For $a=frac 14$, $f$ is continuous and differentiable.












0












$begingroup$


Let $f(x,y) =
begin{cases}
frac{xy}{(x^2+y^2)^a}, & text{if $(x,y)neq (0,0)$ } \[2ex]
0, & text{if $(x,y)= (0,0)$}
end{cases}$



Then which one of the following is TRUE for $f$ at the point $(0,0)$?



(A) For $a=1$, $f$ is continuous but not differentiable.



(B) For $a=frac 12$, $f$ is continuous and differentiable.



(C) For $a=frac 14$, $f$ is continuous and differentiable.



(D) For $a=frac 34$, $f$ is neither continuous nor differentiable



My attempt:



(A) For $a=1$ $f(x,y) =
begin{cases}
frac{xy}{(x^2+y^2)}, & text{if $(x,y)neq (0,0)$ } \[2ex]
0, & text{if $(x,y)= (0,0)$}
end{cases}$



is not continuous at $(0,0)$ as if we put $y=mx$ then $f(x,mx) =
begin{cases}
frac{m}{(1+m^2)}, & text{if $xneq 0$ } \[2ex]
0, & text{if $x= 0$}
end{cases}$



not continuous at $0$. So, (A) is not true.



(B) For $a=frac 12$ $f(x,y) =
begin{cases}
frac{xy}{(x^2+y^2)^{frac 12}}, & text{if $(x,y)neq (0,0)$ } \[2ex]
0, & text{if $(x,y)= (0,0)$}
end{cases}$



is not differentiable at $(0,0)$ as $frac {f(h,k)-f(0,0)}{sqrt{h^2+k^2}}=
frac{hk}{(h^2+k^2)}$ where $(h,k) to (0,0)$ if we put $k=mh$ then



$f(h,mh) =frac{m}{(1+m^2)} neq (0,0) $



Hence, $f$ is not differentiable at $(0,0)$. So, (B) is not true.



(C) For $a=frac 14$ $f(x,y) =
begin{cases}
frac{xy}{(x^2+y^2)^{frac 14}}, & text{if $(x,y)neq (0,0)$ } \[2ex]
0, & text{if $(x,y)= (0,0)$}
end{cases}$



Then, $f(x,y) =frac{xsqrt y}{(frac{x^2}{y^2}+1)^{frac 14}}$ Can we prove continuity of $f$ from here?



Observe, $f$ is differentiable at $(0,0)$ as $frac {f(h,k)-f(0,0)}{sqrt{h^2+k^2}}=
frac{hk}{(h^2+k^2)^frac 34}=frac{h}{(frac{h^2}{k^{4/3}}+k^{2/3})^{frac 34}} to (0,0)$ where $(h,k) to (0,0)$. So we can directly say that $f$ is differentiable and hence is continuous.



(D) For $a=frac 34$ $f(x,y) =
begin{cases}
frac{xy}{(x^2+y^2)^{frac 34}}, & text{if $(x,y)neq (0,0)$ } \[2ex]
0, & text{if $(x,y)= (0,0)$}
end{cases}$



From the observation of (C) we get $f$ is continuous at $(0,0)$ so (D) is also false.




So, my basic question is although we know that (C) will be the correct option can we prove $f$ is continuous directly there? See the highlighted portion of part (C) under My attempt section.







real-analysis calculus analysis multivariable-calculus alternative-proof







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Let $f(x,y) =
    begin{cases}
    frac{xy}{(x^2+y^2)^a}, & text{if $(x,y)neq (0,0)$ } \[2ex]
    0, & text{if $(x,y)= (0,0)$}
    end{cases}$



    Then which one of the following is TRUE for $f$ at the point $(0,0)$?



    (A) For $a=1$, $f$ is continuous but not differentiable.



    (B) For $a=frac 12$, $f$ is continuous and differentiable.



    (C) For $a=frac 14$, $f$ is continuous and differentiable.



    (D) For $a=frac 34$, $f$ is neither continuous nor differentiable



    My attempt:



    (A) For $a=1$ $f(x,y) =
    begin{cases}
    frac{xy}{(x^2+y^2)}, & text{if $(x,y)neq (0,0)$ } \[2ex]
    0, & text{if $(x,y)= (0,0)$}
    end{cases}$



    is not continuous at $(0,0)$ as if we put $y=mx$ then $f(x,mx) =
    begin{cases}
    frac{m}{(1+m^2)}, & text{if $xneq 0$ } \[2ex]
    0, & text{if $x= 0$}
    end{cases}$



    not continuous at $0$. So, (A) is not true.



