How do I solve this limit? The Next CEO of Stack Overflowrational limit problemHow to evaluate the following limit? $limlimits_xtoinftyxleft(fracpi2-arctan xright).$Limit only using squeeze theorem $ lim_(x,y)to(0,2) x,arctanleft(frac1y-2right)$How to calculate this limit at infinity?How can I use the Limit Laws to solve this limit?How do I even start approaching this limit?how can I find this limitHow to solve this limit without L'Hospital?How to solve this limit when direct substitution fails. Why do this work?L'Hôpital's rule - How solve this limit question

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How do I solve this limit?



How do I solve this limit?



The Next CEO of Stack Overflowrational limit problemHow to evaluate the following limit? $limlimits_xtoinftyxleft(fracpi2-arctan xright).$Limit only using squeeze theorem $ lim_(x,y)to(0,2) x,arctanleft(frac1y-2right)$How to calculate this limit at infinity?How can I use the Limit Laws to solve this limit?How do I even start approaching this limit?how can I find this limitHow to solve this limit without L'Hospital?How to solve this limit when direct substitution fails. Why do this work?L'Hôpital's rule - How solve this limit question










1












$begingroup$


How do I solve:



$$
lim_x to 0left[1 - xarctanleft(nxright)right]^, 1/x^2
$$



I know the answer is $e^-n$ for every n > 1, but for the life of me I have no idea how to actually get to that answer. Am I supposed to use a common limit or any theorem? Also, why isn't the limit equal to 1?



Thanks.






calculus limits







share|cite|improve this question









New contributor




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







$endgroup$











  • $begingroup$
    I accidentally put k instead of n, sorry.
    $endgroup$
    – radoo
    2 hours ago










  • $begingroup$
    You can take the logarithm of the function and use L'Hopitals rule
    $endgroup$
    – Nimish
    2 hours ago















1












$begingroup$


How do I solve:



$$
lim_x to 0left[1 - xarctanleft(nxright)right]^, 1/x^2
$$



I know the answer is $e^-n$ for every n > 1, but for the life of me I have no idea how to actually get to that answer. Am I supposed to use a common limit or any theorem? Also, why isn't the limit equal to 1?



Thanks.






calculus limits







share|cite|improve this question









New contributor




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







$endgroup$











  • $begingroup$
    I accidentally put k instead of n, sorry.
    $endgroup$
    – radoo
    2 hours ago










  • $begingroup$
    You can take the logarithm of the function and use L'Hopitals rule
    $endgroup$
    – Nimish
    2 hours ago













1












1








1





$begingroup$


How do I solve:



$$
lim_x to 0left[1 - xarctanleft(nxright)right]^, 1/x^2
$$



I know the answer is $e^-n$ for every n > 1, but for the life of me I have no idea how to actually get to that answer. Am I supposed to use a common limit or any theorem? Also, why isn't the limit equal to 1?



Thanks.






calculus limits







share|cite|improve this question









New contributor




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







$endgroup$




How do I solve:



$$
lim_x to 0left[1 - xarctanleft(nxright)right]^, 1/x^2
$$



I know the answer is $e^-n$ for every n > 1, but for the life of me I have no idea how to actually get to that answer. Am I supposed to use a common limit or any theorem? Also, why isn't the limit equal to 1?



Thanks.






calculus limits




calculus limits






share|cite|improve this question









New contributor




radoo 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 question









New contributor




radoo 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 question




share|cite|improve this question








edited 1 hour ago









Felix Marin

68.8k7109146




68.8k7109146






New contributor




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









asked 2 hours ago









radooradoo

84




84




New contributor




radoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





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






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











  • $begingroup$
    I accidentally put k instead of n, sorry.
    $endgroup$
    – radoo
    2 hours ago










  • $begingroup$
    You can take the logarithm of the function and use L'Hopitals rule
    $endgroup$
    – Nimish
    2 hours ago
















  • $begingroup$
    I accidentally put k instead of n, sorry.
    $endgroup$
    – radoo
    2 hours ago










