Limit $lim_{(x, y) to (0, infty)} (xy)$












4












$begingroup$


$$lim_{(x,y) to (0, infty)} (xy) = [xtofrac{1}{x}Rightarrow xto infty] = lim_{(x,y)to(infty, infty)} (frac{y}{x}) = [x = rcostheta, y = rsintheta] = lim_{rtoinfty}frac{rsintheta}{rcostheta} =lim_{rtoinfty}tantheta$$



Therefore, limit does not exist. Is substitution in the beginning viable here?






limits multivariable-calculus







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$endgroup$








  • 2




    $begingroup$
    Not really. If $xto 0^+$ (from one side) then $frac{1}{x}to +infty$, but if $xto 0$ oscillating around zero, then the limit $1/x$ does not exists.
    $endgroup$
    – A.Γ.
    Dec 10 '18 at 16:29










  • $begingroup$
    Change $y$ to $1/y$.
    $endgroup$
    – BPP
    Dec 10 '18 at 16:29










  • $begingroup$
    You're assuming $xto 0^+$.
    $endgroup$
    – Shaun
    Dec 10 '18 at 16:30










  • $begingroup$
    This limit doesn't exist or $0cdotinfty$ would not be an indeterminate form.
    $endgroup$
    – egreg
    Dec 10 '18 at 17:05
















4












$begingroup$


$$lim_{(x,y) to (0, infty)} (xy) = [xtofrac{1}{x}Rightarrow xto infty] = lim_{(x,y)to(infty, infty)} (frac{y}{x}) = [x = rcostheta, y = rsintheta] = lim_{rtoinfty}frac{rsintheta}{rcostheta} =lim_{rtoinfty}tantheta$$



Therefore, limit does not exist. Is substitution in the beginning viable here?






limits multivariable-calculus







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Not really. If $xto 0^+$ (from one side) then $frac{1}{x}to +infty$, but if $xto 0$ oscillating around zero, then the limit $1/x$ does not exists.
    $endgroup$
    – A.Γ.
    Dec 10 '18 at 16:29










  • $begingroup$
    Change $y$ to $1/y$.
    $endgroup$
    – BPP
    Dec 10 '18 at 16:29










  • $begingroup$
    You're assuming $xto 0^+$.
    $endgroup$
    – Shaun
    Dec 10 '18 at 16:30










  • $begingroup$
    This limit doesn't exist or $0cdotinfty$ would not be an indeterminate form.
    $endgroup$
    – egreg
    Dec 10 '18 at 17:05














4












4








4





$begingroup$


$$lim_{(x,y) to (0, infty)} (xy) = [xtofrac{1}{x}Rightarrow xto infty] = lim_{(x,y)to(infty, infty)} (frac{y}{x}) = [x = rcostheta, y = rsintheta] = lim_{rtoinfty}frac{rsintheta}{rcostheta} =lim_{rtoinfty}tantheta$$



Therefore, limit does not exist. Is substitution in the beginning viable here?






limits multivariable-calculus







share|cite|improve this question











$endgroup$




$$lim_{(x,y) to (0, infty)} (xy) = [xtofrac{1}{x}Rightarrow xto infty] = lim_{(x,y)to(infty, infty)} (frac{y}{x}) = [x = rcostheta, y = rsintheta] = lim_{rtoinfty}frac{rsintheta}{rcostheta} =lim_{rtoinfty}tantheta$$



Therefore, limit does not exist. Is substitution in the beginning viable here?






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 Dec 10 '18 at 17:02









Martin Sleziak

44.7k10118272




44.7k10118272










asked Dec 10 '18 at 16:23









H. BaoH. Bao

474




474








  • 2




    $begingroup$
    Not really. If $xto 0^+$ (from one side) then $frac{1}{x}to +infty$, but if $xto 0$ oscillating around zero, then the limit $1/x$ does not exists.
    $endgroup$
    – A.Γ.
    Dec 10 '18 at 16:29










  • $begingroup$
    Change $y$ to $1/y$.
    $endgroup$
    – BPP
    Dec 10 '18 at 16:29










  • $begingroup$
    You're assuming $xto 0^+$.
    $endgroup$
    – Shaun
    Dec 10 '18 at 16:30










  • $begingroup$
    This limit doesn't exist or $0cdotinfty$ would not be an indeterminate form.
    $endgroup$
    – egreg
    Dec 10 '18 at 17:05














  • 2




    $begingroup$
    Not really. If $xto 0^+$ (from one side) then $frac{1}{x}to +infty$, but if $xto 0$ oscillating around zero, then the limit $1/x$ does not exists.
    $endgroup$
    – A.Γ.
    Dec 10 '18 at 16:29










