This polynomial limit doesn't exist but I can't seem to prove it.

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$$lim_(x,y)to(0,0)fracx^2+y^2+5xyx-y$$



I've tried approximating using parabolas and even use polar coordinates but it seems to always result in $0$.
The limit seems simple so I must be forgetting something basic.
Hints? Thank you.






limits







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    up vote
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    down vote

    favorite












    $$lim_(x,y)to(0,0)fracx^2+y^2+5xyx-y$$



    I've tried approximating using parabolas and even use polar coordinates but it seems to always result in $0$.
    The limit seems simple so I must be forgetting something basic.
    Hints? Thank you.






    limits







    share|cite|improve this question









    New contributor




    João Simões is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.





















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      $$lim_(x,y)to(0,0)fracx^2+y^2+5xyx-y$$



      I've tried approximating using parabolas and even use polar coordinates but it seems to always result in $0$.
      The limit seems simple so I must be forgetting something basic.
      Hints? Thank you.






      limits







      share|cite|improve this question









      New contributor




      João Simões is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      $$lim_(x,y)to(0,0)fracx^2+y^2+5xyx-y$$



      I've tried approximating using parabolas and even use polar coordinates but it seems to always result in $0$.
      The limit seems simple so I must be forgetting something basic.
      Hints? Thank you.






      limits




      limits






      share|cite|improve this question









      New contributor




      João Simões is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      João Simões is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









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      edited 2 hours ago









      Robert Howard

      1,417622




      1,417622






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      asked 2 hours ago









      João Simões

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          2 Answers
          2






          active

          oldest

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          up vote
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          accepted










          The quotient blows up near the line $x=y$ no matter how close to $(0,0)$ you get, so the limit does not exist.






          share|cite|improve this answer






















          • Small point: It's not a polynomial.
            – zhw.
            1 hour ago










          • @zhw: Right, fixed.
            – Henning Makholm
            32 mins ago

















          up vote
          3
          down vote













          Write
          beginalign
          dfracx^2+y^2+5xyx-y &= x-y +dfrac7xyx-y \
          &=x-y+yleft(dfrac7x-7y+7yx-yright) \
          &=x-y+7y+dfrac7y^2x-y=x+6y+dfrac7y^2x-y.
          endalign
          To see that the limit fails to exist, consider $2$ paths: the first one is $(x,y) = (2t,t)$,and the limit is $0$. And the second path is $(x,y) = (t^2+t,t)$, and the limit is $7$. Thus you have $2$ different values of the limit, hence it does not exist.






          share|cite|improve this answer






















          • You might want to use beginalign and endalign for readability. $(+1)$ by the way.
            – Mattos
            1 hour ago











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          The quotient blows up near the line $x=y$ no matter how close to $(0,0)$ you get, so the limit does not exist.






          share|cite|improve this answer






















          • Small point: It's not a polynomial.
            – zhw.
            1 hour ago










          • @zhw: Right, fixed.
            – Henning Makholm
            32 mins ago














          up vote
          2
          down vote



          accepted










          The quotient blows up near the line $x=y$ no matter how close to $(0,0)$ you get, so the limit does not exist.






          share|cite|improve this answer






















          • Small point: It's not a polynomial.
            – zhw.
            1 hour ago










          • @zhw: Right, fixed.
            – Henning Makholm
            32 mins ago












          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          The quotient blows up near the line $x=y$ no matter how close to $(0,0)$ you get, so the limit does not exist.