    (B) For $a=frac 12$ $f(x,y) =
    begin{cases}
    frac{xy}{(x^2+y^2)^{frac 12}}, & text{if $(x,y)neq (0,0)$ } \[2ex]
    0, & text{if $(x,y)= (0,0)$}
    end{cases}$



    is not differentiable at $(0,0)$ as $frac {f(h,k)-f(0,0)}{sqrt{h^2+k^2}}=
    frac{hk}{(h^2+k^2)}$     where $(h,k) to (0,0)$ if we put $k=mh$ then



    $f(h,mh) =frac{m}{(1+m^2)} neq (0,0) $



    Hence, $f$ is not differentiable at $(0,0)$. So, (B) is not true.



    (C) For $a=frac 14$ $f(x,y) =
    begin{cases}
    frac{xy}{(x^2+y^2)^{frac 14}}, & text{if $(x,y)neq (0,0)$ } \[2ex]
    0, & text{if $(x,y)= (0,0)$}
    end{cases}$



    Then, $f(x,y) =frac{xsqrt y}{(frac{x^2}{y^2}+1)^{frac 14}}$ Can we prove continuity of $f$ from here?



    Observe, $f$ is differentiable at $(0,0)$ as $frac {f(h,k)-f(0,0)}{sqrt{h^2+k^2}}=
    frac{hk}{(h^2+k^2)^frac 34}=frac{h}{(frac{h^2}{k^{4/3}}+k^{2/3})^{frac 34}} to (0,0)$     where $(h,k) to (0,0)$. So we can directly say that $f$ is differentiable and hence is continuous.



    (D) For $a=frac 34$ $f(x,y) =
    begin{cases}
    frac{xy}{(x^2+y^2)^{frac 34}}, & text{if $(x,y)neq (0,0)$ } \[2ex]
    0, & text{if $(x,y)= (0,0)$}
    end{cases}$



    From the observation of (C) we get $f$ is continuous at $(0,0)$ so (D) is also false.




    So, my basic question is although we know that (C) will be the correct option can we prove $f$ is continuous directly there? See the highlighted portion of part (C) under My attempt section.







    real-analysis calculus analysis multivariable-calculus alternative-proof







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $f(x,y) =
      begin{cases}
      frac{xy}{(x^2+y^2)^a}, & text{if $(x,y)neq (0,0)$ } \[2ex]
      0, & text{if $(x,y)= (0,0)$}
      end{cases}$



      Then which one of the following is TRUE for $f$ at the point $(0,0)$?



      (A) For $a=1$, $f$ is continuous but not differentiable.



      (B) For $a=frac 12$, $f$ is continuous and differentiable.



      (C) For $a=frac 14$, $f$ is continuous and differentiable.



      (D) For $a=frac 34$, $f$ is neither continuous nor differentiable



      My attempt:



      (A) For $a=1$ $f(x,y) =
      begin{cases}
      frac{xy}{(x^2+y^2)}, & text{if $(x,y)neq (0,0)$ } \[2ex]
      0, & text{if $(x,y)= (0,0)$}
      end{cases}$



      is not continuous at $(0,0)$ as if we put $y=mx$ then $f(x,mx) =
      begin{cases}
      frac{m}{(1+m^2)}, & text{if $xneq 0$ } \[2ex]
      0, & text{if $x= 0$}
      end{cases}$



      not continuous at $0$. So, (A) is not true.



      (B) For $a=frac 12$ $f(x,y) =
      begin{cases}
      frac{xy}{(x^2+y^2)^{frac 12}}, & text{if $(x,y)neq (0,0)$ } \[2ex]
      0, & text{if $(x,y)= (0,0)$}
      end{cases}$



      is not differentiable at $(0,0)$ as $frac {f(h,k)-f(0,0)}{sqrt{h^2+k^2}}=
      frac{hk}{(h^2+k^2)}$       where $(h,k) to (0,0)$ if we put $k=mh$ then



      $f(h,mh) =frac{m}{(1+m^2)} neq (0,0) $



      Hence, $f$ is not differentiable at $(0,0)$. So, (B) is not true.



      (C) For $a=frac 14$ $f(x,y) =
      begin{cases}
      frac{xy}{(x^2+y^2)^{frac 14}}, & text{if $(x,y)neq (0,0)$ } \[2ex]
      0, & text{if $(x,y)= (0,0)$}
      end{cases}$



      Then, $f(x,y) =frac{xsqrt y}{(frac{x^2}{y^2}+1)^{frac 14}}$ Can we prove continuity of $f$ from here?



      Observe, $f$ is differentiable at $(0,0)$ as $frac {f(h,k)-f(0,0)}{sqrt{h^2+k^2}}=
      frac{hk}{(h^2+k^2)^frac 34}=frac{h}{(frac{h^2}{k^{4/3}}+k^{2/3})^{frac 34}} to (0,0)$       where $(h,k) to (0,0)$. So we can directly say that $f$ is differentiable and hence is continuous.



