Find existence of limit












3














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 limits







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  • 3




    "I got stuck as" has a mistake with the $r$ in the denominator.
    – Teepeemm
    Dec 13 at 20:31
















3














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 limits







share|cite|improve this question




















  • 3




    "I got stuck as" has a mistake with the $r$ in the denominator.
    – Teepeemm
    Dec 13 at 20:31














3












3








3







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 limits







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 limits




calculus limits






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edited Dec 16 at 11:54









Jam

4,94711431




4,94711431










asked Dec 13 at 14:32









Kashmira

463




463








  • 3




    "I got stuck as" has a mistake with the $r$ in the denominator.
    – Teepeemm
    Dec 13 at 20:31














  • 3




    "I got stuck as" has a mistake with the $r$ in the denominator.
    – Teepeemm
    Dec 13 at 20:31








3




3




"I got stuck as" has a mistake with the $r$ in the denominator.
– Teepeemm
Dec 13 at 20:31




"I got stuck as" has a mistake with the $r$ in the denominator.
– Teepeemm
Dec 13 at 20:31










4 Answers
4






active

oldest

votes


















8














Note that



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






share|cite|improve this answer





























    4














    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.



    EDIT:
    As suggested in the comments, this is an expansion as to why you can seperate the limits.



    $$lim_{(x,y)to(0,0)}{bigg(frac{1}{|x|}+frac{1}{|y|}bigg)}=lim_{(x,y)to(0,0)}{bigg(frac{1}{|x|}bigg)}+lim_{(x,y)to(0,0)}{bigg(frac{1}{|y|}bigg)}$$



    Examining the LHS we see that the first limit is independent of y, and the second independent of x. As a result we can rewrite the equation as follows:



    $$=lim_{xto0}{bigg(frac{1}{|x|}bigg)}+lim_{yto0}{bigg(frac{1}{|y|}bigg)}$$



    As stated above, the magnitudes mean that both limits tend to infinity as x and y tend to zero from either direction, so their sum is necessarily also infinity.






    share|cite|improve this answer



















    • 2




      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
      Dec 13 at 14:44










    • Thank you for the advice, I'll add an edit expanding on that.
      – M.M.
      Dec 14 at 10:06



















    2














    Hint:



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






    share|cite|improve this answer





























      1














      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









        8














        Note that



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






        share|cite|improve this answer


























          8














          Note that



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






          share|cite|improve this answer
























            8












            8








            8






            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 Dec 13 at 14:37









            gimusi

            1




            1























                4














                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.



                EDIT:
                As suggested in the comments, this is an expansion as to why you can seperate the limits.



                $$lim_{(x,y)to(0,0)}{bigg(frac{1}{|x|}+frac{1}{|y|}bigg)}=lim_{(x,y)to(0,0)}{bigg(frac{1}{|x|}bigg)}+lim_{(x,y)to(0,0)}{bigg(frac{1}{|y|}bigg)}$$



                Examining the LHS we see that the first limit is independent of y, and the second independent of x. As a result we can rewrite the equation as follows:



                $$=lim_{xto0}{bigg(frac{1}{|x|}bigg)}+lim_{yto0}{bigg(frac{1}{|y|}bigg)}$$



                As stated above, the magnitudes mean that both limits tend to infinity as x and y tend to zero from either direction, so their sum is necessarily also infinity.






                share|cite|improve this answer



















                • 2




                  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
                  Dec 13 at 14:44










                • Thank you for the advice, I'll add an edit expanding on that.
                  – M.M.
                  Dec 14 at 10:06
















                4














                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.



                EDIT:
                As suggested in the comments, this is an expansion as to why you can seperate the limits.



                $$lim_{(x,y)to(0,0)}{bigg(frac{1}{|x|}+frac{1}{|y|}bigg)}=lim_{(x,y)to(0,0)}{bigg(frac{1}{|x|}bigg)}+lim_{(x,y)to(0,0)}{bigg(frac{1}{|y|}bigg)}$$



                Examining the LHS we see that the first limit is independent of y, and the second independent of x. As a result we can rewrite the equation as follows:



                $$=lim_{xto0}{bigg(frac{1}{|x|}bigg)}+lim_{yto0}{bigg(frac{1}{|y|}bigg)}$$



                As stated above, the magnitudes mean that both limits tend to infinity as x and y tend to zero from either direction, so their sum is necessarily also infinity.






                share|cite|improve this answer



















                • 2




                  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
                  Dec 13 at 14:44










                • Thank you for the advice, I'll add an edit expanding on that.
                  – M.M.
                  Dec 14 at 10:06














                4












                4








                4






                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.



