Existence of the Limit of a Two-Variable Function












2












$begingroup$


This problem is an example in my calculus textbook. Let:
$$
f(x,y)=frac{y^2sin^2(x)}{x^4+y^4}
$$
My textbook says that
$$
lim_{(x,y)to(0,0)}f(x,y) text{ does not exist.}
$$
Questions:




  1. How do we know the limit does not exist?


  2. In general, suppose the limit of a function exists but we do not know the value of such limit, how do we find it?







limits multivariable-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Typically, to prove that the limit does not exist, people find distinct sublimits. If you know the limit exists and just want to find it, then you just need to consider an appropriate (read easy) sublimit.
    $endgroup$
    – Git Gud
    Jan 18 at 0:49










  • $begingroup$
    @GitGud Does this mean that proving a limit does not exist, as in this question, might be very difficult if we cannot identify the proper sublimits, which more or less boils down to trial & error?
    $endgroup$
    – A Slow Learner
    Jan 18 at 0:59










  • $begingroup$
    There are strategies you can use. For rational functions in two variables you can try looking at sublimits of the form $tmapsto (t, kt)$, $tmapsto (t, kt^2)$, $tmapsto (t, kt^3)$ and so on, with varying values of $k$. If you're unable to find different sublimits, then perhaps the sublimit exists and you can try to focus on proving it does exist. When you don't have a rational function, you can probably use Taylor's theorem to get a similarly behaved rational function.
    $endgroup$
    – Git Gud
    Jan 18 at 8:38
















2












$begingroup$


This problem is an example in my calculus textbook. Let:
$$
f(x,y)=frac{y^2sin^2(x)}{x^4+y^4}
$$
My textbook says that
$$
lim_{(x,y)to(0,0)}f(x,y) text{ does not exist.}
$$
Questions:




  1. How do we know the limit does not exist?


  2. In general, suppose the limit of a function exists but we do not know the value of such limit, how do we find it?







limits multivariable-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Typically, to prove that the limit does not exist, people find distinct sublimits. If you know the limit exists and just want to find it, then you just need to consider an appropriate (read easy) sublimit.
    $endgroup$
    – Git Gud
    Jan 18 at 0:49










  • $begingroup$
    @GitGud Does this mean that proving a limit does not exist, as in this question, might be very difficult if we cannot identify the proper sublimits, which more or less boils down to trial & error?
    $endgroup$
    – A Slow Learner
    Jan 18 at 0:59










  • $begingroup$
    There are strategies you can use. For rational functions in two variables you can try looking at sublimits of the form $tmapsto (t, kt)$, $tmapsto (t, kt^2)$, $tmapsto (t, kt^3)$ and so on, with varying values of $k$. If you're unable to find different sublimits, then perhaps the sublimit exists and you can try to focus on proving it does exist. When you don't have a rational function, you can probably use Taylor's theorem to get a similarly behaved rational function.
    $endgroup$
    – Git Gud
    Jan 18 at 8:38














2












2








2





$begingroup$


This problem is an example in my calculus textbook. Let:
$$
f(x,y)=frac{y^2sin^2(x)}{x^4+y^4}
$$
My textbook says that
$$
lim_{(x,y)to(0,0)}f(x,y) text{ does not exist.}
$$
Questions:




  1. How do we know the limit does not exist?


  2. In general, suppose the limit of a function exists but we do not know the value of such limit, how do we find it?







limits multivariable-calculus







share|cite|improve this question











$endgroup$




This problem is an example in my calculus textbook. Let:
$$
f(x,y)=frac{y^2sin^2(x)}{x^4+y^4}
$$
My textbook says that
$$
lim_{(x,y)to(0,0)}f(x,y) text{ does not exist.}
$$
Questions:




  1. How do we know the limit does not exist?


  2. In general, suppose the limit of a function exists but we do not know the value of such limit, how do we find it?







limits multivariable-calculus




limits multivariable-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 0:51









Git Gud

28.9k1050101




28.9k1050101










asked Jan 18 at 0:37









A Slow LearnerA Slow Learner

468313




468313












  • $begingroup$
    Typically, to prove that the limit does not exist, people find distinct sublimits. If you know the limit exists and just want to find it, then you just need to consider an appropriate (read easy) sublimit.
    $endgroup$
    – Git Gud
    Jan 18 at 0:49










