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












2












$begingroup$


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











$endgroup$

















    2












    $begingroup$


    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











    $endgroup$















      2












      2








      2





      $begingroup$


      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











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 4 at 16:01









      mathcounterexamples.net

      26.9k22158




      26.9k22158










      asked Jan 4 at 15:31









      PoujhPoujh

      6111516




      6111516






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          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









          $endgroup$














            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061759%2ffirst-and-second-differentiability-of-the-piecewise-function-x4-sin-frac1x%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            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









            $endgroup$


















              2












              $begingroup$

              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









              $endgroup$
















                2












                2








                2





                $begingroup$

                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









                $endgroup$



                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









                mathcounterexamples.netmathcounterexamples.net

                26.9k22158




                26.9k22158






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061759%2ffirst-and-second-differentiability-of-the-piecewise-function-x4-sin-frac1x%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Probability when a professor distributes a quiz and homework assignment to a class of n students.

                    Image
                    0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
                    Read more

                    Aardman Animations

                    Image
                    Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
                    Read more

                    Are they similar matrix

                    Image
                    0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
                    Read more