Limit $lim_{(x, y) to (infty, infty)} frac{x+sqrt{y}}{x^2+y}$












2












$begingroup$


Show whether the limit exists and find it, or prove that it does not.
$$lim_{(x, y) to(infty,infty)}frac{x+sqrt{y}}{x^2+y}$$
WolframAlpha shows that limit does not exist, however, I do fail to conclude so.
$$lim_{(x, y) to(infty,infty)}frac{x+sqrt{y}}{x^2+y} = [x=rcostheta, y = rsintheta] = lim_{rtoinfty}frac{rcostheta+sqrt{rsintheta}}{r^2cos^2theta+rsintheta} = lim_{rtoinfty}frac{costhetafrac{sqrt{sintheta}}{sqrt{r}}}{rcos^2theta+sintheta} = 0.$$
Having gotten the exact results for whatever the substitution is made (such as $y = x, y = x^2, [x = t^2, y = t])$, my conclusion is that limit does exist and equals $0.$



Did I miss something?






limits multivariable-calculus







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$endgroup$












  • $begingroup$
    Can you link to the WA calculation, or provide the exact formula you typed in there?
    $endgroup$
    – Barry Cipra
    Dec 15 '18 at 13:53
















2












$begingroup$


Show whether the limit exists and find it, or prove that it does not.
$$lim_{(x, y) to(infty,infty)}frac{x+sqrt{y}}{x^2+y}$$
WolframAlpha shows that limit does not exist, however, I do fail to conclude so.
$$lim_{(x, y) to(infty,infty)}frac{x+sqrt{y}}{x^2+y} = [x=rcostheta, y = rsintheta] = lim_{rtoinfty}frac{rcostheta+sqrt{rsintheta}}{r^2cos^2theta+rsintheta} = lim_{rtoinfty}frac{costhetafrac{sqrt{sintheta}}{sqrt{r}}}{rcos^2theta+sintheta} = 0.$$
Having gotten the exact results for whatever the substitution is made (such as $y = x, y = x^2, [x = t^2, y = t])$, my conclusion is that limit does exist and equals $0.$



Did I miss something?






limits multivariable-calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    Can you link to the WA calculation, or provide the exact formula you typed in there?
    $endgroup$
    – Barry Cipra
    Dec 15 '18 at 13:53














2












2








2





$begingroup$


Show whether the limit exists and find it, or prove that it does not.
$$lim_{(x, y) to(infty,infty)}frac{x+sqrt{y}}{x^2+y}$$
WolframAlpha shows that limit does not exist, however, I do fail to conclude so.
$$lim_{(x, y) to(infty,infty)}frac{x+sqrt{y}}{x^2+y} = [x=rcostheta, y = rsintheta] = lim_{rtoinfty}frac{rcostheta+sqrt{rsintheta}}{r^2cos^2theta+rsintheta} = lim_{rtoinfty}frac{costhetafrac{sqrt{sintheta}}{sqrt{r}}}{rcos^2theta+sintheta} = 0.$$
Having gotten the exact results for whatever the substitution is made (such as $y = x, y = x^2, [x = t^2, y = t])$, my conclusion is that limit does exist and equals $0.$



Did I miss something?






limits multivariable-calculus







share|cite|improve this question









$endgroup$




Show whether the limit exists and find it, or prove that it does not.
$$lim_{(x, y) to(infty,infty)}frac{x+sqrt{y}}{x^2+y}$$
WolframAlpha shows that limit does not exist, however, I do fail to conclude so.
$$lim_{(x, y) to(infty,infty)}frac{x+sqrt{y}}{x^2+y} = [x=rcostheta, y = rsintheta] = lim_{rtoinfty}frac{rcostheta+sqrt{rsintheta}}{r^2cos^2theta+rsintheta} = lim_{rtoinfty}frac{costhetafrac{sqrt{sintheta}}{sqrt{r}}}{rcos^2theta+sintheta} = 0.$$
Having gotten the exact results for whatever the substitution is made (such as $y = x, y = x^2, [x = t^2, y = t])$, my conclusion is that limit does exist and equals $0.$