  • $begingroup$
    @GitGud Does this mean that proving a limit does not exist, as in this question, might be very difficult if we cannot identify the proper sublimits, which more or less boils down to trial & error?
    $endgroup$
    – A Slow Learner
    Jan 18 at 0:59










  • $begingroup$
    There are strategies you can use. For rational functions in two variables you can try looking at sublimits of the form $tmapsto (t, kt)$, $tmapsto (t, kt^2)$, $tmapsto (t, kt^3)$ and so on, with varying values of $k$. If you're unable to find different sublimits, then perhaps the sublimit exists and you can try to focus on proving it does exist. When you don't have a rational function, you can probably use Taylor's theorem to get a similarly behaved rational function.
    $endgroup$
    – Git Gud
    Jan 18 at 8:38


















  • $begingroup$
    Typically, to prove that the limit does not exist, people find distinct sublimits. If you know the limit exists and just want to find it, then you just need to consider an appropriate (read easy) sublimit.
    $endgroup$
    – Git Gud
    Jan 18 at 0:49










  • $begingroup$
    @GitGud Does this mean that proving a limit does not exist, as in this question, might be very difficult if we cannot identify the proper sublimits, which more or less boils down to trial & error?
    $endgroup$
    – A Slow Learner
    Jan 18 at 0:59










  • $begingroup$
    There are strategies you can use. For rational functions in two variables you can try looking at sublimits of the form $tmapsto (t, kt)$, $tmapsto (t, kt^2)$, $tmapsto (t, kt^3)$ and so on, with varying values of $k$. If you're unable to find different sublimits, then perhaps the sublimit exists and you can try to focus on proving it does exist. When you don't have a rational function, you can probably use Taylor's theorem to get a similarly behaved rational function.
    $endgroup$
    – Git Gud
    Jan 18 at 8:38
















$begingroup$
Typically, to prove that the limit does not exist, people find distinct sublimits. If you know the limit exists and just want to find it, then you just need to consider an appropriate (read easy) sublimit.
$endgroup$
– Git Gud
Jan 18 at 0:49




$begingroup$
Typically, to prove that the limit does not exist, people find distinct sublimits. If you know the limit exists and just want to find it, then you just need to consider an appropriate (read easy) sublimit.
$endgroup$
– Git Gud
Jan 18 at 0:49












$begingroup$
@GitGud Does this mean that proving a limit does not exist, as in this question, might be very difficult if we cannot identify the proper sublimits, which more or less boils down to trial & error?
$endgroup$
– A Slow Learner
Jan 18 at 0:59




$begingroup$
@GitGud Does this mean that proving a limit does not exist, as in this question, might be very difficult if we cannot identify the proper sublimits, which more or less boils down to trial & error?
$endgroup$
– A Slow Learner
Jan 18 at 0:59












$begingroup$
There are strategies you can use. For rational functions in two variables you can try looking at sublimits of the form $tmapsto (t, kt)$, $tmapsto (t, kt^2)$, $tmapsto (t, kt^3)$ and so on, with varying values of $k$. If you're unable to find different sublimits, then perhaps the sublimit exists and you can try to focus on proving it does exist. When you don't have a rational function, you can probably use Taylor's theorem to get a similarly behaved rational function.
$endgroup$
– Git Gud
Jan 18 at 8:38




$begingroup$
There are strategies you can use. For rational functions in two variables you can try looking at sublimits of the form $tmapsto (t, kt)$, $tmapsto (t, kt^2)$, $tmapsto (t, kt^3)$ and so on, with varying values of $k$. If you're unable to find different sublimits, then perhaps the sublimit exists and you can try to focus on proving it does exist. When you don't have a rational function, you can probably use Taylor's theorem to get a similarly behaved rational function.
$endgroup$
– Git Gud
Jan 18 at 8:38










