How to prove $lim_{(x,y) to (0,0)} frac{xy}{x^4+y^4}$ does not exist












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$$lim_{(x,y)to(0,0)} frac{xy}{x^4+y^4}$$



Anyone can teach me how to solve this question? I have tried so many times but still unable to solve it.






calculus limits







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  • 3




    $begingroup$
    Work out what happens as $xto0$ with $y$ held fixed at zero. Then work out what happens as you approach the origin along the curve $y=x^4$. Then write up what you have found, and post it as an answer.
    $endgroup$
    – Gerry Myerson
    Dec 2 '18 at 3:39










  • $begingroup$
    Try to choose a suitable path depending on some variable constant.
    $endgroup$
    – Anik Bhowmick
    Dec 2 '18 at 4:00










  • $begingroup$
    Well the fact that the expression has transposable variables (i.e. $f(x,y) equiv f(y,x)$) means that you just need to find $lim_{xto 0} f(x,x)$, or more precisely: $$lim_{xto 0}{2x^{-2}}$$ I think
    $endgroup$
    – Rhys Hughes
    Dec 2 '18 at 4:21


















1












$begingroup$


$$lim_{(x,y)to(0,0)} frac{xy}{x^4+y^4}$$



Anyone can teach me how to solve this question? I have tried so many times but still unable to solve it.






calculus limits







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Work out what happens as $xto0$ with $y$ held fixed at zero. Then work out what happens as you approach the origin along the curve $y=x^4$. Then write up what you have found, and post it as an answer.
    $endgroup$
    – Gerry Myerson
    Dec 2 '18 at 3:39










  • $begingroup$
    Try to choose a suitable path depending on some variable constant.
    $endgroup$
    – Anik Bhowmick
    Dec 2 '18 at 4:00










  • $begingroup$
    Well the fact that the expression has transposable variables (i.e. $f(x,y) equiv f(y,x)$) means that you just need to find $lim_{xto 0} f(x,x)$, or more precisely: $$lim_{xto 0}{2x^{-2}}$$ I think
    $endgroup$
    – Rhys Hughes
    Dec 2 '18 at 4:21
















1












1








1





$begingroup$


$$lim_{(x,y)to(0,0)} frac{xy}{x^4+y^4}$$



Anyone can teach me how to solve this question? I have tried so many times but still unable to solve it.






calculus limits







share|cite|improve this question











$endgroup$




$$lim_{(x,y)to(0,0)} frac{xy}{x^4+y^4}$$



Anyone can teach me how to solve this question? I have tried so many times but still unable to solve it.






calculus limits




calculus limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 4:20









Tianlalu

3,08121038




3,08121038










asked Dec 2 '18 at 3:35









AlvinAlvin

61




61








  • 3




    $begingroup$
    Work out what happens as $xto0$ with $y$ held fixed at zero. Then work out what happens as you approach the origin along the curve $y=x^4$. Then write up what you have found, and post it as an answer.
    $endgroup$
    – Gerry Myerson
    Dec 2 '18 at 3:39










  • $begingroup$
    Try to choose a suitable path depending on some variable constant.
    $endgroup$
    – Anik Bhowmick
    Dec 2 '18 at 4:00










  • $begingroup$
    Well the fact that the expression has transposable variables (i.e. $f(x,y) equiv f(y,x)$) means that you just need to find $lim_{xto 0} f(x,x)$, or more precisely: $$lim_{xto 0}{2x^{-2}}$$ I think
    $endgroup$
    – Rhys Hughes
    Dec 2 '18 at 4:21
















  • 3




    $begingroup$
    Work out what happens as $xto0$ with $y$ held fixed at zero. Then work out what happens as you approach the origin along the curve $y=x^4$. Then write up what you have found, and post it as an answer.
    $endgroup$
    – Gerry Myerson
    Dec 2 '18 at 3:39










  • $begingroup$
    Try to choose a suitable path depending on some variable constant.
    $endgroup$
    – Anik Bhowmick
    Dec 2 '18 at 4:00










  • $begingroup$
    Well the fact that the expression has transposable variables (i.e. $f(x,y) equiv f(y,x)$) means that you just need to find $lim_{xto 0} f(x,x)$, or more precisely: $$lim_{xto 0}{2x^{-2}}$$ I think
    $endgroup$
    – Rhys Hughes
    Dec 2 '18 at 4:21










3




3




$begingroup$
Work out what happens as $xto0$ with $y$ held fixed at zero. Then work out what happens as you approach the origin along the curve $y=x^4$. Then write up what you have found, and post it as an answer.
$endgroup$
– Gerry Myerson
Dec 2 '18 at 3:39




$begingroup$
Work out what happens as $xto0$ with $y$ held fixed at zero. Then work out what happens as you approach the origin along the curve $y=x^4$. Then write up what you have found, and post it as an answer.
$endgroup$
– Gerry Myerson
Dec 2 '18 at 3:39












$begingroup$
Try to choose a suitable path depending on some variable constant.
$endgroup$
– Anik Bhowmick
Dec 2 '18 at 4:00




$begingroup$
Try to choose a suitable path depending on some variable constant.
$endgroup$
– Anik Bhowmick
Dec 2 '18 at 4:00












$begingroup$
Well the fact that the expression has transposable variables (i.e. $f(x,y) equiv f(y,x)$) means that you just need to find $lim_{xto 0} f(x,x)$, or more precisely: $$lim_{xto 0}{2x^{-2}}$$ I think
$endgroup$
– Rhys Hughes
Dec 2 '18 at 4:21






