Find limit of function of two variables (if exist)












1












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Find $$
lim_{(x,0)to (pi,0)}frac{3sin^3(x+y)}{y^2+(x-pi)^2}.
$$



My attempt was to show that that limit does not exist. But for any pairs(checked by me) of $(x_n,y_n)to(pi,0)$ this function tends to $0$ (when $nto infty$) (for example I checked $(x_n,y_n)=(frac{1}{n}+pi,frac{1}{n}).$ I would be grateful for your hints and ideas.






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    1












    $begingroup$


    Find $$
    lim_{(x,0)to (pi,0)}frac{3sin^3(x+y)}{y^2+(x-pi)^2}.
    $$



    My attempt was to show that that limit does not exist. But for any pairs(checked by me) of $(x_n,y_n)to(pi,0)$ this function tends to $0$ (when $nto infty$) (for example I checked $(x_n,y_n)=(frac{1}{n}+pi,frac{1}{n}).$ I would be grateful for your hints and ideas.






    limits







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    $endgroup$















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      $begingroup$


      Find $$
      lim_{(x,0)to (pi,0)}frac{3sin^3(x+y)}{y^2+(x-pi)^2}.
      $$



      My attempt was to show that that limit does not exist. But for any pairs(checked by me) of $(x_n,y_n)to(pi,0)$ this function tends to $0$ (when $nto infty$) (for example I checked $(x_n,y_n)=(frac{1}{n}+pi,frac{1}{n}).$ I would be grateful for your hints and ideas.






      limits







      share|cite|improve this question









      $endgroup$




      Find $$
      lim_{(x,0)to (pi,0)}frac{3sin^3(x+y)}{y^2+(x-pi)^2}.
      $$



      My attempt was to show that that limit does not exist. But for any pairs(checked by me) of $(x_n,y_n)to(pi,0)$ this function tends to $0$ (when $nto infty$) (for example I checked $(x_n,y_n)=(frac{1}{n}+pi,frac{1}{n}).$ I would be grateful for your hints and ideas.






      limits




      limits






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      asked Jan 28 at 22:54









      akapakap

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          $begingroup$

          I would expect the limit to exist and be zero.




          • Note that $sin (tpmpi)=pmsin t$. So $sin (x+y)=-sin (x-pi+y)$.


          • We also have, for small $t$, $|sin t|leq t$.


          • And, $(a+b)^3leq 4a^3+4b^3$.



          Thus, for $(x,y)$ close to $(pi,0)$,
          begin{align}
          left|frac{3sin^3(x+y)}{y^2+(x-pi)^2}right|
          &=left|frac{3sin^3(x-pi+y)}{y^2+(x-pi)^2}right|
          leqleft|frac{3(x-pi+y)^3}{y^2+(x-pi)^2}right|
          leqleft|frac{12(x-pi)^3+12y^3}{y^2+(x-pi)^2}right|\ \
          &leqleft|frac{12(x-pi)^3 }{y^2+(x-pi)^2}right|+left|frac{ 12y^3}{y^2+(x-pi)^2}right|\ \
          &leq12|x-pi|+12|y|,
          end{align}
          using that
          $$
          left|frac{ (x-pi)^2 }{y^2+(x-pi)^2}right|leq1, left|frac{ y^2 }{y^2+(x-pi)^2}right|leq1.
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So that's mean that limit is equal $0$. (Maybe one more comment, smal correction - without 12 in last line(in modules)). So wolframalpha.com/input/… wolfram make mistake?
            $endgroup$
            – akap
            Jan 28 at 23:20








          • 1




            $begingroup$
            Yes and yes. I think whatever algorithm it uses is getting confused with the sine.
            $endgroup$
            – Martin Argerami
            Jan 28 at 23:28












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          1 Answer
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          1 Answer
          1






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          active

          oldest

          votes









          1












          $begingroup$

          I would expect the limit to exist and be zero.




          • Note that $sin (tpmpi)=pmsin t$. So $sin (x+y)=-sin (x-pi+y)$.


          • We also have, for small $t$, $|sin t|leq t$.


          • And, $(a+b)^3leq 4a^3+4b^3$.



          Thus, for $(x,y)$ close to $(pi,0)$,
          begin{align}
          left|frac{3sin^3(x+y)}{y^2+(x-pi)^2}right|
          &=left|frac{3sin^3(x-pi+y)}{y^2+(x-pi)^2}right|
          leqleft|frac{3(x-pi+y)^3}{y^2+(x-pi)^2}right|
          leqleft|frac{12(x-pi)^3+12y^3}{y^2+(x-pi)^2}right|\ \
          &leqleft|frac{12(x-pi)^3 }{y^2+(x-pi)^2}right|+left|frac{ 12y^3}{y^2+(x-pi)^2}right|\ \
          &leq12|x-pi|+12|y|,
          end{align}
          using that
          $$
          left|frac{ (x-pi)^2 }{y^2+(x-pi)^2}right|leq1, left|frac{ y^2 }{y^2+(x-pi)^2}right|leq1.
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So that's mean that limit is equal $0$. (Maybe one more comment, smal correction - without 12 in last line(in modules)). So wolframalpha.com/input/… wolfram make mistake?
            $endgroup$
            – akap
            Jan 28 at 23:20








          • 1




            $begingroup$
            Yes and yes. I think whatever algorithm it uses is getting confused with the sine.
            $endgroup$
            – Martin Argerami
            Jan 28 at 23:28
















          1












          $begingroup$

          I would expect the limit to exist and be zero.




          • Note that $sin (tpmpi)=pmsin t$. So $sin (x+y)=-sin (x-pi+y)$.


          • We also have, for small $t$, $|sin t|leq t$.


          • And, $(a+b)^3leq 4a^3+4b^3$.