2 Answers
2






active

oldest

votes


















1












$begingroup$

Consider the curves $gamma_{1} = (t,t)$ and $gamma_{2}(t) = (t,t^{2})$. Thus we have
begin{align*}f(gamma_{1}(t)) = frac{t^{2}sin^{2}(t)}{t^{4} + t^{4}} = frac{sin^{2}(t)}{2t^{2}} Longrightarrow lim_{trightarrow 0}f(gamma_{1}(t)) = lim_{trightarrow 0}left[frac{1}{2}left(frac{sin(t)}{t}right)^{2}right] = frac{1}{2}
end{align*}



On the other hand, we have
begin{align*}
f(gamma_{2}(t)) = frac{t^{8}sin^{2}(t)}{t^{4} + t^{8}} = frac{t^{4}sin^{2}(t)}{1 + t^{4}} Longrightarrow lim_{trightarrow 0}f(gamma_{2}(t)) = lim_{trightarrow 0}frac{t^{4}sin^{2}(t)}{1+t^{4}} = frac{0times 0}{1+0} = 0
end{align*}



If the given limit existed, we should have $displaystylelim_{trightarrow0}f(gamma_{1}(t)) = lim_{trightarrow 0}f(gamma_{2}(t))$. Hence $displaystylelim_{textbf{x}rightarrowtextbf{0}} f(textbf{x})$ does not exist. Hope this helps.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! What is your general strategy for finding these two curves as in this example?
    $endgroup$
    – A Slow Learner
    Jan 18 at 1:00












  • $begingroup$
    As previously mentioned, there is no general strategy. In this particular case, such choices were made because they "break the symmetry".
    $endgroup$
    – user1337
    Jan 18 at 1:07










  • $begingroup$
    For this example, could you explain to me more how the two curves "break the symmetry"? I never heard of that before.
    $endgroup$
    – A Slow Learner
    Jan 18 at 1:08










  • $begingroup$
    Sorry, I am not an English native speaker. What I actually meant is: choose two curves whose limits of $f(gamma_{k}(t))$ are different as $trightarrow 0$.
    $endgroup$
    – user1337
    Jan 18 at 1:12












  • $begingroup$
    What if finding such two curves is difficult? For example, what do you do if the limit doesn't exist but all the curves you have found produce the same limit? Is keeping trying our only option?
    $endgroup$
    – A Slow Learner
    Jan 18 at 1:15



















0












$begingroup$

1: Use the approximation (for small $x), sin(x)approx x$. Then $f(x,y)approx frac{x^2y^2}{x^4+y^4}$. Now consider two cases:



(a) $x=y$ then the function $=frac{1}{2}$.



(b) $x=0$ or $y=0$ where the function $=0$, when the other variable $ne 0$.



Therefore the limit does not exist.



2: There is no general solution. Each problem must be evaluated on its own.






share|cite|improve this answer









$endgroup$














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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Consider the curves $gamma_{1} = (t,t)$ and $gamma_{2}(t) = (t,t^{2})$. Thus we have
    begin{align*}f(gamma_{1}(t)) = frac{t^{2}sin^{2}(t)}{t^{4} + t^{4}} = frac{sin^{2}(t)}{2t^{2}} Longrightarrow lim_{trightarrow 0}f(gamma_{1}(t)) = lim_{trightarrow 0}left[frac{1}{2}left(frac{sin(t)}{t}right)^{2}right] = frac{1}{2}
    end{align*}



    On the other hand, we have
    begin{align*}
    f(gamma_{2}(t)) = frac{t^{8}sin^{2}(t)}{t^{4} + t^{8}} = frac{t^{4}sin^{2}(t)}{1 + t^{4}} Longrightarrow lim_{trightarrow 0}f(gamma_{2}(t)) = lim_{trightarrow 0}frac{t^{4}sin^{2}(t)}{1+t^{4}} = frac{0times 0}{1+0} = 0
    end{align*}



    If the given limit existed, we should have $displaystylelim_{trightarrow0}f(gamma_{1}(t)) = lim_{trightarrow 0}f(gamma_{2}(t))$. Hence $displaystylelim_{textbf{x}rightarrowtextbf{0}} f(textbf{x})$ does not exist. Hope this helps.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you! What is your general strategy for finding these two curves as in this example?
      $endgroup$
      – A Slow Learner
      Jan 18 at 1:00