Did I miss something?






limits multivariable-calculus




limits multivariable-calculus






share|cite|improve this question













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asked Dec 15 '18 at 12:54









H. BaoH. Bao

474




474












  • $begingroup$
    Can you link to the WA calculation, or provide the exact formula you typed in there?
    $endgroup$
    – Barry Cipra
    Dec 15 '18 at 13:53


















  • $begingroup$
    Can you link to the WA calculation, or provide the exact formula you typed in there?
    $endgroup$
    – Barry Cipra
    Dec 15 '18 at 13:53
















$begingroup$
Can you link to the WA calculation, or provide the exact formula you typed in there?
$endgroup$
– Barry Cipra
Dec 15 '18 at 13:53




$begingroup$
Can you link to the WA calculation, or provide the exact formula you typed in there?
$endgroup$
– Barry Cipra
Dec 15 '18 at 13:53










4 Answers
4






active

oldest

votes


















1












$begingroup$

By $x=u$ and $y=v^2$ the limit becomes



$$lim_{(x, y) to (infty, infty)} frac{x+sqrt{y}}{x^2+y}=lim_{(u,v) to (infty, infty)} frac{u+v}{u^2+v^2}=0$$



indeed for example by polar coordinates



$$frac{u+v}{u^2+v^2}=frac1r(cos theta +sin theta)to 0$$






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    For $x,ygt0$, we have



    $$begin{align}
    {x+sqrt yover x^2+y}
    &={xover x^2+y}+{sqrt yover x^2+y}\
    &le{xover x^2}+{sqrt yover y}\
    &={1over x}+{1oversqrt y}\
    &to0+0
    end{align}$$



    The key step here uses the fact that a smaller denominator makes for a larger fraction.






    share|cite|improve this answer











    $endgroup$





















      2












      $begingroup$

      It is enough to observe that, if $ygeq 0$,
      $$
      x^2 + y geq frac{1}{2} (|x| +sqrt{y})^2,
      $$
      so that
      $$
      left|frac{x+sqrt{y}}{x^2+y}right|
      leqfrac{|x|+sqrt{y}}{x^2+y}leq frac{2}{|x|+sqrt{y}}.
      $$






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        Since $(x,y)rightarrow (infty,infty),$ assume $x,y>0.$ $$0<frac{x+sqrt{y}}{x^2+y}=frac{xsqrt {y}left(frac{1}{sqrt y}+frac 1xright)}{xsqrt {y}left(frac{x}{sqrt y}+frac {sqrt y}{x}right)}leq
        frac{frac{1}{sqrt y}+frac 1x}{2} rightarrow 0$$
        as $frac ab + frac ba geq 2$ for $a,b>0.$ Thus by the squeeze theorem is our limit $0.$






        share|cite|improve this answer











        $endgroup$













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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          By $x=u$ and $y=v^2$ the limit becomes



          $$lim_{(x, y) to (infty, infty)} frac{x+sqrt{y}}{x^2+y}=lim_{(u,v) to (infty, infty)} frac{u+v}{u^2+v^2}=0$$



          indeed for example by polar coordinates



          $$frac{u+v}{u^2+v^2}=frac1r(cos theta +sin theta)to 0$$






          share|cite|improve this answer









          $endgroup$


















            1












            $begingroup$

            By $x=u$ and $y=v^2$ the limit becomes



            $$lim_{(x, y) to (infty, infty)} frac{x+sqrt{y}}{x^2+y}=lim_{(u,v) to (infty, infty)} frac{u+v}{u^2+v^2}=0$$



