Is the function $h(x,y)=frac{x^3y^3}{x^4+y^6}$ if $(x,y)=(0,0)$ ; $h(0,0) = 0$ continuous at $(0,0)$?












0












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I think it is, because I cannot find a set such that the limit through it is different from $0$. However, polar coordinates don't simplify the way and I'm not able to bound the function. Any hints on how to proceed?






limits multivariable-calculus







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    8 minutes were enough to make you understand the post below and to be sure that it needed no correction and that no other later post would surpass it? How?
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    Jan 4 at 9:52


















0












$begingroup$


I think it is, because I cannot find a set such that the limit through it is different from $0$. However, polar coordinates don't simplify the way and I'm not able to bound the function. Any hints on how to proceed?






limits multivariable-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    8 minutes were enough to make you understand the post below and to be sure that it needed no correction and that no other later post would surpass it? How?
    $endgroup$
    – Did
    Jan 4 at 9:52
















0












0








0





$begingroup$


I think it is, because I cannot find a set such that the limit through it is different from $0$. However, polar coordinates don't simplify the way and I'm not able to bound the function. Any hints on how to proceed?






limits multivariable-calculus







share|cite|improve this question











$endgroup$




I think it is, because I cannot find a set such that the limit through it is different from $0$. However, polar coordinates don't simplify the way and I'm not able to bound the function. Any hints on how to proceed?






limits multivariable-calculus




limits multivariable-calculus






share|cite|improve this question















share|cite|improve this question













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edited Jan 4 at 9:37









José Carlos Santos

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165k22132235










asked Jan 4 at 9:28









SevenSeven

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  • $begingroup$
    8 minutes were enough to make you understand the post below and to be sure that it needed no correction and that no other later post would surpass it? How?
    $endgroup$
    – Did
    Jan 4 at 9:52




















  • $begingroup$
    8 minutes were enough to make you understand the post below and to be sure that it needed no correction and that no other later post would surpass it? How?
    $endgroup$
    – Did
    Jan 4 at 9:52


















$begingroup$
8 minutes were enough to make you understand the post below and to be sure that it needed no correction and that no other later post would surpass it? How?
$endgroup$
– Did
Jan 4 at 9:52






$begingroup$
8 minutes were enough to make you understand the post below and to be sure that it needed no correction and that no other later post would surpass it? How?
$endgroup$
– Did
Jan 4 at 9:52












2 Answers
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$begingroup$

The function is continuous at $(0,0)$. It is enough to consider $x,y>0$. [Because $(|x|,|y|) to 0$ as $(x,y) to 0$ and $|h(x,y)|=frac {|x|^{3} |y|^{3}} {|x|^{4}+|y|^{6}}$]. Since $(x^{2}-y^{3})^{2} geq 0$ we get $x^{4}+y^{6} geq 2x^{2}y^{3}$ so $ 0<frac {x^{3}y^{3}} {x^{4}+y^{6}} leqfrac {x^{3}y^{3}} {2x^{2}y{3}}=frac x 2$ and $frac x 2$ tends to $0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Several absolute values are lacking from your post. Please correct it.
    $endgroup$
    – Did
    Jan 4 at 9:51










  • $begingroup$
    @Did I did say in the second sentence that it is enough to consider $x,y >0$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 4 at 9:53










  • $begingroup$
    Bad idea. To justify that it is, takes more time and more thinking than to add absolute values.
    $endgroup$
    – Did
    Jan 4 at 9:55










  • $begingroup$
    @Did I don't agree. $h(x,y) to 0$ iff $|h(x,y)| to 0$. What is $|h(x,y)|$?
    $endgroup$
    – Kavi Rama Murthy
    Jan 4 at 9:56










  • $begingroup$
    What? Of course, $|h(x,y)|=frac{|x|^3|y|^3}{x^4+y^6}$. 1. Not obvious? 2. Your point being?
    $endgroup$
    – Did
    Jan 4 at 9:58



















1












$begingroup$

By $text{HM}letext{GM}$,



$dfrac{2}{frac{1}{x^4}+frac{1}{y^6}}lesqrt{x^4cdot y^6}=x^2cdot y^3impliesdfrac{x^4y^6}{x^4+y^6}ledfrac{x^2cdot y^3}{2}implies h(x,y)=dfrac{x^3y^3}{x^4+y^6}ledfrac{x}{2}to0text{ as }xto0$



Hence $h(x,y)$ is continuous at $(0,0).$






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    2 Answers
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    2 Answers
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    5












    $begingroup$

    The function is continuous at $(0,0)$. It is enough to consider $x,y>0$. [Because $(|x|,|y|) to 0$ as $(x,y) to 0$ and $|h(x,y)|=frac {|x|^{3} |y|^{3}} {|x|^{4}+|y|^{6}}$]. Since $(x^{2}-y^{3})^{2} geq 0$ we get $x^{4}+y^{6} geq 2x^{2}y^{3}$ so $ 0<frac {x^{3}y^{3}} {x^{4}+y^{6}} leqfrac {x^{3}y^{3}} {2x^{2}y{3}}=frac x 2$ and $frac x 2$ tends to $0$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Several absolute values are lacking from your post. Please correct it.
      $endgroup$
      – Did
      Jan 4 at 9:51