      (D) For $a=frac 34$ $f(x,y) =
      begin{cases}
      frac{xy}{(x^2+y^2)^{frac 34}}, & text{if $(x,y)neq (0,0)$ } \[2ex]
      0, & text{if $(x,y)= (0,0)$}
      end{cases}$



      From the observation of (C) we get $f$ is continuous at $(0,0)$ so (D) is also false.




      So, my basic question is although we know that (C) will be the correct option can we prove $f$ is continuous directly there? See the highlighted portion of part (C) under My attempt section.







      real-analysis calculus analysis multivariable-calculus alternative-proof







      share|cite|improve this question









      $endgroup$




      Let $f(x,y) =
      begin{cases}
      frac{xy}{(x^2+y^2)^a}, & text{if $(x,y)neq (0,0)$ } \[2ex]
      0, & text{if $(x,y)= (0,0)$}
      end{cases}$



      Then which one of the following is TRUE for $f$ at the point $(0,0)$?



      (A) For $a=1$, $f$ is continuous but not differentiable.



      (B) For $a=frac 12$, $f$ is continuous and differentiable.



      (C) For $a=frac 14$, $f$ is continuous and differentiable.



      (D) For $a=frac 34$, $f$ is neither continuous nor differentiable



      My attempt:



      (A) For $a=1$ $f(x,y) =
      begin{cases}
      frac{xy}{(x^2+y^2)}, & text{if $(x,y)neq (0,0)$ } \[2ex]
      0, & text{if $(x,y)= (0,0)$}
      end{cases}$



      is not continuous at $(0,0)$ as if we put $y=mx$ then $f(x,mx) =
      begin{cases}
      frac{m}{(1+m^2)}, & text{if $xneq 0$ } \[2ex]
      0, & text{if $x= 0$}
      end{cases}$



      not continuous at $0$. So, (A) is not true.



      (B) For $a=frac 12$ $f(x,y) =
      begin{cases}
      frac{xy}{(x^2+y^2)^{frac 12}}, & text{if $(x,y)neq (0,0)$ } \[2ex]
      0, & text{if $(x,y)= (0,0)$}
      end{cases}$



      is not differentiable at $(0,0)$ as $frac {f(h,k)-f(0,0)}{sqrt{h^2+k^2}}=
      frac{hk}{(h^2+k^2)}$       where $(h,k) to (0,0)$ if we put $k=mh$ then



      $f(h,mh) =frac{m}{(1+m^2)} neq (0,0) $



      Hence, $f$ is not differentiable at $(0,0)$. So, (B) is not true.



      (C) For $a=frac 14$ $f(x,y) =
      begin{cases}
      frac{xy}{(x^2+y^2)^{frac 14}}, & text{if $(x,y)neq (0,0)$ } \[2ex]
      0, & text{if $(x,y)= (0,0)$}
      end{cases}$



      Then, $f(x,y) =frac{xsqrt y}{(frac{x^2}{y^2}+1)^{frac 14}}$ Can we prove continuity of $f$ from here?



      Observe, $f$ is differentiable at $(0,0)$ as $frac {f(h,k)-f(0,0)}{sqrt{h^2+k^2}}=
      frac{hk}{(h^2+k^2)^frac 34}=frac{h}{(frac{h^2}{k^{4/3}}+k^{2/3})^{frac 34}} to (0,0)$       where $(h,k) to (0,0)$. So we can directly say that $f$ is differentiable and hence is continuous.



      (D) For $a=frac 34$ $f(x,y) =
      begin{cases}
      frac{xy}{(x^2+y^2)^{frac 34}}, & text{if $(x,y)neq (0,0)$ } \[2ex]
      0, & text{if $(x,y)= (0,0)$}
      end{cases}$



      From the observation of (C) we get $f$ is continuous at $(0,0)$ so (D) is also false.




      So, my basic question is although we know that (C) will be the correct option can we prove $f$ is continuous directly there? See the highlighted portion of part (C) under My attempt section.







      real-analysis calculus analysis multivariable-calculus alternative-proof




      real-analysis calculus analysis multivariable-calculus alternative-proof






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 7 at 5:52









      GimgimGimgim

      1869




      1869






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Recall that
          $$|xy|leq frac{(x^2+y^2)}{2}$$
          and so
          $$frac{|xy|}{(x^2+y^2)^{1/4}}leq frac{(x^2+y^2)^{3/4}}{2}to 0$$
          as $(x,y)to (0,0).$
          Hence you have continuity at $(0,0).$






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            For $a=frac 1 4$ the derivative exist and is $0$: by definition of derivative this means $ frac {xy} {(x^{2}+y^{2})^{3/4}} to 0$ which follows easily from the fact that $2|xy | leq x^{2}+y^{2}$.






