Can I prove that a 2-variable limit does not exists if the limit on a curve is infinity?












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Consider a 2 variable function $f(x,y)$ and the limit



$$lim_{(x,y)to (0,0)} f(x,y)$$



If I find two continuous functions $gamma_1(t)$ and $gamma_2(t)$ such that $gamma_1(0)=gamma_2(0)=(0,0)$ and



$$lim_{tto 0} f(gamma_1(t))=l in mathbb{R} ,,,,,,,,,,, and ,,,,,,,,,, lim_{tto 0} f(gamma_2(t))=+infty $$



Can I conclude that the original limit $lim_{(x,y)to (0,0)} f(x,y)$ does not exist?



I'm not sure about this because of the infinity in one of the two limits above










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  • 2




    $begingroup$
    yes of course it doesn't exist !
    $endgroup$
    – Surb
    Dec 6 '18 at 22:36
















1












$begingroup$


Consider a 2 variable function $f(x,y)$ and the limit



$$lim_{(x,y)to (0,0)} f(x,y)$$



If I find two continuous functions $gamma_1(t)$ and $gamma_2(t)$ such that $gamma_1(0)=gamma_2(0)=(0,0)$ and



$$lim_{tto 0} f(gamma_1(t))=l in mathbb{R} ,,,,,,,,,,, and ,,,,,,,,,, lim_{tto 0} f(gamma_2(t))=+infty $$



Can I conclude that the original limit $lim_{(x,y)to (0,0)} f(x,y)$ does not exist?



I'm not sure about this because of the infinity in one of the two limits above










share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    yes of course it doesn't exist !
    $endgroup$
    – Surb
    Dec 6 '18 at 22:36














1












1








1





$begingroup$


Consider a 2 variable function $f(x,y)$ and the limit



$$lim_{(x,y)to (0,0)} f(x,y)$$



If I find two continuous functions $gamma_1(t)$ and $gamma_2(t)$ such that $gamma_1(0)=gamma_2(0)=(0,0)$ and



$$lim_{tto 0} f(gamma_1(t))=l in mathbb{R} ,,,,,,,,,,, and ,,,,,,,,,, lim_{tto 0} f(gamma_2(t))=+infty $$



Can I conclude that the original limit $lim_{(x,y)to (0,0)} f(x,y)$ does not exist?



I'm not sure about this because of the infinity in one of the two limits above










share|cite|improve this question









$endgroup$




Consider a 2 variable function $f(x,y)$ and the limit



$$lim_{(x,y)to (0,0)} f(x,y)$$



If I find two continuous functions $gamma_1(t)$ and $gamma_2(t)$ such that $gamma_1(0)=gamma_2(0)=(0,0)$ and



$$lim_{tto 0} f(gamma_1(t))=l in mathbb{R} ,,,,,,,,,,, and ,,,,,,,,,, lim_{tto 0} f(gamma_2(t))=+infty $$



Can I conclude that the original limit $lim_{(x,y)to (0,0)} f(x,y)$ does not exist?



I'm not sure about this because of the infinity in one of the two limits above







calculus limits multivariable-calculus infinity






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asked Dec 6 '18 at 22:34









GianolepoGianolepo

770918




770918








  • 2




    $begingroup$
    yes of course it doesn't exist !
    $endgroup$
    – Surb
    Dec 6 '18 at 22:36














  • 2




    $begingroup$
    yes of course it doesn't exist !
    $endgroup$
    – Surb
    Dec 6 '18 at 22:36








2




2




$begingroup$
yes of course it doesn't exist !
$endgroup$
– Surb
Dec 6 '18 at 22:36




$begingroup$
yes of course it doesn't exist !
$endgroup$
– Surb
Dec 6 '18 at 22:36










1 Answer
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$begingroup$

Yes of course it suffices to find, at least, two different limits finite or infinite, for different paths as for example for



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



we have




  • $frac{xy}{x^3-y^3} to 0$ for $x=0$


  • $frac{xy}{x^3-y^3} to +infty$ for $y=-x=t to 0^+$







share|cite|improve this answer











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    $begingroup$

    Yes of course it suffices to find, at least, two different limits finite or infinite, for different paths as for example for



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



    we have




    • $frac{xy}{x^3-y^3} to 0$ for $x=0$


    • $frac{xy}{x^3-y^3} to +infty$ for $y=-x=t to 0^+$







    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Yes of course it suffices to find, at least, two different limits finite or infinite, for different paths as for example for



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



      we have




      • $frac{xy}{x^3-y^3} to 0$ for $x=0$


      • $frac{xy}{x^3-y^3} to +infty$ for $y=-x=t to 0^+$







      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Yes of course it suffices to find, at least, two different limits finite or infinite, for different paths as for example for



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



        we have




        • $frac{xy}{x^3-y^3} to 0$ for $x=0$


        • $frac{xy}{x^3-y^3} to +infty$ for $y=-x=t to 0^+$







        share|cite|improve this answer











        $endgroup$



        Yes of course it suffices to find, at least, two different limits finite or infinite, for different paths as for example for



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



        we have




        • $frac{xy}{x^3-y^3} to 0$ for $x=0$


        • $frac{xy}{x^3-y^3} to +infty$ for $y=-x=t to 0^+$








        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 6 '18 at 23:04

























        answered Dec 6 '18 at 22:53









        gimusigimusi

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