Limit of a real valued function of two variables.












-1














$lim_{(x,y)to(0,0)} frac {x²y}{x⁴+y²} = $?
If we take $y = mx^2$,
it comes out to be $frac {m^2}{1+m^2}$ which depends on $m$, so limit doesn't exist.



How to prove the non existence of limit by using polar form
What I tried –



$lim_{rto 0}frac{ r³cos²(t)sin(t)}{r³cos²(t) + rsin(t)}$



I don't know how to proceed.



I know that limit exists if the result I get is independent of $t$.






calculus







share|cite|improve this question





























    -1














    $lim_{(x,y)to(0,0)} frac {x²y}{x⁴+y²} = $?
    If we take $y = mx^2$,
    it comes out to be $frac {m^2}{1+m^2}$ which depends on $m$, so limit doesn't exist.



    How to prove the non existence of limit by using polar form
    What I tried –



    $lim_{rto 0}frac{ r³cos²(t)sin(t)}{r³cos²(t) + rsin(t)}$



    I don't know how to proceed.



    I know that limit exists if the result I get is independent of $t$.






    calculus







    share|cite|improve this question



























      -1












      -1








      -1







      $lim_{(x,y)to(0,0)} frac {x²y}{x⁴+y²} = $?
      If we take $y = mx^2$,
      it comes out to be $frac {m^2}{1+m^2}$ which depends on $m$, so limit doesn't exist.



      How to prove the non existence of limit by using polar form
      What I tried –



      $lim_{rto 0}frac{ r³cos²(t)sin(t)}{r³cos²(t) + rsin(t)}$



      I don't know how to proceed.



      I know that limit exists if the result I get is independent of $t$.






      calculus







      share|cite|improve this question















      $lim_{(x,y)to(0,0)} frac {x²y}{x⁴+y²} = $?
      If we take $y = mx^2$,
      it comes out to be $frac {m^2}{1+m^2}$ which depends on $m$, so limit doesn't exist.



      How to prove the non existence of limit by using polar form
      What I tried –



      $lim_{rto 0}frac{ r³cos²(t)sin(t)}{r³cos²(t) + rsin(t)}$



      I don't know how to proceed.



      I know that limit exists if the result I get is independent of $t$.






      calculus




      calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 29 '18 at 4:24









      Thomas Shelby

      1,667216




      1,667216










      asked Nov 29 '18 at 3:20









      Mathsaddict

      2608




      2608






















          1 Answer
          1






          active

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          2














          Polar form? Don't go polar in terms of $x$ and $y$; substitute $u=x^2$ first. Then it's $lim_{(u,y)to (0,0),uge 0} frac{uy}{u^2+y^2}$. Convert that to polar coordinates (the $uge 0$ restriction becomes an angle restriction $-frac{pi}{2}le thetale frac{pi}{2}$) and work from there.



          After all, the point of polar coordinates in this sort of question is to make the denominator nice.






          share|cite|improve this answer





















          • I did as you explained, and I got sin(t)cos(t), which depends on t. But, without converting x^2 to u, why do i get 0 in polar form? Limit should be same.
            – Mathsaddict
            Nov 29 '18 at 8:07












          • The limit along any straight-line path in x and y is zero. Converting to polar right away and looking at constant angles looks only at these straight-line paths - but you already noted that following a parabolic path gets you something else. It's just not enough on its own.
            – jmerry
            Nov 29 '18 at 8:28











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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2














          Polar form? Don't go polar in terms of $x$ and $y$; substitute $u=x^2$ first. Then it's $lim_{(u,y)to (0,0),uge 0} frac{uy}{u^2+y^2}$. Convert that to polar coordinates (the $uge 0$ restriction becomes an angle restriction $-frac{pi}{2}le thetale frac{pi}{2}$) and work from there.