                EDIT:
                As suggested in the comments, this is an expansion as to why you can seperate the limits.



                $$lim_{(x,y)to(0,0)}{bigg(frac{1}{|x|}+frac{1}{|y|}bigg)}=lim_{(x,y)to(0,0)}{bigg(frac{1}{|x|}bigg)}+lim_{(x,y)to(0,0)}{bigg(frac{1}{|y|}bigg)}$$



                Examining the LHS we see that the first limit is independent of y, and the second independent of x. As a result we can rewrite the equation as follows:



                $$=lim_{xto0}{bigg(frac{1}{|x|}bigg)}+lim_{yto0}{bigg(frac{1}{|y|}bigg)}$$



                As stated above, the magnitudes mean that both limits tend to infinity as x and y tend to zero from either direction, so their sum is necessarily also infinity.






                share|cite|improve this answer














                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.



                EDIT:
                As suggested in the comments, this is an expansion as to why you can seperate the limits.



                $$lim_{(x,y)to(0,0)}{bigg(frac{1}{|x|}+frac{1}{|y|}bigg)}=lim_{(x,y)to(0,0)}{bigg(frac{1}{|x|}bigg)}+lim_{(x,y)to(0,0)}{bigg(frac{1}{|y|}bigg)}$$



                Examining the LHS we see that the first limit is independent of y, and the second independent of x. As a result we can rewrite the equation as follows:



                $$=lim_{xto0}{bigg(frac{1}{|x|}bigg)}+lim_{yto0}{bigg(frac{1}{|y|}bigg)}$$



                As stated above, the magnitudes mean that both limits tend to infinity as x and y tend to zero from either direction, so their sum is necessarily also infinity.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 14 at 10:12

























                answered Dec 13 at 14:35









                M.M.

                637




                637








                • 2




                  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
                  Dec 13 at 14:44










                • Thank you for the advice, I'll add an edit expanding on that.
                  – M.M.
                  Dec 14 at 10:06














                • 2




                  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
                  Dec 13 at 14:44










                • Thank you for the advice, I'll add an edit expanding on that.
                  – M.M.
                  Dec 14 at 10:06








                2




                2




                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
                Dec 13 at 14:44




                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
                Dec 13 at 14:44












                Thank you for the advice, I'll add an edit expanding on that.
                – M.M.
                Dec 14 at 10:06




                Thank you for the advice, I'll add an edit expanding on that.
                – M.M.
                Dec 14 at 10:06











                2














                Hint:



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






                share|cite|improve this answer


























                  2














                  Hint:



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






                  share|cite|improve this answer
























                    2












                    2








                    2






                    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 Dec 13 at 14:34









                    5xum

                    89.5k393161




                    89.5k393161























                        1














                        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




























                          1














                          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


























                            1












                            1








                            1






                            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 Dec 13 at 17:05

























                            answered Dec 13 at 16:57









                            Peter Szilas

                            10.6k2720




                            10.6k2720






























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                                Алькесар

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                                Муниципалитет Алькесар Alquézar Герб 42°10′16″ с. ш. 0°01′27″ в. д. H G Я O L Страна Испания   Испания Автономное сообщество Арагон Провинция Уэска Комарка Сомонтано-де-Барбастро Алькальд Хосе Марияно Альтемир Ласкорс (ИСРП) История и география Площадь 32,36 км² Высота НУМ 660 м Часовой пояс UTC+1, летом UTC+2 Население Население 313 человек ( 2010 ) Плотность 9,92 чел./км² Этнохороним alquezarano/-a Цифровые идентификаторы Телефонный код +34 974 Почтовые индексы 22112 Автомобильный код HU alquezar.org   (исп.) Показать/скрыть карты Алькесар Алькесар Алькесар  Аудио, фото и видео на Викискладе Алькесар (исп.  Alquézar ) — муниципалитет в Испании, входит в провинцию Уэска, в составе автономного сообщества Арагон. Муниципалитет находится в составе района (комарки) Сомонтано-де-Барбастро. Занимает площадь 32,36 км². Население — 321 челове...
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