            indeed for example by polar coordinates



            $$frac{u+v}{u^2+v^2}=frac1r(cos theta +sin theta)to 0$$






            share|cite|improve this answer









            $endgroup$
















              1












              1








              1





              $begingroup$

              By $x=u$ and $y=v^2$ the limit becomes



              $$lim_{(x, y) to (infty, infty)} frac{x+sqrt{y}}{x^2+y}=lim_{(u,v) to (infty, infty)} frac{u+v}{u^2+v^2}=0$$



              indeed for example by polar coordinates



              $$frac{u+v}{u^2+v^2}=frac1r(cos theta +sin theta)to 0$$






              share|cite|improve this answer









              $endgroup$



              By $x=u$ and $y=v^2$ the limit becomes



              $$lim_{(x, y) to (infty, infty)} frac{x+sqrt{y}}{x^2+y}=lim_{(u,v) to (infty, infty)} frac{u+v}{u^2+v^2}=0$$



              indeed for example by polar coordinates



              $$frac{u+v}{u^2+v^2}=frac1r(cos theta +sin theta)to 0$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 15 '18 at 13:46









              gimusigimusi

              92.8k84494




              92.8k84494























                  3












                  $begingroup$

                  For $x,ygt0$, we have



                  $$begin{align}
                  {x+sqrt yover x^2+y}
                  &={xover x^2+y}+{sqrt yover x^2+y}\
                  &le{xover x^2}+{sqrt yover y}\
                  &={1over x}+{1oversqrt y}\
                  &to0+0
                  end{align}$$



                  The key step here uses the fact that a smaller denominator makes for a larger fraction.






                  share|cite|improve this answer











                  $endgroup$


















                    3












                    $begingroup$

                    For $x,ygt0$, we have



                    $$begin{align}
                    {x+sqrt yover x^2+y}
                    &={xover x^2+y}+{sqrt yover x^2+y}\
                    &le{xover x^2}+{sqrt yover y}\
                    &={1over x}+{1oversqrt y}\
                    &to0+0
                    end{align}$$



                    The key step here uses the fact that a smaller denominator makes for a larger fraction.






                    share|cite|improve this answer











                    $endgroup$
















                      3












                      3








                      3





                      $begingroup$

                      For $x,ygt0$, we have



                      $$begin{align}
                      {x+sqrt yover x^2+y}
                      &={xover x^2+y}+{sqrt yover x^2+y}\
                      &le{xover x^2}+{sqrt yover y}\
                      &={1over x}+{1oversqrt y}\
                      &to0+0
                      end{align}$$



                      The key step here uses the fact that a smaller denominator makes for a larger fraction.






                      share|cite|improve this answer











                      $endgroup$



                      For $x,ygt0$, we have



                      $$begin{align}
                      {x+sqrt yover x^2+y}
                      &={xover x^2+y}+{sqrt yover x^2+y}\
                      &le{xover x^2}+{sqrt yover y}\
                      &={1over x}+{1oversqrt y}\
                      &to0+0
                      end{align}$$



                      The key step here uses the fact that a smaller denominator makes for a larger fraction.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 15 '18 at 14:54

























                      answered Dec 15 '18 at 13:52









                      Barry CipraBarry Cipra

                      59.6k653126




                      59.6k653126























                          2












                          $begingroup$

                          It is enough to observe that, if $ygeq 0$,
                          $$
                          x^2 + y geq frac{1}{2} (|x| +sqrt{y})^2,
                          $$
                          so that
                          $$
                          left|frac{x+sqrt{y}}{x^2+y}right|
                          leqfrac{|x|+sqrt{y}}{x^2+y}leq frac{2}{|x|+sqrt{y}}.
                          $$






                          share|cite|improve this answer









                          $endgroup$


















                            2












                            $begingroup$

                            It is enough to observe that, if $ygeq 0$,
                            $$
                            x^2 + y geq frac{1}{2} (|x| +sqrt{y})^2,
                            $$
                            so that
                            $$
                            left|frac{x+sqrt{y}}{x^2+y}right|
                            leqfrac{|x|+sqrt{y}}{x^2+y}leq frac{2}{|x|+sqrt{y}}.
                            $$






