your results are correct.



            However your explanation on the fact that $12x^2sin(frac{1}{x})-sin(frac{1}{x})-6xcos(frac{1}{x})$ doesn't have a limit at $0$ is a bit fuzzy. For example $g(x)=sin(frac{1}{x})-sin(frac{1}{x})=0$ do have a limit at $0$ while $sin(frac{1}{x})$ don't. The sum of two maps that don't have a limit at a point may have a limit!



            You should say that both $12x^2sin(frac{1}{x})$ and $6xcos(frac{1}{x})$ converges towards $0$ at $0$ while $sin(frac{1}{x})$ doesn't have a limit. Hence the sum of the three terms can't have a limit at zero.






            share|cite|improve this answer


























              2














              Your results are correct.



              However your explanation on the fact that $12x^2sin(frac{1}{x})-sin(frac{1}{x})-6xcos(frac{1}{x})$ doesn't have a limit at $0$ is a bit fuzzy. For example $g(x)=sin(frac{1}{x})-sin(frac{1}{x})=0$ do have a limit at $0$ while $sin(frac{1}{x})$ don't. The sum of two maps that don't have a limit at a point may have a limit!



              You should say that both $12x^2sin(frac{1}{x})$ and $6xcos(frac{1}{x})$ converges towards $0$ at $0$ while $sin(frac{1}{x})$ doesn't have a limit. Hence the sum of the three terms can't have a limit at zero.






              share|cite|improve this answer
























                2












                2








                2






                Your results are correct.



                However your explanation on the fact that $12x^2sin(frac{1}{x})-sin(frac{1}{x})-6xcos(frac{1}{x})$ doesn't have a limit at $0$ is a bit fuzzy. For example $g(x)=sin(frac{1}{x})-sin(frac{1}{x})=0$ do have a limit at $0$ while $sin(frac{1}{x})$ don't. The sum of two maps that don't have a limit at a point may have a limit!



                You should say that both $12x^2sin(frac{1}{x})$ and $6xcos(frac{1}{x})$ converges towards $0$ at $0$ while $sin(frac{1}{x})$ doesn't have a limit. Hence the sum of the three terms can't have a limit at zero.






                share|cite|improve this answer












                Your results are correct.



                However your explanation on the fact that $12x^2sin(frac{1}{x})-sin(frac{1}{x})-6xcos(frac{1}{x})$ doesn't have a limit at $0$ is a bit fuzzy. For example $g(x)=sin(frac{1}{x})-sin(frac{1}{x})=0$ do have a limit at $0$ while $sin(frac{1}{x})$ don't. The sum of two maps that don't have a limit at a point may have a limit!



                You should say that both $12x^2sin(frac{1}{x})$ and $6xcos(frac{1}{x})$ converges towards $0$ at $0$ while $sin(frac{1}{x})$ doesn't have a limit. Hence the sum of the three terms can't have a limit at zero.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 4 at 16:07









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