    • $begingroup$
      As previously mentioned, there is no general strategy. In this particular case, such choices were made because they "break the symmetry".
      $endgroup$
      – user1337
      Jan 18 at 1:07










    • $begingroup$
      For this example, could you explain to me more how the two curves "break the symmetry"? I never heard of that before.
      $endgroup$
      – A Slow Learner
      Jan 18 at 1:08










    • $begingroup$
      Sorry, I am not an English native speaker. What I actually meant is: choose two curves whose limits of $f(gamma_{k}(t))$ are different as $trightarrow 0$.
      $endgroup$
      – user1337
      Jan 18 at 1:12












    • $begingroup$
      What if finding such two curves is difficult? For example, what do you do if the limit doesn't exist but all the curves you have found produce the same limit? Is keeping trying our only option?
      $endgroup$
      – A Slow Learner
      Jan 18 at 1:15
















    1












    $begingroup$

    Consider the curves $gamma_{1} = (t,t)$ and $gamma_{2}(t) = (t,t^{2})$. Thus we have
    begin{align*}f(gamma_{1}(t)) = frac{t^{2}sin^{2}(t)}{t^{4} + t^{4}} = frac{sin^{2}(t)}{2t^{2}} Longrightarrow lim_{trightarrow 0}f(gamma_{1}(t)) = lim_{trightarrow 0}left[frac{1}{2}left(frac{sin(t)}{t}right)^{2}right] = frac{1}{2}
    end{align*}



    On the other hand, we have
    begin{align*}
    f(gamma_{2}(t)) = frac{t^{8}sin^{2}(t)}{t^{4} + t^{8}} = frac{t^{4}sin^{2}(t)}{1 + t^{4}} Longrightarrow lim_{trightarrow 0}f(gamma_{2}(t)) = lim_{trightarrow 0}frac{t^{4}sin^{2}(t)}{1+t^{4}} = frac{0times 0}{1+0} = 0
    end{align*}



    If the given limit existed, we should have $displaystylelim_{trightarrow0}f(gamma_{1}(t)) = lim_{trightarrow 0}f(gamma_{2}(t))$. Hence $displaystylelim_{textbf{x}rightarrowtextbf{0}} f(textbf{x})$ does not exist. Hope this helps.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you! What is your general strategy for finding these two curves as in this example?
      $endgroup$
      – A Slow Learner
      Jan 18 at 1:00












    • $begingroup$
      As previously mentioned, there is no general strategy. In this particular case, such choices were made because they "break the symmetry".
      $endgroup$
      – user1337
      Jan 18 at 1:07










    • $begingroup$
      For this example, could you explain to me more how the two curves "break the symmetry"? I never heard of that before.
      $endgroup$
      – A Slow Learner
      Jan 18 at 1:08










    • $begingroup$
      Sorry, I am not an English native speaker. What I actually meant is: choose two curves whose limits of $f(gamma_{k}(t))$ are different as $trightarrow 0$.
      $endgroup$
      – user1337
      Jan 18 at 1:12












    • $begingroup$
      What if finding such two curves is difficult? For example, what do you do if the limit doesn't exist but all the curves you have found produce the same limit? Is keeping trying our only option?
      $endgroup$
      – A Slow Learner
      Jan 18 at 1:15














    1












    1








    1





    $begingroup$

    Consider the curves $gamma_{1} = (t,t)$ and $gamma_{2}(t) = (t,t^{2})$. Thus we have
    begin{align*}f(gamma_{1}(t)) = frac{t^{2}sin^{2}(t)}{t^{4} + t^{4}} = frac{sin^{2}(t)}{2t^{2}} Longrightarrow lim_{trightarrow 0}f(gamma_{1}(t)) = lim_{trightarrow 0}left[frac{1}{2}left(frac{sin(t)}{t}right)^{2}right] = frac{1}{2}
    end{align*}