                            share|cite|improve this answer









                            $endgroup$
















                              2












                              2








                              2





                              $begingroup$

                              It is enough to observe that, if $ygeq 0$,
                              $$
                              x^2 + y geq frac{1}{2} (|x| +sqrt{y})^2,
                              $$
                              so that
                              $$
                              left|frac{x+sqrt{y}}{x^2+y}right|
                              leqfrac{|x|+sqrt{y}}{x^2+y}leq frac{2}{|x|+sqrt{y}}.
                              $$






                              share|cite|improve this answer









                              $endgroup$



                              It is enough to observe that, if $ygeq 0$,
                              $$
                              x^2 + y geq frac{1}{2} (|x| +sqrt{y})^2,
                              $$
                              so that
                              $$
                              left|frac{x+sqrt{y}}{x^2+y}right|
                              leqfrac{|x|+sqrt{y}}{x^2+y}leq frac{2}{|x|+sqrt{y}}.
                              $$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 15 '18 at 13:14









                              RigelRigel

                              11.2k11320




                              11.2k11320























                                  0












                                  $begingroup$

                                  Since $(x,y)rightarrow (infty,infty),$ assume $x,y>0.$ $$0<frac{x+sqrt{y}}{x^2+y}=frac{xsqrt {y}left(frac{1}{sqrt y}+frac 1xright)}{xsqrt {y}left(frac{x}{sqrt y}+frac {sqrt y}{x}right)}leq
                                  frac{frac{1}{sqrt y}+frac 1x}{2} rightarrow 0$$
                                  as $frac ab + frac ba geq 2$ for $a,b>0.$ Thus by the squeeze theorem is our limit $0.$






                                  share|cite|improve this answer











                                  $endgroup$


















                                    0












                                    $begingroup$

                                    Since $(x,y)rightarrow (infty,infty),$ assume $x,y>0.$ $$0<frac{x+sqrt{y}}{x^2+y}=frac{xsqrt {y}left(frac{1}{sqrt y}+frac 1xright)}{xsqrt {y}left(frac{x}{sqrt y}+frac {sqrt y}{x}right)}leq
                                    frac{frac{1}{sqrt y}+frac 1x}{2} rightarrow 0$$
                                    as $frac ab + frac ba geq 2$ for $a,b>0.$ Thus by the squeeze theorem is our limit $0.$






                                    share|cite|improve this answer











                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      Since $(x,y)rightarrow (infty,infty),$ assume $x,y>0.$ $$0<frac{x+sqrt{y}}{x^2+y}=frac{xsqrt {y}left(frac{1}{sqrt y}+frac 1xright)}{xsqrt {y}left(frac{x}{sqrt y}+frac {sqrt y}{x}right)}leq
                                      frac{frac{1}{sqrt y}+frac 1x}{2} rightarrow 0$$
                                      as $frac ab + frac ba geq 2$ for $a,b>0.$ Thus by the squeeze theorem is our limit $0.$






                                      share|cite|improve this answer











                                      $endgroup$



                                      Since $(x,y)rightarrow (infty,infty),$ assume $x,y>0.$ $$0<frac{x+sqrt{y}}{x^2+y}=frac{xsqrt {y}left(frac{1}{sqrt y}+frac 1xright)}{xsqrt {y}left(frac{x}{sqrt y}+frac {sqrt y}{x}right)}leq
                                      frac{frac{1}{sqrt y}+frac 1x}{2} rightarrow 0$$
                                      as $frac ab + frac ba geq 2$ for $a,b>0.$ Thus by the squeeze theorem is our limit $0.$







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Dec 17 '18 at 3:06

























                                      answered Dec 15 '18 at 13:38









                                      user376343user376343

                                      3,7883827




                                      3,7883827






























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