          Thus, for $(x,y)$ close to $(pi,0)$,
          begin{align}
          left|frac{3sin^3(x+y)}{y^2+(x-pi)^2}right|
          &=left|frac{3sin^3(x-pi+y)}{y^2+(x-pi)^2}right|
          leqleft|frac{3(x-pi+y)^3}{y^2+(x-pi)^2}right|
          leqleft|frac{12(x-pi)^3+12y^3}{y^2+(x-pi)^2}right|\ \
          &leqleft|frac{12(x-pi)^3 }{y^2+(x-pi)^2}right|+left|frac{ 12y^3}{y^2+(x-pi)^2}right|\ \
          &leq12|x-pi|+12|y|,
          end{align}
          using that
          $$
          left|frac{ (x-pi)^2 }{y^2+(x-pi)^2}right|leq1, left|frac{ y^2 }{y^2+(x-pi)^2}right|leq1.
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So that's mean that limit is equal $0$. (Maybe one more comment, smal correction - without 12 in last line(in modules)). So wolframalpha.com/input/… wolfram make mistake?
            $endgroup$
            – akap
            Jan 28 at 23:20








          • 1




            $begingroup$
            Yes and yes. I think whatever algorithm it uses is getting confused with the sine.
            $endgroup$
            – Martin Argerami
            Jan 28 at 23:28














          1












          1








          1





          $begingroup$

          I would expect the limit to exist and be zero.




          • Note that $sin (tpmpi)=pmsin t$. So $sin (x+y)=-sin (x-pi+y)$.


          • We also have, for small $t$, $|sin t|leq t$.


          • And, $(a+b)^3leq 4a^3+4b^3$.



          Thus, for $(x,y)$ close to $(pi,0)$,
          begin{align}
          left|frac{3sin^3(x+y)}{y^2+(x-pi)^2}right|
          &=left|frac{3sin^3(x-pi+y)}{y^2+(x-pi)^2}right|
          leqleft|frac{3(x-pi+y)^3}{y^2+(x-pi)^2}right|
          leqleft|frac{12(x-pi)^3+12y^3}{y^2+(x-pi)^2}right|\ \
          &leqleft|frac{12(x-pi)^3 }{y^2+(x-pi)^2}right|+left|frac{ 12y^3}{y^2+(x-pi)^2}right|\ \
          &leq12|x-pi|+12|y|,
          end{align}
          using that
          $$
          left|frac{ (x-pi)^2 }{y^2+(x-pi)^2}right|leq1, left|frac{ y^2 }{y^2+(x-pi)^2}right|leq1.
          $$






          share|cite|improve this answer











          $endgroup$



          I would expect the limit to exist and be zero.




          • Note that $sin (tpmpi)=pmsin t$. So $sin (x+y)=-sin (x-pi+y)$.


          • We also have, for small $t$, $|sin t|leq t$.


          • And, $(a+b)^3leq 4a^3+4b^3$.



          Thus, for $(x,y)$ close to $(pi,0)$,
          begin{align}
          left|frac{3sin^3(x+y)}{y^2+(x-pi)^2}right|
          &=left|frac{3sin^3(x-pi+y)}{y^2+(x-pi)^2}right|
          leqleft|frac{3(x-pi+y)^3}{y^2+(x-pi)^2}right|
          leqleft|frac{12(x-pi)^3+12y^3}{y^2+(x-pi)^2}right|\ \
          &leqleft|frac{12(x-pi)^3 }{y^2+(x-pi)^2}right|+left|frac{ 12y^3}{y^2+(x-pi)^2}right|\ \
          &leq12|x-pi|+12|y|,
          end{align}
          using that
          $$
          left|frac{ (x-pi)^2 }{y^2+(x-pi)^2}right|leq1, left|frac{ y^2 }{y^2+(x-pi)^2}right|leq1.
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 28 at 23:25

























          answered Jan 28 at 23:11









          Martin ArgeramiMartin Argerami

          129k1184184




          129k1184184












          • $begingroup$
            So that's mean that limit is equal $0$. (Maybe one more comment, smal correction - without 12 in last line(in modules)). So wolframalpha.com/input/… wolfram make mistake?
            $endgroup$
            – akap
            Jan 28 at 23:20








          • 1




            $begingroup$
            Yes and yes. I think whatever algorithm it uses is getting confused with the sine.
            $endgroup$
            – Martin Argerami
            Jan 28 at 23:28


















          • $begingroup$
            So that's mean that limit is equal $0$. (Maybe one more comment, smal correction - without 12 in last line(in modules)). So wolframalpha.com/input/… wolfram make mistake?
            $endgroup$
            – akap
            Jan 28 at 23:20








          • 1




            $begingroup$
            Yes and yes. I think whatever algorithm it uses is getting confused with the sine.
            $endgroup$
            – Martin Argerami
            Jan 28 at 23:28
















          $begingroup$
          So that's mean that limit is equal $0$. (Maybe one more comment, smal correction - without 12 in last line(in modules)). So wolframalpha.com/input/… wolfram make mistake?
          $endgroup$
          – akap
          Jan 28 at 23:20






          $begingroup$
          So that's mean that limit is equal $0$. (Maybe one more comment, smal correction - without 12 in last line(in modules)). So wolframalpha.com/input/… wolfram make mistake?
          $endgroup$
          – akap
          Jan 28 at 23:20






          1




          1




          $begingroup$
          Yes and yes. I think whatever algorithm it uses is getting confused with the sine.
          $endgroup$
          – Martin Argerami
          Jan 28 at 23:28




          $begingroup$
          Yes and yes. I think whatever algorithm it uses is getting confused with the sine.
          $endgroup$
          – Martin Argerami
          Jan 28 at 23:28


















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