$begingroup$
Well the fact that the expression has transposable variables (i.e. $f(x,y) equiv f(y,x)$) means that you just need to find $lim_{xto 0} f(x,x)$, or more precisely: $$lim_{xto 0}{2x^{-2}}$$ I think
$endgroup$
– Rhys Hughes
Dec 2 '18 at 4:21












1 Answer
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Suppose the limit $lim_{(x,y)to(0,0)} frac{xy}{x^4+y^4}$ exits then in some deleted nbd of $(0,0)$ the function $f:Bbb R^2-{(0,0)}rightarrow Bbb R,f(x,y)=frac{xy}{x^4+y^4}$ will be bounded . But notice that on the set $S:={(x,y)in Bbb R^2-{(0,0)}: x=y}$ the function $f$ is of the form $frac{x^2}{2x^4}=frac{1}{2x^2}$ i.e. in every deleted nbd of $(0,0)$ the function is unbounded.



$lim_{(x,y)rightarrow (0,0)} f(x,y)=limplies$ for $epsilon=1$ there is a $delta>0$ such that $0<x^2+y^2<deltaimplies |f(x,y)-l|<1implies l-1<f(x,y)<l+1 , forall (x,y)$ with $0<x^2+y^2<delta$ i.e. $f$ is bounded in a deleted nbd of $(0,0)$.






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    $begingroup$

    Suppose the limit $lim_{(x,y)to(0,0)} frac{xy}{x^4+y^4}$ exits then in some deleted nbd of $(0,0)$ the function $f:Bbb R^2-{(0,0)}rightarrow Bbb R,f(x,y)=frac{xy}{x^4+y^4}$ will be bounded . But notice that on the set $S:={(x,y)in Bbb R^2-{(0,0)}: x=y}$ the function $f$ is of the form $frac{x^2}{2x^4}=frac{1}{2x^2}$ i.e. in every deleted nbd of $(0,0)$ the function is unbounded.



    $lim_{(x,y)rightarrow (0,0)} f(x,y)=limplies$ for $epsilon=1$ there is a $delta>0$ such that $0<x^2+y^2<deltaimplies |f(x,y)-l|<1implies l-1<f(x,y)<l+1 , forall (x,y)$ with $0<x^2+y^2<delta$ i.e. $f$ is bounded in a deleted nbd of $(0,0)$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Suppose the limit $lim_{(x,y)to(0,0)} frac{xy}{x^4+y^4}$ exits then in some deleted nbd of $(0,0)$ the function $f:Bbb R^2-{(0,0)}rightarrow Bbb R,f(x,y)=frac{xy}{x^4+y^4}$ will be bounded . But notice that on the set $S:={(x,y)in Bbb R^2-{(0,0)}: x=y}$ the function $f$ is of the form $frac{x^2}{2x^4}=frac{1}{2x^2}$ i.e. in every deleted nbd of $(0,0)$ the function is unbounded.



      $lim_{(x,y)rightarrow (0,0)} f(x,y)=limplies$ for $epsilon=1$ there is a $delta>0$ such that $0<x^2+y^2<deltaimplies |f(x,y)-l|<1implies l-1<f(x,y)<l+1 , forall (x,y)$ with $0<x^2+y^2<delta$ i.e. $f$ is bounded in a deleted nbd of $(0,0)$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Suppose the limit $lim_{(x,y)to(0,0)} frac{xy}{x^4+y^4}$ exits then in some deleted nbd of $(0,0)$ the function $f:Bbb R^2-{(0,0)}rightarrow Bbb R,f(x,y)=frac{xy}{x^4+y^4}$ will be bounded . But notice that on the set $S:={(x,y)in Bbb R^2-{(0,0)}: x=y}$ the function $f$ is of the form $frac{x^2}{2x^4}=frac{1}{2x^2}$ i.e. in every deleted nbd of $(0,0)$ the function is unbounded.



        $lim_{(x,y)rightarrow (0,0)} f(x,y)=limplies$ for $epsilon=1$ there is a $delta>0$ such that $0<x^2+y^2<deltaimplies |f(x,y)-l|<1implies l-1<f(x,y)<l+1 , forall (x,y)$ with $0<x^2+y^2<delta$ i.e. $f$ is bounded in a deleted nbd of $(0,0)$.






        share|cite|improve this answer











        $endgroup$



        Suppose the limit $lim_{(x,y)to(0,0)} frac{xy}{x^4+y^4}$ exits then in some deleted nbd of $(0,0)$ the function $f:Bbb R^2-{(0,0)}rightarrow Bbb R,f(x,y)=frac{xy}{x^4+y^4}$ will be bounded . But notice that on the set $S:={(x,y)in Bbb R^2-{(0,0)}: x=y}$ the function $f$ is of the form $frac{x^2}{2x^4}=frac{1}{2x^2}$ i.e. in every deleted nbd of $(0,0)$ the function is unbounded.



        $lim_{(x,y)rightarrow (0,0)} f(x,y)=limplies$ for $epsilon=1$ there is a $delta>0$ such that $0<x^2+y^2<deltaimplies |f(x,y)-l|<1implies l-1<f(x,y)<l+1 , forall (x,y)$ with $0<x^2+y^2<delta$ i.e. $f$ is bounded in a deleted nbd of $(0,0)$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 2 '18 at 4:45

























        answered Dec 2 '18 at 4:35









        UserSUserS

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