Find existence of limit











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Show that $left(frac{1}{|x|} +frac{1}{|y |}right)$ tends to infinity as $(x,y)$ tends to $(0,0)$.
I have used $x=rcos alpha$ and $y=rsin alpha$, $r>0$.But I got stuck as $left(frac{1}{|cos alpha|} +frac{1}{ |rsin alpha|}right) left(frac{1}{r}right)$ lies between $frac{1}{r}$ and infinity.






calculus







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    up vote
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    Show that $left(frac{1}{|x|} +frac{1}{|y |}right)$ tends to infinity as $(x,y)$ tends to $(0,0)$.
    I have used $x=rcos alpha$ and $y=rsin alpha$, $r>0$.But I got stuck as $left(frac{1}{|cos alpha|} +frac{1}{ |rsin alpha|}right) left(frac{1}{r}right)$ lies between $frac{1}{r}$ and infinity.






    calculus







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      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Show that $left(frac{1}{|x|} +frac{1}{|y |}right)$ tends to infinity as $(x,y)$ tends to $(0,0)$.
      I have used $x=rcos alpha$ and $y=rsin alpha$, $r>0$.But I got stuck as $left(frac{1}{|cos alpha|} +frac{1}{ |rsin alpha|}right) left(frac{1}{r}right)$ lies between $frac{1}{r}$ and infinity.






      calculus







      share|cite|improve this question















      Show that $left(frac{1}{|x|} +frac{1}{|y |}right)$ tends to infinity as $(x,y)$ tends to $(0,0)$.
      I have used $x=rcos alpha$ and $y=rsin alpha$, $r>0$.But I got stuck as $left(frac{1}{|cos alpha|} +frac{1}{ |rsin alpha|}right) left(frac{1}{r}right)$ lies between $frac{1}{r}$ and infinity.






      calculus




      calculus






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      edited 3 hours ago









      user376343

      2,6912821




      2,6912821










      asked 3 hours ago









      Kashmira

      223




      223






















          4 Answers
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          up vote
          5
          down vote













          Note that



          $$frac1{|x|}+frac1{|y|}ge frac1{|x|}to infty$$






          share|cite|improve this answer




























            up vote
            2
            down vote













            Hint:



            Use the fact that if $|(x,y)| < epsilon$, then $|x|<epsilon$ and $|y|<epsilon$






            share|cite|improve this answer




























              up vote
              2
              down vote













              Take the individual limits of each component. x and y are tending towards zero (the direction doesn't matter due to the magnitudes) so $frac{1}{|x|}$ and $frac{1}{|y|}$ both tend to infinity, and as a result so does their sum.






              share|cite|improve this answer








              New contributor




              M.M. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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              • You are right, but a few more words about why taking "individual limits of each component" can replace $(x,y)to (0,0)$ would be helpful in the present circumstance.
                – hardmath
                3 hours ago


















              up vote
              0
              down vote













              We have:



              $(x,y) rightarrow 0.$



              Let $epsilon_n =1/n$, $n$ positive integer, be given.



              Then $|(x^2+y^2)^{1/2}| < delta_n$ $(=epsilon_n)$ implies



              $|(x^2+y^2)^{1/2}| < epsilon_n =1/n$, or



              $n=1/epsilon_n lt dfrac{1}{(x^2+y^2)^{1/2}}.$



              Note:



              $(x^2+y^2)^{1/2} ge |x|$, and
              $(x^2+y^2)^{1/2} ge |y|.$



              Then



              $dfrac{2}{(x^2+y^2)^{1/2}} le dfrac{1}{|x|}+dfrac{1}{|y|}.$



              And finally :



              $|(x^2+y^2)^{1/2}| lt delta_n$ implies



              $2n =2/epsilon_n lt dfrac{2}{(x^2+y^2)^{1/2}} le $



              $dfrac{1}{|x|} +dfrac{1}{|y|}.$






              share|cite|improve this answer























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                4 Answers
                4






                active

                oldest

                votes








                4 Answers
                4






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes








                up vote
                5
                down vote













                Note that



                $$frac1{|x|}+frac1{|y|}ge frac1{|x|}to infty$$






                share|cite|improve this answer

























                  up vote
                  5
                  down vote













                  Note that



                  $$frac1{|x|}+frac1{|y|}ge frac1{|x|}to infty$$






                  share|cite|improve this answer























                    up vote
                    5
                    down vote










                    up vote
                    5
                    down vote









                    Note that



                    $$frac1{|x|}+frac1{|y|}ge frac1{|x|}to infty$$






                    share|cite|improve this answer












                    Note that



                    $$frac1{|x|}+frac1{|y|}ge frac1{|x|}to infty$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 3 hours ago









                    gimusi

                    91.9k84495




                    91.9k84495






















                        up vote
                        2
                        down vote













                        Hint:



                        Use the fact that if $|(x,y)| < epsilon$, then $|x|<epsilon$ and $|y|<epsilon$






                        share|cite|improve this answer

























                          up vote
                          2
                          down vote













                          Hint:



                          Use the fact that if $|(x,y)| < epsilon$, then $|x|<epsilon$ and $|y|<epsilon$






                          share|cite|improve this answer























                            up vote
                            2
                            down vote










                            up vote
                            2
                            down vote









                            Hint:



                            Use the fact that if $|(x,y)| < epsilon$, then $|x|<epsilon$ and $|y|<epsilon$






                            share|cite|improve this answer












                            Hint:



                            Use the fact that if $|(x,y)| < epsilon$, then $|x|<epsilon$ and $|y|<epsilon$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 3 hours ago









                            5xum

                            89.4k393161




                            89.4k393161






















                                up vote
                                2
                                down vote













                                Take the individual limits of each component. x and y are tending towards zero (the direction doesn't matter due to the magnitudes) so $frac{1}{|x|}$ and $frac{1}{|y|}$ both tend to infinity, and as a result so does their sum.






                                share|cite|improve this answer








                                New contributor




                                M.M. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.


















                                • You are right, but a few more words about why taking "individual limits of each component" can replace $(x,y)to (0,0)$ would be helpful in the present circumstance.
                                  – hardmath
                                  3 hours ago















                                up vote
                                2
                                down vote













                                Take the individual limits of each component. x and y are tending towards zero (the direction doesn't matter due to the magnitudes) so $frac{1}{|x|}$ and $frac{1}{|y|}$ both tend to infinity, and as a result so does their sum.






                                share|cite|improve this answer








                                New contributor




                                M.M. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.


















                                • You are right, but a few more words about why taking "individual limits of each component" can replace $(x,y)to (0,0)$ would be helpful in the present circumstance.
                                  – hardmath
                                  3 hours ago













                                up vote
                                2
                                down vote










                                up vote
                                2
                                down vote









                                Take the individual limits of each component. x and y are tending towards zero (the direction doesn't matter due to the magnitudes) so $frac{1}{|x|}$ and $frac{1}{|y|}$ both tend to infinity, and as a result so does their sum.






                                share|cite|improve this answer








                                New contributor




                                M.M. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.









                                Take the individual limits of each component. x and y are tending towards zero (the direction doesn't matter due to the magnitudes) so $frac{1}{|x|}$ and $frac{1}{|y|}$ both tend to infinity, and as a result so does their sum.







                                share|cite|improve this answer








                                New contributor




                                M.M. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.









                                share|cite|improve this answer



                                share|cite|improve this answer






                                New contributor




                                M.M. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                                answered 3 hours ago









                                M.M.

                                212




                                212




                                New contributor




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                                Check out our Code of Conduct.





                                New contributor





                                M.M. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.






                                M.M. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                Check out our Code of Conduct.












                                • You are right, but a few more words about why taking "individual limits of each component" can replace $(x,y)to (0,0)$ would be helpful in the present circumstance.
                                  – hardmath
                                  3 hours ago


















                                • You are right, but a few more words about why taking "individual limits of each component" can replace $(x,y)to (0,0)$ would be helpful in the present circumstance.
                                  – hardmath
                                  3 hours ago
















                                You are right, but a few more words about why taking "individual limits of each component" can replace $(x,y)to (0,0)$ would be helpful in the present circumstance.
                                – hardmath
                                3 hours ago




                                You are right, but a few more words about why taking "individual limits of each component" can replace $(x,y)to (0,0)$ would be helpful in the present circumstance.
                                – hardmath
                                3 hours ago










                                up vote
                                0
                                down vote













                                We have:



                                $(x,y) rightarrow 0.$



                                Let $epsilon_n =1/n$, $n$ positive integer, be given.



