Proving a Limit Does Not Exist












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enter image description here



I'm not even sure how to approach this. I try factoring out $xy$ in the numerator to get $xy(x^2 - y^2)$ but I don't think that gets me anywhere with the denominator.






calculus limits multivariable-calculus







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    2












    $begingroup$


    enter image description here



    I'm not even sure how to approach this. I try factoring out $xy$ in the numerator to get $xy(x^2 - y^2)$ but I don't think that gets me anywhere with the denominator.






    calculus limits multivariable-calculus







    share|cite











    $endgroup$















      2












      2








      2





      $begingroup$


      enter image description here



      I'm not even sure how to approach this. I try factoring out $xy$ in the numerator to get $xy(x^2 - y^2)$ but I don't think that gets me anywhere with the denominator.






      calculus limits multivariable-calculus







      share|cite











      $endgroup$




      enter image description here



      I'm not even sure how to approach this. I try factoring out $xy$ in the numerator to get $xy(x^2 - y^2)$ but I don't think that gets me anywhere with the denominator.






      calculus limits multivariable-calculus




      calculus limits multivariable-calculus






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      share|cite













      share|cite




      share|cite








      edited 38 mins ago









      Thomas Shelby

      3,1771524




      3,1771524










      asked 50 mins ago









      krauser126krauser126

      394




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          3 Answers
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          $begingroup$

          HINT:



          What happens if the limit is taken along $y=2x$? What happens when the limit is taken along $y=0$? Are these equal? If not, what can one conclude?






          share|cite|improve this answer









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            Putting $x=y$ your expression vanishes, and for$x=2y$, the limit will be $1/3$. Therefore the limit does not exist






            share|cite|improve this answer









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              2












              $begingroup$

              Let's approach the limit along the line $y=mx.$



              $begin{align}
              &lim_{(x,y)to (0,0)}dfrac{x^3y-xy^3}{x^4+2y^4}\
              &=lim_{xto 0}dfrac{x^3mx-x(mx)^3}{x^4+2(mx)^4}\
              &=lim_{xto 0}dfrac{x^4m-m^3x^4}{x^4+2m^4x^4}\
              &=lim_{xto 0}dfrac{m-m^3}{1+2m^4}\
              &=dfrac{m-m^3}{1+2m^4}\
              end{align}$



              So what can you conclude about the limit ?






              share|cite|improve this answer











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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                5












                $begingroup$

                HINT:



                What happens if the limit is taken along $y=2x$? What happens when the limit is taken along $y=0$? Are these equal? If not, what can one conclude?






                share|cite|improve this answer









                $endgroup$


















                  5












                  $begingroup$

                  HINT:



                  What happens if the limit is taken along $y=2x$? What happens when the limit is taken along $y=0$? Are these equal? If not, what can one conclude?






                  share|cite|improve this answer









                  $endgroup$
















                    5












                    5








                    5





                    $begingroup$

                    HINT:



                    What happens if the limit is taken along $y=2x$? What happens when the limit is taken along $y=0$? Are these equal? If not, what can one conclude?






                    share|cite|improve this answer









                    $endgroup$



                    HINT:



                    What happens if the limit is taken along $y=2x$? What happens when the limit is taken along $y=0$? Are these equal? If not, what can one conclude?







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 46 mins ago









                    Mark ViolaMark Viola

                    132k1275173




                    132k1275173























                        2












                        $begingroup$

                        Putting $x=y$ your expression vanishes, and for$x=2y$, the limit will be $1/3$. Therefore the limit does not exist






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          Putting $x=y$ your expression vanishes, and for$x=2y$, the limit will be $1/3$. Therefore the limit does not exist






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            Putting $x=y$ your expression vanishes, and for$x=2y$, the limit will be $1/3$. Therefore the limit does not exist






                            share|cite|improve this answer









                            $endgroup$



                            Putting $x=y$ your expression vanishes, and for$x=2y$, the limit will be $1/3$. Therefore the limit does not exist







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 46 mins ago









                            HAMIDINE SOUMAREHAMIDINE SOUMARE

                            71729




                            71729























                                2












                                $begingroup$

                                Let's approach the limit along the line $y=mx.$



                                $begin{align}
                                &lim_{(x,y)to (0,0)}dfrac{x^3y-xy^3}{x^4+2y^4}\
                                &=lim_{xto 0}dfrac{x^3mx-x(mx)^3}{x^4+2(mx)^4}\
                                &=lim_{xto 0}dfrac{x^4m-m^3x^4}{x^4+2m^4x^4}\
                                &=lim_{xto 0}dfrac{m-m^3}{1+2m^4}\
                                &=dfrac{m-m^3}{1+2m^4}\
                                end{align}$



                                So what can you conclude about the limit ?






                                share|cite|improve this answer











                                $endgroup$


















                                  2












                                  $begingroup$

                                  Let's approach the limit along the line $y=mx.$



                                  $begin{align}
                                  &lim_{(x,y)to (0,0)}dfrac{x^3y-xy^3}{x^4+2y^4}\
                                  &=lim_{xto 0}dfrac{x^3mx-x(mx)^3}{x^4+2(mx)^4}\
                                  &=lim_{xto 0}dfrac{x^4m-m^3x^4}{x^4+2m^4x^4}\
                                  &=lim_{xto 0}dfrac{m-m^3}{1+2m^4}\
                                  &=dfrac{m-m^3}{1+2m^4}\
                                  end{align}$



                                  So what can you conclude about the limit ?






                                  share|cite|improve this answer











                                  $endgroup$
















                                    2












                                    2








                                    2





                                    $begingroup$

                                    Let's approach the limit along the line $y=mx.$



                                    $begin{align}
                                    &lim_{(x,y)to (0,0)}dfrac{x^3y-xy^3}{x^4+2y^4}\
                                    &=lim_{xto 0}dfrac{x^3mx-x(mx)^3}{x^4+2(mx)^4}\
                                    &=lim_{xto 0}dfrac{x^4m-m^3x^4}{x^4+2m^4x^4}\
                                    &=lim_{xto 0}dfrac{m-m^3}{1+2m^4}\
                                    &=dfrac{m-m^3}{1+2m^4}\
                                    end{align}$



                                    So what can you conclude about the limit ?






                                    share|cite|improve this answer











                                    $endgroup$



                                    Let's approach the limit along the line $y=mx.$



                                    $begin{align}
                                    &lim_{(x,y)to (0,0)}dfrac{x^3y-xy^3}{x^4+2y^4}\
                                    &=lim_{xto 0}dfrac{x^3mx-x(mx)^3}{x^4+2(mx)^4}\
                                    &=lim_{xto 0}dfrac{x^4m-m^3x^4}{x^4+2m^4x^4}\
                                    &=lim_{xto 0}dfrac{m-m^3}{1+2m^4}\
                                    &=dfrac{m-m^3}{1+2m^4}\
                                    end{align}$



                                    So what can you conclude about the limit ?







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited 10 mins ago









                                    Javi maxwell

                                    878




                                    878










                                    answered 41 mins ago









                                    Thomas ShelbyThomas Shelby

                                    3,1771524




                                    3,1771524






























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