Why doesn't this limit exist? $lim_{(x,y)to(0,0)} frac{x^2y^3}{(x^2-y^2)^2}$











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$$lim_{(x,y)to(0,0)} frac{x^2y^3}{(x^2-y^2)^2}$$



WolframAlpha says it doesn't exist, but I don't know how to prove it. Are there any directionals that give values different from zero?



Thank you all










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    Hint: The thing inside the limit is undefined on two whole lines through $0$.
    – user3482749
    Nov 28 at 22:27










  • $x=0$ or $y=0$ or $x=y$ all have $infty$ as limits For $x=0$, limit could be $pm infty$. It looks like getting $0$ as a limit is harder.
    – herb steinberg
    Nov 28 at 22:30

















up vote
-1
down vote

favorite












$$lim_{(x,y)to(0,0)} frac{x^2y^3}{(x^2-y^2)^2}$$



WolframAlpha says it doesn't exist, but I don't know how to prove it. Are there any directionals that give values different from zero?



Thank you all










share|cite|improve this question




















  • 1




    Hint: The thing inside the limit is undefined on two whole lines through $0$.
    – user3482749
    Nov 28 at 22:27










  • $x=0$ or $y=0$ or $x=y$ all have $infty$ as limits For $x=0$, limit could be $pm infty$. It looks like getting $0$ as a limit is harder.
    – herb steinberg
    Nov 28 at 22:30















up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











$$lim_{(x,y)to(0,0)} frac{x^2y^3}{(x^2-y^2)^2}$$



WolframAlpha says it doesn't exist, but I don't know how to prove it. Are there any directionals that give values different from zero?



Thank you all










share|cite|improve this question















$$lim_{(x,y)to(0,0)} frac{x^2y^3}{(x^2-y^2)^2}$$



WolframAlpha says it doesn't exist, but I don't know how to prove it. Are there any directionals that give values different from zero?



Thank you all







calculus limits






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edited Nov 28 at 22:40









Blue

47.3k870149




47.3k870149










asked Nov 28 at 22:25









Miguel Torres

1




1








  • 1




    Hint: The thing inside the limit is undefined on two whole lines through $0$.
    – user3482749
    Nov 28 at 22:27










  • $x=0$ or $y=0$ or $x=y$ all have $infty$ as limits For $x=0$, limit could be $pm infty$. It looks like getting $0$ as a limit is harder.
    – herb steinberg
    Nov 28 at 22:30
















  • 1




    Hint: The thing inside the limit is undefined on two whole lines through $0$.
    – user3482749
    Nov 28 at 22:27










  • $x=0$ or $y=0$ or $x=y$ all have $infty$ as limits For $x=0$, limit could be $pm infty$. It looks like getting $0$ as a limit is harder.
    – herb steinberg
    Nov 28 at 22:30










1




1




Hint: The thing inside the limit is undefined on two whole lines through $0$.
– user3482749
Nov 28 at 22:27




Hint: The thing inside the limit is undefined on two whole lines through $0$.
– user3482749
Nov 28 at 22:27












$x=0$ or $y=0$ or $x=y$ all have $infty$ as limits For $x=0$, limit could be $pm infty$. It looks like getting $0$ as a limit is harder.
– herb steinberg
Nov 28 at 22:30






$x=0$ or $y=0$ or $x=y$ all have $infty$ as limits For $x=0$, limit could be $pm infty$. It looks like getting $0$ as a limit is harder.
– herb steinberg
Nov 28 at 22:30












1 Answer
1






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HINT



We have that




  • $x=0 implies frac{x^2y^3}{(x^2-y^2)^2}=0$


  • $x=tquad y=t+t^2 quad tto 0^+$ $$frac{x^2y^3}{(x^2-y^2)^2}=frac{t^2(t+t^2)^3}{(t^2-(t+t^2)^2)^2}=frac{t^5+3t^6+3t^7+t^8}{(-2t^3-t^4)^2}to ?$$






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    1 Answer
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    active

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    1 Answer
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    active

    oldest

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    active

    oldest

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    active

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    up vote
    1
    down vote













    HINT



    We have that




    • $x=0 implies frac{x^2y^3}{(x^2-y^2)^2}=0$


    • $x=tquad y=t+t^2 quad tto 0^+$ $$frac{x^2y^3}{(x^2-y^2)^2}=frac{t^2(t+t^2)^3}{(t^2-(t+t^2)^2)^2}=frac{t^5+3t^6+3t^7+t^8}{(-2t^3-t^4)^2}to ?$$






    share|cite|improve this answer

























      up vote
      1
      down vote













      HINT



      We have that




      • $x=0 implies frac{x^2y^3}{(x^2-y^2)^2}=0$


      • $x=tquad y=t+t^2 quad tto 0^+$ $$frac{x^2y^3}{(x^2-y^2)^2}=frac{t^2(t+t^2)^3}{(t^2-(t+t^2)^2)^2}=frac{t^5+3t^6+3t^7+t^8}{(-2t^3-t^4)^2}to ?$$






      share|cite|improve this answer























        up vote
        1
        down vote










        up vote
        1
        down vote









        HINT



        We have that




        • $x=0 implies frac{x^2y^3}{(x^2-y^2)^2}=0$


        • $x=tquad y=t+t^2 quad tto 0^+$ $$frac{x^2y^3}{(x^2-y^2)^2}=frac{t^2(t+t^2)^3}{(t^2-(t+t^2)^2)^2}=frac{t^5+3t^6+3t^7+t^8}{(-2t^3-t^4)^2}to ?$$






        share|cite|improve this answer












        HINT



        We have that




        • $x=0 implies frac{x^2y^3}{(x^2-y^2)^2}=0$


        • $x=tquad y=t+t^2 quad tto 0^+$ $$frac{x^2y^3}{(x^2-y^2)^2}=frac{t^2(t+t^2)^3}{(t^2-(t+t^2)^2)^2}=frac{t^5+3t^6+3t^7+t^8}{(-2t^3-t^4)^2}to ?$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 at 22:32









        gimusi

        1




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