    On the other hand, we have
    begin{align*}
    f(gamma_{2}(t)) = frac{t^{8}sin^{2}(t)}{t^{4} + t^{8}} = frac{t^{4}sin^{2}(t)}{1 + t^{4}} Longrightarrow lim_{trightarrow 0}f(gamma_{2}(t)) = lim_{trightarrow 0}frac{t^{4}sin^{2}(t)}{1+t^{4}} = frac{0times 0}{1+0} = 0
    end{align*}



    If the given limit existed, we should have $displaystylelim_{trightarrow0}f(gamma_{1}(t)) = lim_{trightarrow 0}f(gamma_{2}(t))$. Hence $displaystylelim_{textbf{x}rightarrowtextbf{0}} f(textbf{x})$ does not exist. Hope this helps.






    share|cite|improve this answer









    $endgroup$



    Consider the curves $gamma_{1} = (t,t)$ and $gamma_{2}(t) = (t,t^{2})$. Thus we have
    begin{align*}f(gamma_{1}(t)) = frac{t^{2}sin^{2}(t)}{t^{4} + t^{4}} = frac{sin^{2}(t)}{2t^{2}} Longrightarrow lim_{trightarrow 0}f(gamma_{1}(t)) = lim_{trightarrow 0}left[frac{1}{2}left(frac{sin(t)}{t}right)^{2}right] = frac{1}{2}
    end{align*}



    On the other hand, we have
    begin{align*}
    f(gamma_{2}(t)) = frac{t^{8}sin^{2}(t)}{t^{4} + t^{8}} = frac{t^{4}sin^{2}(t)}{1 + t^{4}} Longrightarrow lim_{trightarrow 0}f(gamma_{2}(t)) = lim_{trightarrow 0}frac{t^{4}sin^{2}(t)}{1+t^{4}} = frac{0times 0}{1+0} = 0
    end{align*}



    If the given limit existed, we should have $displaystylelim_{trightarrow0}f(gamma_{1}(t)) = lim_{trightarrow 0}f(gamma_{2}(t))$. Hence $displaystylelim_{textbf{x}rightarrowtextbf{0}} f(textbf{x})$ does not exist. Hope this helps.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 18 at 0:48









    user1337user1337

    48210




    48210












    • $begingroup$
      Thank you! What is your general strategy for finding these two curves as in this example?
      $endgroup$
      – A Slow Learner
      Jan 18 at 1:00












    • $begingroup$
      As previously mentioned, there is no general strategy. In this particular case, such choices were made because they "break the symmetry".
      $endgroup$
      – user1337
      Jan 18 at 1:07










    • $begingroup$
      For this example, could you explain to me more how the two curves "break the symmetry"? I never heard of that before.
      $endgroup$
      – A Slow Learner
      Jan 18 at 1:08










    • $begingroup$
      Sorry, I am not an English native speaker. What I actually meant is: choose two curves whose limits of $f(gamma_{k}(t))$ are different as $trightarrow 0$.
      $endgroup$
      – user1337
      Jan 18 at 1:12












    • $begingroup$
      What if finding such two curves is difficult? For example, what do you do if the limit doesn't exist but all the curves you have found produce the same limit? Is keeping trying our only option?
      $endgroup$
      – A Slow Learner
      Jan 18 at 1:15


















    • $begingroup$
      Thank you! What is your general strategy for finding these two curves as in this example?
      $endgroup$
      – A Slow Learner
      Jan 18 at 1:00












    • $begingroup$
      As previously mentioned, there is no general strategy. In this particular case, such choices were made because they "break the symmetry".
      $endgroup$
      – user1337
      Jan 18 at 1:07










    • $begingroup$
      For this example, could you explain to me more how the two curves "break the symmetry"? I never heard of that before.
      $endgroup$
      – A Slow Learner
      Jan 18 at 1:08










    • $begingroup$
      Sorry, I am not an English native speaker. What I actually meant is: choose two curves whose limits of $f(gamma_{k}(t))$ are different as $trightarrow 0$.
      $endgroup$
      – user1337
      Jan 18 at 1:12