    • $begingroup$
      @Did I did say in the second sentence that it is enough to consider $x,y >0$.
      $endgroup$
      – Kavi Rama Murthy
      Jan 4 at 9:53










    • $begingroup$
      Bad idea. To justify that it is, takes more time and more thinking than to add absolute values.
      $endgroup$
      – Did
      Jan 4 at 9:55










    • $begingroup$
      @Did I don't agree. $h(x,y) to 0$ iff $|h(x,y)| to 0$. What is $|h(x,y)|$?
      $endgroup$
      – Kavi Rama Murthy
      Jan 4 at 9:56










    • $begingroup$
      What? Of course, $|h(x,y)|=frac{|x|^3|y|^3}{x^4+y^6}$. 1. Not obvious? 2. Your point being?
      $endgroup$
      – Did
      Jan 4 at 9:58
















    5












    $begingroup$

    The function is continuous at $(0,0)$. It is enough to consider $x,y>0$. [Because $(|x|,|y|) to 0$ as $(x,y) to 0$ and $|h(x,y)|=frac {|x|^{3} |y|^{3}} {|x|^{4}+|y|^{6}}$]. Since $(x^{2}-y^{3})^{2} geq 0$ we get $x^{4}+y^{6} geq 2x^{2}y^{3}$ so $ 0<frac {x^{3}y^{3}} {x^{4}+y^{6}} leqfrac {x^{3}y^{3}} {2x^{2}y{3}}=frac x 2$ and $frac x 2$ tends to $0$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Several absolute values are lacking from your post. Please correct it.
      $endgroup$
      – Did
      Jan 4 at 9:51










    • $begingroup$
      @Did I did say in the second sentence that it is enough to consider $x,y >0$.
      $endgroup$
      – Kavi Rama Murthy
      Jan 4 at 9:53










    • $begingroup$
      Bad idea. To justify that it is, takes more time and more thinking than to add absolute values.
      $endgroup$
      – Did
      Jan 4 at 9:55










    • $begingroup$
      @Did I don't agree. $h(x,y) to 0$ iff $|h(x,y)| to 0$. What is $|h(x,y)|$?
      $endgroup$
      – Kavi Rama Murthy
      Jan 4 at 9:56










    • $begingroup$
      What? Of course, $|h(x,y)|=frac{|x|^3|y|^3}{x^4+y^6}$. 1. Not obvious? 2. Your point being?
      $endgroup$
      – Did
      Jan 4 at 9:58














    5












    5








    5





    $begingroup$

    The function is continuous at $(0,0)$. It is enough to consider $x,y>0$. [Because $(|x|,|y|) to 0$ as $(x,y) to 0$ and $|h(x,y)|=frac {|x|^{3} |y|^{3}} {|x|^{4}+|y|^{6}}$]. Since $(x^{2}-y^{3})^{2} geq 0$ we get $x^{4}+y^{6} geq 2x^{2}y^{3}$ so $ 0<frac {x^{3}y^{3}} {x^{4}+y^{6}} leqfrac {x^{3}y^{3}} {2x^{2}y{3}}=frac x 2$ and $frac x 2$ tends to $0$.






    share|cite|improve this answer











    $endgroup$



    The function is continuous at $(0,0)$. It is enough to consider $x,y>0$. [Because $(|x|,|y|) to 0$ as $(x,y) to 0$ and $|h(x,y)|=frac {|x|^{3} |y|^{3}} {|x|^{4}+|y|^{6}}$]. Since $(x^{2}-y^{3})^{2} geq 0$ we get $x^{4}+y^{6} geq 2x^{2}y^{3}$ so $ 0<frac {x^{3}y^{3}} {x^{4}+y^{6}} leqfrac {x^{3}y^{3}} {2x^{2}y{3}}=frac x 2$ and $frac x 2$ tends to $0$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 4 at 10:02

























    answered Jan 4 at 9:40









    Kavi Rama MurthyKavi Rama Murthy

    64.5k42665




    64.5k42665












    • $begingroup$
      Several absolute values are lacking from your post. Please correct it.
      $endgroup$
      – Did
      Jan 4 at 9:51










    • $begingroup$
      @Did I did say in the second sentence that it is enough to consider $x,y >0$.
      $endgroup$
      – Kavi Rama Murthy
      Jan 4 at 9:53










    • $begingroup$
      Bad idea. To justify that it is, takes more time and more thinking than to add absolute values.
      $endgroup$
      – Did
      Jan 4 at 9:55










    • $begingroup$
      @Did I don't agree. $h(x,y) to 0$ iff $|h(x,y)| to 0$. What is $|h(x,y)|$?
      $endgroup$
      – Kavi Rama Murthy
      Jan 4 at 9:56










    • $begingroup$
      What? Of course, $|h(x,y)|=frac{|x|^3|y|^3}{x^4+y^6}$. 1. Not obvious? 2. Your point being?
      $endgroup$
      – Did
      Jan 4 at 9:58


