          share|cite|improve this answer














          The quotient blows up near the line $x=y$ no matter how close to $(0,0)$ you get, so the limit does not exist.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 33 mins ago

























          answered 2 hours ago









          Henning Makholm

          234k16299531




          234k16299531











          • Small point: It's not a polynomial.
            – zhw.
            1 hour ago










          • @zhw: Right, fixed.
            – Henning Makholm
            32 mins ago
















          • Small point: It's not a polynomial.
            – zhw.
            1 hour ago










          • @zhw: Right, fixed.
            – Henning Makholm
            32 mins ago















          Small point: It's not a polynomial.
          – zhw.
          1 hour ago




          Small point: It's not a polynomial.
          – zhw.
          1 hour ago












          @zhw: Right, fixed.
          – Henning Makholm
          32 mins ago




          @zhw: Right, fixed.
          – Henning Makholm
          32 mins ago










          up vote
          3
          down vote













          Write
          beginalign
          dfracx^2+y^2+5xyx-y &= x-y +dfrac7xyx-y \
          &=x-y+yleft(dfrac7x-7y+7yx-yright) \
          &=x-y+7y+dfrac7y^2x-y=x+6y+dfrac7y^2x-y.
          endalign
          To see that the limit fails to exist, consider $2$ paths: the first one is $(x,y) = (2t,t)$,and the limit is $0$. And the second path is $(x,y) = (t^2+t,t)$, and the limit is $7$. Thus you have $2$ different values of the limit, hence it does not exist.






          share|cite|improve this answer






















          • You might want to use beginalign and endalign for readability. $(+1)$ by the way.
            – Mattos
            1 hour ago















          up vote
          3
          down vote













          Write
          beginalign
          dfracx^2+y^2+5xyx-y &= x-y +dfrac7xyx-y \
          &=x-y+yleft(dfrac7x-7y+7yx-yright) \
          &=x-y+7y+dfrac7y^2x-y=x+6y+dfrac7y^2x-y.
          endalign
          To see that the limit fails to exist, consider $2$ paths: the first one is $(x,y) = (2t,t)$,and the limit is $0$. And the second path is $(x,y) = (t^2+t,t)$, and the limit is $7$. Thus you have $2$ different values of the limit, hence it does not exist.






          share|cite|improve this answer






















          • You might want to use beginalign and endalign for readability. $(+1)$ by the way.
            – Mattos
            1 hour ago













          up vote
          3
          down vote










          up vote
          3
          down vote









          Write
          beginalign
          dfracx^2+y^2+5xyx-y &= x-y +dfrac7xyx-y \
          &=x-y+yleft(dfrac7x-7y+7yx-yright) \
          &=x-y+7y+dfrac7y^2x-y=x+6y+dfrac7y^2x-y.
          endalign
          To see that the limit fails to exist, consider $2$ paths: the first one is $(x,y) = (2t,t)$,and the limit is $0$. And the second path is $(x,y) = (t^2+t,t)$, and the limit is $7$. Thus you have $2$ different values of the limit, hence it does not exist.






          share|cite|improve this answer














          Write
          beginalign
          dfracx^2+y^2+5xyx-y &= x-y +dfrac7xyx-y \
          &=x-y+yleft(dfrac7x-7y+7yx-yright) \
          &=x-y+7y+dfrac7y^2x-y=x+6y+dfrac7y^2x-y.
          endalign
          To see that the limit fails to exist, consider $2$ paths: the first one is $(x,y) = (2t,t)$,and the limit is $0$. And the second path is $(x,y) = (t^2+t,t)$, and the limit is $7$. Thus you have $2$ different values of the limit, hence it does not exist.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 1 hour ago









          Rócherz

          2,2962518




          2,2962518










          answered 2 hours ago









          DeepSea

          69.9k54386




          69.9k54386











          • You might want to use beginalign and endalign for readability. $(+1)$ by the way.
            – Mattos
            1 hour ago

















          • You might want to use beginalign and endalign for readability. $(+1)$ by the way.
            – Mattos
            1 hour ago
















          You might want to use beginalign and endalign for readability. $(+1)$ by the way.
          – Mattos
          1 hour ago





          You might want to use beginalign and endalign for readability. $(+1)$ by the way.
          – Mattos
          1 hour ago











          João Simões is a new contributor. Be nice, and check out our Code of Conduct.









           

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