          After all, the point of polar coordinates in this sort of question is to make the denominator nice.






          share|cite|improve this answer





















          • I did as you explained, and I got sin(t)cos(t), which depends on t. But, without converting x^2 to u, why do i get 0 in polar form? Limit should be same.
            – Mathsaddict
            Nov 29 '18 at 8:07












          • The limit along any straight-line path in x and y is zero. Converting to polar right away and looking at constant angles looks only at these straight-line paths - but you already noted that following a parabolic path gets you something else. It's just not enough on its own.
            – jmerry
            Nov 29 '18 at 8:28
















          2














          Polar form? Don't go polar in terms of $x$ and $y$; substitute $u=x^2$ first. Then it's $lim_{(u,y)to (0,0),uge 0} frac{uy}{u^2+y^2}$. Convert that to polar coordinates (the $uge 0$ restriction becomes an angle restriction $-frac{pi}{2}le thetale frac{pi}{2}$) and work from there.



          After all, the point of polar coordinates in this sort of question is to make the denominator nice.






          share|cite|improve this answer





















          • I did as you explained, and I got sin(t)cos(t), which depends on t. But, without converting x^2 to u, why do i get 0 in polar form? Limit should be same.
            – Mathsaddict
            Nov 29 '18 at 8:07












          • The limit along any straight-line path in x and y is zero. Converting to polar right away and looking at constant angles looks only at these straight-line paths - but you already noted that following a parabolic path gets you something else. It's just not enough on its own.
            – jmerry
            Nov 29 '18 at 8:28














          2












          2








          2






          Polar form? Don't go polar in terms of $x$ and $y$; substitute $u=x^2$ first. Then it's $lim_{(u,y)to (0,0),uge 0} frac{uy}{u^2+y^2}$. Convert that to polar coordinates (the $uge 0$ restriction becomes an angle restriction $-frac{pi}{2}le thetale frac{pi}{2}$) and work from there.



          After all, the point of polar coordinates in this sort of question is to make the denominator nice.






          share|cite|improve this answer












          Polar form? Don't go polar in terms of $x$ and $y$; substitute $u=x^2$ first. Then it's $lim_{(u,y)to (0,0),uge 0} frac{uy}{u^2+y^2}$. Convert that to polar coordinates (the $uge 0$ restriction becomes an angle restriction $-frac{pi}{2}le thetale frac{pi}{2}$) and work from there.



          After all, the point of polar coordinates in this sort of question is to make the denominator nice.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 29 '18 at 6:52









          jmerry

          2,076210




          2,076210












          • I did as you explained, and I got sin(t)cos(t), which depends on t. But, without converting x^2 to u, why do i get 0 in polar form? Limit should be same.
            – Mathsaddict
            Nov 29 '18 at 8:07












          • The limit along any straight-line path in x and y is zero. Converting to polar right away and looking at constant angles looks only at these straight-line paths - but you already noted that following a parabolic path gets you something else. It's just not enough on its own.
            – jmerry
            Nov 29 '18 at 8:28


















          • I did as you explained, and I got sin(t)cos(t), which depends on t. But, without converting x^2 to u, why do i get 0 in polar form? Limit should be same.
            – Mathsaddict
            Nov 29 '18 at 8:07












          • The limit along any straight-line path in x and y is zero. Converting to polar right away and looking at constant angles looks only at these straight-line paths - but you already noted that following a parabolic path gets you something else. It's just not enough on its own.
            – jmerry
            Nov 29 '18 at 8:28
















          I did as you explained, and I got sin(t)cos(t), which depends on t. But, without converting x^2 to u, why do i get 0 in polar form? Limit should be same.
          – Mathsaddict
          Nov 29 '18 at 8:07






          I did as you explained, and I got sin(t)cos(t), which depends on t. But, without converting x^2 to u, why do i get 0 in polar form? Limit should be same.
          – Mathsaddict
          Nov 29 '18 at 8:07














          The limit along any straight-line path in x and y is zero. Converting to polar right away and looking at constant angles looks only at these straight-line paths - but you already noted that following a parabolic path gets you something else. It's just not enough on its own.
          – jmerry
          Nov 29 '18 at 8:28




          The limit along any straight-line path in x and y is zero. Converting to polar right away and looking at constant angles looks only at these straight-line paths - but you already noted that following a parabolic path gets you something else. It's just not enough on its own.
          – jmerry
          Nov 29 '18 at 8:28


















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