    • $begingroup$
      What if finding such two curves is difficult? For example, what do you do if the limit doesn't exist but all the curves you have found produce the same limit? Is keeping trying our only option?
      $endgroup$
      – A Slow Learner
      Jan 18 at 1:15
















    $begingroup$
    Thank you! What is your general strategy for finding these two curves as in this example?
    $endgroup$
    – A Slow Learner
    Jan 18 at 1:00






    $begingroup$
    Thank you! What is your general strategy for finding these two curves as in this example?
    $endgroup$
    – A Slow Learner
    Jan 18 at 1:00














    $begingroup$
    As previously mentioned, there is no general strategy. In this particular case, such choices were made because they "break the symmetry".
    $endgroup$
    – user1337
    Jan 18 at 1:07




    $begingroup$
    As previously mentioned, there is no general strategy. In this particular case, such choices were made because they "break the symmetry".
    $endgroup$
    – user1337
    Jan 18 at 1:07












    $begingroup$
    For this example, could you explain to me more how the two curves "break the symmetry"? I never heard of that before.
    $endgroup$
    – A Slow Learner
    Jan 18 at 1:08




    $begingroup$
    For this example, could you explain to me more how the two curves "break the symmetry"? I never heard of that before.
    $endgroup$
    – A Slow Learner
    Jan 18 at 1:08












    $begingroup$
    Sorry, I am not an English native speaker. What I actually meant is: choose two curves whose limits of $f(gamma_{k}(t))$ are different as $trightarrow 0$.
    $endgroup$
    – user1337
    Jan 18 at 1:12






    $begingroup$
    Sorry, I am not an English native speaker. What I actually meant is: choose two curves whose limits of $f(gamma_{k}(t))$ are different as $trightarrow 0$.
    $endgroup$
    – user1337
    Jan 18 at 1:12














    $begingroup$
    What if finding such two curves is difficult? For example, what do you do if the limit doesn't exist but all the curves you have found produce the same limit? Is keeping trying our only option?
    $endgroup$
    – A Slow Learner
    Jan 18 at 1:15




    $begingroup$
    What if finding such two curves is difficult? For example, what do you do if the limit doesn't exist but all the curves you have found produce the same limit? Is keeping trying our only option?
    $endgroup$
    – A Slow Learner
    Jan 18 at 1:15











    0












    $begingroup$

    1: Use the approximation (for small $x), sin(x)approx x$. Then $f(x,y)approx frac{x^2y^2}{x^4+y^4}$. Now consider two cases:



    (a) $x=y$ then the function $=frac{1}{2}$.



    (b) $x=0$ or $y=0$ where the function $=0$, when the other variable $ne 0$.



    Therefore the limit does not exist.



    2: There is no general solution. Each problem must be evaluated on its own.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      1: Use the approximation (for small $x), sin(x)approx x$. Then $f(x,y)approx frac{x^2y^2}{x^4+y^4}$. Now consider two cases:



      (a) $x=y$ then the function $=frac{1}{2}$.



      (b) $x=0$ or $y=0$ where the function $=0$, when the other variable $ne 0$.



      Therefore the limit does not exist.



      2: There is no general solution. Each problem must be evaluated on its own.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        1: Use the approximation (for small $x), sin(x)approx x$. Then $f(x,y)approx frac{x^2y^2}{x^4+y^4}$. Now consider two cases:



        (a) $x=y$ then the function $=frac{1}{2}$.



        (b) $x=0$ or $y=0$ where the function $=0$, when the other variable $ne 0$.



        Therefore the limit does not exist.



        2: There is no general solution. Each problem must be evaluated on its own.






        share|cite|improve this answer









        $endgroup$



        1: Use the approximation (for small $x), sin(x)approx x$. Then $f(x,y)approx frac{x^2y^2}{x^4+y^4}$. Now consider two cases:



        (a) $x=y$ then the function $=frac{1}{2}$.



        (b) $x=0$ or $y=0$ where the function $=0$, when the other variable $ne 0$.



        Therefore the limit does not exist.



        2: There is no general solution. Each problem must be evaluated on its own.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 18 at 0:56









        herb steinbergherb steinberg

        3,1982311




        3,1982311






























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