  • $begingroup$
    Change $y$ to $1/y$.
    $endgroup$
    – BPP
    Dec 10 '18 at 16:29










  • $begingroup$
    You're assuming $xto 0^+$.
    $endgroup$
    – Shaun
    Dec 10 '18 at 16:30










  • $begingroup$
    This limit doesn't exist or $0cdotinfty$ would not be an indeterminate form.
    $endgroup$
    – egreg
    Dec 10 '18 at 17:05








2




2




$begingroup$
Not really. If $xto 0^+$ (from one side) then $frac{1}{x}to +infty$, but if $xto 0$ oscillating around zero, then the limit $1/x$ does not exists.
$endgroup$
– A.Γ.
Dec 10 '18 at 16:29




$begingroup$
Not really. If $xto 0^+$ (from one side) then $frac{1}{x}to +infty$, but if $xto 0$ oscillating around zero, then the limit $1/x$ does not exists.
$endgroup$
– A.Γ.
Dec 10 '18 at 16:29












$begingroup$
Change $y$ to $1/y$.
$endgroup$
– BPP
Dec 10 '18 at 16:29




$begingroup$
Change $y$ to $1/y$.
$endgroup$
– BPP
Dec 10 '18 at 16:29












$begingroup$
You're assuming $xto 0^+$.
$endgroup$
– Shaun
Dec 10 '18 at 16:30




$begingroup$
You're assuming $xto 0^+$.
$endgroup$
– Shaun
Dec 10 '18 at 16:30












$begingroup$
This limit doesn't exist or $0cdotinfty$ would not be an indeterminate form.
$endgroup$
– egreg
Dec 10 '18 at 17:05




$begingroup$
This limit doesn't exist or $0cdotinfty$ would not be an indeterminate form.
$endgroup$
– egreg
Dec 10 '18 at 17:05










2 Answers
2






active

oldest

votes


















2












$begingroup$

More simply we have that as $tto infty$




  • $x=frac1tto 0 quad y=tto infty$


$$xy=frac1t cdot t to 1$$




  • $x=frac1tto 0 quad y=t^2to infty$


$$xy=frac1t cdot t^2=t to infty$$



and therefore the limit doesn't exist.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    The best way of saying things is this : $lim_{(x,y) to (0,infty)} xy$ exists and equals $L$, if and only if for every subsequence $x_n to 0$ and $y_n to infty$ we have $x_ny_n to L$, where the latter is convergence of a sequence, in which case we have the $epsilon-N$ definition of convergence.



    Of course, one sees that taking $x_n = frac 1n$ and $y_n = kn$ for any $k in mathbb R_{> 0}$, we have $x_n to 0$, $y_n to infty$ and $x_ny_n to k$. Therefore, the limit in question does not exist, since different choices of subsequences give different limit values.




    As to what you have done, unfortunately as $x to 0$ it is not true that $frac 1x$ goes to $infty$, since $x$ can approach $0$ from below, in which case $frac 1x$ assumes negative values and cannot converge to any positive value, let alone positive infinity.



    Furthermore, the part where you set $(x,y) = (r cos theta,r sin theta)$ : The point is that change of variable in limits only works in certain situations. In particular, since the variable $theta$ has no particular limit as $(x,y) to (0,infty)$, I do not think that this change of variable is correct.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      More simply we have that as $tto infty$




      • $x=frac1tto 0 quad y=tto infty$


      $$xy=frac1t cdot t to 1$$




      • $x=frac1tto 0 quad y=t^2to infty$


      $$xy=frac1t cdot t^2=t to infty$$



      and therefore the limit doesn't exist.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        More simply we have that as $tto infty$




        • $x=frac1tto 0 quad y=tto infty$


        $$xy=frac1t cdot t to 1$$




        • $x=frac1tto 0 quad y=t^2to infty$


        $$xy=frac1t cdot t^2=t to infty$$



        and therefore the limit doesn't exist.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          More simply we have that as $tto infty$




          • $x=frac1tto 0 quad y=tto infty$


          $$xy=frac1t cdot t to 1$$




          • $x=frac1tto 0 quad y=t^2to infty$


          $$xy=frac1t cdot t^2=t to infty$$



          and therefore the limit doesn't exist.






          share|cite|improve this answer









          $endgroup$



          More simply we have that as $tto infty$




          • $x=frac1tto 0 quad y=tto infty$


          $$xy=frac1t cdot t to 1$$




          • $x=frac1tto 0 quad y=t^2to infty$


          $$xy=frac1t cdot t^2=t to infty$$



          and therefore the limit doesn't exist.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 17:00









          gimusigimusi

          92.8k84494




          92.8k84494























              3












              $begingroup$

              The best way of saying things is this : $lim_{(x,y) to (0,infty)} xy$ exists and equals $L$, if and only if for every subsequence $x_n to 0$ and $y_n to infty$ we have $x_ny_n to L$, where the latter is convergence of a sequence, in which case we have the $epsilon-N$ definition of convergence.