            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%2f3064698%2ffor-a-frac-14-f-is-continuous-and-differentiable%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              Recall that
              $$|xy|leq frac{(x^2+y^2)}{2}$$
              and so
              $$frac{|xy|}{(x^2+y^2)^{1/4}}leq frac{(x^2+y^2)^{3/4}}{2}to 0$$
              as $(x,y)to (0,0).$
              Hence you have continuity at $(0,0).$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Recall that
                $$|xy|leq frac{(x^2+y^2)}{2}$$
                and so
                $$frac{|xy|}{(x^2+y^2)^{1/4}}leq frac{(x^2+y^2)^{3/4}}{2}to 0$$
                as $(x,y)to (0,0).$
                Hence you have continuity at $(0,0).$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Recall that
                  $$|xy|leq frac{(x^2+y^2)}{2}$$
                  and so
                  $$frac{|xy|}{(x^2+y^2)^{1/4}}leq frac{(x^2+y^2)^{3/4}}{2}to 0$$
                  as $(x,y)to (0,0).$
                  Hence you have continuity at $(0,0).$






                  share|cite|improve this answer









                  $endgroup$



                  Recall that
                  $$|xy|leq frac{(x^2+y^2)}{2}$$
                  and so
                  $$frac{|xy|}{(x^2+y^2)^{1/4}}leq frac{(x^2+y^2)^{3/4}}{2}to 0$$
                  as $(x,y)to (0,0).$
                  Hence you have continuity at $(0,0).$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 7 at 5:59









                  Hello_WorldHello_World

                  4,11621731




                  4,11621731























                      1












                      $begingroup$

                      For $a=frac 1 4$ the derivative exist and is $0$: by definition of derivative this means $ frac {xy} {(x^{2}+y^{2})^{3/4}} to 0$ which follows easily from the fact that $2|xy | leq x^{2}+y^{2}$.






                      share|cite|improve this answer











                      $endgroup$


















                        1












                        $begingroup$

                        For $a=frac 1 4$ the derivative exist and is $0$: by definition of derivative this means $ frac {xy} {(x^{2}+y^{2})^{3/4}} to 0$ which follows easily from the fact that $2|xy | leq x^{2}+y^{2}$.






                        share|cite|improve this answer











                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          For $a=frac 1 4$ the derivative exist and is $0$: by definition of derivative this means $ frac {xy} {(x^{2}+y^{2})^{3/4}} to 0$ which follows easily from the fact that $2|xy | leq x^{2}+y^{2}$.






                          share|cite|improve this answer











                          $endgroup$



                          For $a=frac 1 4$ the derivative exist and is $0$: by definition of derivative this means $ frac {xy} {(x^{2}+y^{2})^{3/4}} to 0$ which follows easily from the fact that $2|xy | leq x^{2}+y^{2}$.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 7 at 7:13

























                          answered Jan 7 at 6:04









                          Kavi Rama MurthyKavi Rama Murthy

                          54.4k32055




                          54.4k32055






























                              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%2f3064698%2ffor-a-frac-14-f-is-continuous-and-differentiable%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

                              An IMO inspired problem

                              Image
                              14 4 This problem from IMO 1988 is said to be one of the most elegant ones in functional equations. Problem : The function $f$ is defined on the set of all positive integers as follows: begin{align} f(1) = 1, f(3) &= 3, f(2n) = f(n), \ f(4n+1) &= 2f(2n+1) - f(n), \ f(4n+3) &= 3 f(2n+1) - 2f(n) end{align} Find the number of $n$ with $f(n) = n, 1 leq n leq 1988$. The main idea towards the solution is realizing that these conditions stand for the fact that $f(n)$ just reverses the digits in the binary representation of the number $n$. So, essentially the solution is finding the number of binary pallindromes $leq 1988_{10}$. My question is the following : How to reformulate the problem so that $f(n)$ reverses the digits of $n$ in its ternary representation. Or even better, can we reformulate it for a
                              Read more

                              Management

                              Image
                              Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
                              Read more

                              Investment

                              Image
                              Investment From Wikipedia, the free encyclopedia Jump to navigation Jump to search This article is about investment in finance. For investment in macroeconomics, see Investment (macroeconomics). For other uses, see Investment (disambiguation). "Invest" redirects here. For the term in meteorology, see Invest (meteorology). This article needs additional citations for verification . Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed. ( February 2017 ) (Learn how and when to remove this template message) In general, to invest is to distribute money in the expectation of some benefit in the future – for example, investment in durable goods, in real estate by the service industry, in factories for manufacturing, in product development, and in research and development. However, this article focuses specifically on investment in financial
                              Read more