                                Then $|(x^2+y^2)^{1/2}| < delta_n$ $(=epsilon_n)$ implies



                                $|(x^2+y^2)^{1/2}| < epsilon_n =1/n$, or



                                $n=1/epsilon_n lt dfrac{1}{(x^2+y^2)^{1/2}}.$



                                Note:



                                $(x^2+y^2)^{1/2} ge |x|$, and
                                $(x^2+y^2)^{1/2} ge |y|.$



                                Then



                                $dfrac{2}{(x^2+y^2)^{1/2}} le dfrac{1}{|x|}+dfrac{1}{|y|}.$



                                And finally :



                                $|(x^2+y^2)^{1/2}| lt delta_n$ implies



                                $2n =2/epsilon_n lt dfrac{2}{(x^2+y^2)^{1/2}} le $



                                $dfrac{1}{|x|} +dfrac{1}{|y|}.$






                                share|cite|improve this answer



























                                  up vote
                                  0
                                  down vote













                                  We have:



                                  $(x,y) rightarrow 0.$



                                  Let $epsilon_n =1/n$, $n$ positive integer, be given.



                                  Then $|(x^2+y^2)^{1/2}| < delta_n$ $(=epsilon_n)$ implies



                                  $|(x^2+y^2)^{1/2}| < epsilon_n =1/n$, or



                                  $n=1/epsilon_n lt dfrac{1}{(x^2+y^2)^{1/2}}.$



                                  Note:



                                  $(x^2+y^2)^{1/2} ge |x|$, and
                                  $(x^2+y^2)^{1/2} ge |y|.$



                                  Then



                                  $dfrac{2}{(x^2+y^2)^{1/2}} le dfrac{1}{|x|}+dfrac{1}{|y|}.$



                                  And finally :



                                  $|(x^2+y^2)^{1/2}| lt delta_n$ implies



                                  $2n =2/epsilon_n lt dfrac{2}{(x^2+y^2)^{1/2}} le $



                                  $dfrac{1}{|x|} +dfrac{1}{|y|}.$






                                  share|cite|improve this answer

























                                    up vote
                                    0
                                    down vote










                                    up vote
                                    0
                                    down vote









                                    We have:



                                    $(x,y) rightarrow 0.$



                                    Let $epsilon_n =1/n$, $n$ positive integer, be given.



                                    Then $|(x^2+y^2)^{1/2}| < delta_n$ $(=epsilon_n)$ implies



                                    $|(x^2+y^2)^{1/2}| < epsilon_n =1/n$, or



                                    $n=1/epsilon_n lt dfrac{1}{(x^2+y^2)^{1/2}}.$



                                    Note:



                                    $(x^2+y^2)^{1/2} ge |x|$, and
                                    $(x^2+y^2)^{1/2} ge |y|.$



                                    Then



                                    $dfrac{2}{(x^2+y^2)^{1/2}} le dfrac{1}{|x|}+dfrac{1}{|y|}.$



                                    And finally :



                                    $|(x^2+y^2)^{1/2}| lt delta_n$ implies



                                    $2n =2/epsilon_n lt dfrac{2}{(x^2+y^2)^{1/2}} le $



                                    $dfrac{1}{|x|} +dfrac{1}{|y|}.$






                                    share|cite|improve this answer














                                    We have:



                                    $(x,y) rightarrow 0.$



                                    Let $epsilon_n =1/n$, $n$ positive integer, be given.



                                    Then $|(x^2+y^2)^{1/2}| < delta_n$ $(=epsilon_n)$ implies



                                    $|(x^2+y^2)^{1/2}| < epsilon_n =1/n$, or



                                    $n=1/epsilon_n lt dfrac{1}{(x^2+y^2)^{1/2}}.$



                                    Note:



                                    $(x^2+y^2)^{1/2} ge |x|$, and
                                    $(x^2+y^2)^{1/2} ge |y|.$



                                    Then



                                    $dfrac{2}{(x^2+y^2)^{1/2}} le dfrac{1}{|x|}+dfrac{1}{|y|}.$



                                    And finally :



                                    $|(x^2+y^2)^{1/2}| lt delta_n$ implies



                                    $2n =2/epsilon_n lt dfrac{2}{(x^2+y^2)^{1/2}} le $



                                    $dfrac{1}{|x|} +dfrac{1}{|y|}.$







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited 1 hour ago

























                                    answered 1 hour ago









                                    Peter Szilas

                                    10.5k2720




                                    10.5k2720






























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