              Of course, one sees that taking $x_n = frac 1n$ and $y_n = kn$ for any $k in mathbb R_{> 0}$, we have $x_n to 0$, $y_n to infty$ and $x_ny_n to k$. Therefore, the limit in question does not exist, since different choices of subsequences give different limit values.




              As to what you have done, unfortunately as $x to 0$ it is not true that $frac 1x$ goes to $infty$, since $x$ can approach $0$ from below, in which case $frac 1x$ assumes negative values and cannot converge to any positive value, let alone positive infinity.



              Furthermore, the part where you set $(x,y) = (r cos theta,r sin theta)$ : The point is that change of variable in limits only works in certain situations. In particular, since the variable $theta$ has no particular limit as $(x,y) to (0,infty)$, I do not think that this change of variable is correct.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                The best way of saying things is this : $lim_{(x,y) to (0,infty)} xy$ exists and equals $L$, if and only if for every subsequence $x_n to 0$ and $y_n to infty$ we have $x_ny_n to L$, where the latter is convergence of a sequence, in which case we have the $epsilon-N$ definition of convergence.



                Of course, one sees that taking $x_n = frac 1n$ and $y_n = kn$ for any $k in mathbb R_{> 0}$, we have $x_n to 0$, $y_n to infty$ and $x_ny_n to k$. Therefore, the limit in question does not exist, since different choices of subsequences give different limit values.




                As to what you have done, unfortunately as $x to 0$ it is not true that $frac 1x$ goes to $infty$, since $x$ can approach $0$ from below, in which case $frac 1x$ assumes negative values and cannot converge to any positive value, let alone positive infinity.



                Furthermore, the part where you set $(x,y) = (r cos theta,r sin theta)$ : The point is that change of variable in limits only works in certain situations. In particular, since the variable $theta$ has no particular limit as $(x,y) to (0,infty)$, I do not think that this change of variable is correct.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  The best way of saying things is this : $lim_{(x,y) to (0,infty)} xy$ exists and equals $L$, if and only if for every subsequence $x_n to 0$ and $y_n to infty$ we have $x_ny_n to L$, where the latter is convergence of a sequence, in which case we have the $epsilon-N$ definition of convergence.



                  Of course, one sees that taking $x_n = frac 1n$ and $y_n = kn$ for any $k in mathbb R_{> 0}$, we have $x_n to 0$, $y_n to infty$ and $x_ny_n to k$. Therefore, the limit in question does not exist, since different choices of subsequences give different limit values.




                  As to what you have done, unfortunately as $x to 0$ it is not true that $frac 1x$ goes to $infty$, since $x$ can approach $0$ from below, in which case $frac 1x$ assumes negative values and cannot converge to any positive value, let alone positive infinity.



                  Furthermore, the part where you set $(x,y) = (r cos theta,r sin theta)$ : The point is that change of variable in limits only works in certain situations. In particular, since the variable $theta$ has no particular limit as $(x,y) to (0,infty)$, I do not think that this change of variable is correct.






                  share|cite|improve this answer









                  $endgroup$



                  The best way of saying things is this : $lim_{(x,y) to (0,infty)} xy$ exists and equals $L$, if and only if for every subsequence $x_n to 0$ and $y_n to infty$ we have $x_ny_n to L$, where the latter is convergence of a sequence, in which case we have the $epsilon-N$ definition of convergence.



                  Of course, one sees that taking $x_n = frac 1n$ and $y_n = kn$ for any $k in mathbb R_{> 0}$, we have $x_n to 0$, $y_n to infty$ and $x_ny_n to k$. Therefore, the limit in question does not exist, since different choices of subsequences give different limit values.




                  As to what you have done, unfortunately as $x to 0$ it is not true that $frac 1x$ goes to $infty$, since $x$ can approach $0$ from below, in which case $frac 1x$ assumes negative values and cannot converge to any positive value, let alone positive infinity.



                  Furthermore, the part where you set $(x,y) = (r cos theta,r sin theta)$ : The point is that change of variable in limits only works in certain situations. In particular, since the variable $theta$ has no particular limit as $(x,y) to (0,infty)$, I do not think that this change of variable is correct.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 10 '18 at 16:52









                  астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

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