    • $begingroup$
      Several absolute values are lacking from your post. Please correct it.
      $endgroup$
      – Did
      Jan 4 at 9:51










    • $begingroup$
      @Did I did say in the second sentence that it is enough to consider $x,y >0$.
      $endgroup$
      – Kavi Rama Murthy
      Jan 4 at 9:53










    • $begingroup$
      Bad idea. To justify that it is, takes more time and more thinking than to add absolute values.
      $endgroup$
      – Did
      Jan 4 at 9:55










    • $begingroup$
      @Did I don't agree. $h(x,y) to 0$ iff $|h(x,y)| to 0$. What is $|h(x,y)|$?
      $endgroup$
      – Kavi Rama Murthy
      Jan 4 at 9:56










    • $begingroup$
      What? Of course, $|h(x,y)|=frac{|x|^3|y|^3}{x^4+y^6}$. 1. Not obvious? 2. Your point being?
      $endgroup$
      – Did
      Jan 4 at 9:58
















    $begingroup$
    Several absolute values are lacking from your post. Please correct it.
    $endgroup$
    – Did
    Jan 4 at 9:51




    $begingroup$
    Several absolute values are lacking from your post. Please correct it.
    $endgroup$
    – Did
    Jan 4 at 9:51












    $begingroup$
    @Did I did say in the second sentence that it is enough to consider $x,y >0$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 4 at 9:53




    $begingroup$
    @Did I did say in the second sentence that it is enough to consider $x,y >0$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 4 at 9:53












    $begingroup$
    Bad idea. To justify that it is, takes more time and more thinking than to add absolute values.
    $endgroup$
    – Did
    Jan 4 at 9:55




    $begingroup$
    Bad idea. To justify that it is, takes more time and more thinking than to add absolute values.
    $endgroup$
    – Did
    Jan 4 at 9:55












    $begingroup$
    @Did I don't agree. $h(x,y) to 0$ iff $|h(x,y)| to 0$. What is $|h(x,y)|$?
    $endgroup$
    – Kavi Rama Murthy
    Jan 4 at 9:56




    $begingroup$
    @Did I don't agree. $h(x,y) to 0$ iff $|h(x,y)| to 0$. What is $|h(x,y)|$?
    $endgroup$
    – Kavi Rama Murthy
    Jan 4 at 9:56












    $begingroup$
    What? Of course, $|h(x,y)|=frac{|x|^3|y|^3}{x^4+y^6}$. 1. Not obvious? 2. Your point being?
    $endgroup$
    – Did
    Jan 4 at 9:58




    $begingroup$
    What? Of course, $|h(x,y)|=frac{|x|^3|y|^3}{x^4+y^6}$. 1. Not obvious? 2. Your point being?
    $endgroup$
    – Did
    Jan 4 at 9:58











    1












    $begingroup$

    By $text{HM}letext{GM}$,



    $dfrac{2}{frac{1}{x^4}+frac{1}{y^6}}lesqrt{x^4cdot y^6}=x^2cdot y^3impliesdfrac{x^4y^6}{x^4+y^6}ledfrac{x^2cdot y^3}{2}implies h(x,y)=dfrac{x^3y^3}{x^4+y^6}ledfrac{x}{2}to0text{ as }xto0$



    Hence $h(x,y)$ is continuous at $(0,0).$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      By $text{HM}letext{GM}$,



      $dfrac{2}{frac{1}{x^4}+frac{1}{y^6}}lesqrt{x^4cdot y^6}=x^2cdot y^3impliesdfrac{x^4y^6}{x^4+y^6}ledfrac{x^2cdot y^3}{2}implies h(x,y)=dfrac{x^3y^3}{x^4+y^6}ledfrac{x}{2}to0text{ as }xto0$



      Hence $h(x,y)$ is continuous at $(0,0).$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        By $text{HM}letext{GM}$,



        $dfrac{2}{frac{1}{x^4}+frac{1}{y^6}}lesqrt{x^4cdot y^6}=x^2cdot y^3impliesdfrac{x^4y^6}{x^4+y^6}ledfrac{x^2cdot y^3}{2}implies h(x,y)=dfrac{x^3y^3}{x^4+y^6}ledfrac{x}{2}to0text{ as }xto0$



        Hence $h(x,y)$ is continuous at $(0,0).$






        share|cite|improve this answer











        $endgroup$



        By $text{HM}letext{GM}$,



        $dfrac{2}{frac{1}{x^4}+frac{1}{y^6}}lesqrt{x^4cdot y^6}=x^2cdot y^3impliesdfrac{x^4y^6}{x^4+y^6}ledfrac{x^2cdot y^3}{2}implies h(x,y)=dfrac{x^3y^3}{x^4+y^6}ledfrac{x}{2}to0text{ as }xto0$



        Hence $h(x,y)$ is continuous at $(0,0).$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 4 at 10:15

























        answered Jan 4 at 10:00









        Arjun BanerjeeArjun Banerjee

        53610




        53610






























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