  • $begingroup$
    You can take the logarithm of the function and use L'Hopitals rule
    $endgroup$
    – Nimish
    2 hours ago















$begingroup$
I accidentally put k instead of n, sorry.
$endgroup$
– radoo
2 hours ago




$begingroup$
I accidentally put k instead of n, sorry.
$endgroup$
– radoo
2 hours ago












$begingroup$
You can take the logarithm of the function and use L'Hopitals rule
$endgroup$
– Nimish
2 hours ago




$begingroup$
You can take the logarithm of the function and use L'Hopitals rule
$endgroup$
– Nimish
2 hours ago










3 Answers
3






active

oldest

votes


















3












$begingroup$

You can do it the following way:



$$
lim_xto 0 (1-xarctan(nx))^1/x^2 = lim_xto 0 e^log((1-xarctan(nx))^1/x^2)
$$



doing some algebra on the exponent:
$$
lim_xto 0 exp(log((1-xarctan(nx))^1/x^2)) = lim_xto 0 expleft(fraclog((1-xarctan(nx))x^2)right)
$$



by limit rules:



$$
lim_xto 0 expleft(fraclog((1-xarctan(nx))x^2)right) =expleft[ lim_xto 0 left(fraclog((1-xarctan(nx))x^2)right)right]
$$



apply L'Hospital's rule and after some algebra you should get:
$$
expleft[- fracn+lim_xto 0 fracarctan(nx)x+n^2lim_xto 0 xarctan(nx)2right]
$$



simplifying



$$
expleft(-fracn+n+n^22right) = e^-n
$$






share|cite|improve this answer









$endgroup$




















    2












    $begingroup$

    Hint: $lim_xto 0 (1-nx)^(1/x)=e^-n$



    Also look at the Taylor expansion of the arctan






    share|cite|improve this answer









    $endgroup$




















      0












      $begingroup$

      Noting that $u=xarctan nxto0$ and $(arctan nx)/xto n$ as $xto 0$, we have



      $$(1-xarctan nx)^1/x^2=((1-xarctan nx)^1/(xarctan nx))^(arctan nx)/x=((1-u)^1/u)^(arctan nx)/xto (e^-1)^n=e^-n$$



      using the general limit property $lim f(x)^g(x)=(lim f(x))^lim g(x)$, provided $lim f(x)$ and $lim g(x)$ both exist (with $lim f(x)ge0$) and are not both $0$.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        You can do it the following way:



        $$
        lim_xto 0 (1-xarctan(nx))^1/x^2 = lim_xto 0 e^log((1-xarctan(nx))^1/x^2)
        $$



        doing some algebra on the exponent:
        $$
        lim_xto 0 exp(log((1-xarctan(nx))^1/x^2)) = lim_xto 0 expleft(fraclog((1-xarctan(nx))x^2)right)
        $$



        by limit rules:



        $$
        lim_xto 0 expleft(fraclog((1-xarctan(nx))x^2)right) =expleft[ lim_xto 0 left(fraclog((1-xarctan(nx))x^2)right)right]
        $$



        apply L'Hospital's rule and after some algebra you should get:
        $$
        expleft[- fracn+lim_xto 0 fracarctan(nx)x+n^2lim_xto 0 xarctan(nx)2right]
        $$



        simplifying



        $$
        expleft(-fracn+n+n^22right) = e^-n
        $$






        share|cite|improve this answer









        $endgroup$

















          3












          $begingroup$

          You can do it the following way:



          $$
          lim_xto 0 (1-xarctan(nx))^1/x^2 = lim_xto 0 e^log((1-xarctan(nx))^1/x^2)
          $$



          doing some algebra on the exponent:
          $$
          lim_xto 0 exp(log((1-xarctan(nx))^1/x^2)) = lim_xto 0 expleft(fraclog((1-xarctan(nx))x^2)right)
          $$



          by limit rules:



          $$
          lim_xto 0 expleft(fraclog((1-xarctan(nx))x^2)right) =expleft[ lim_xto 0 left(fraclog((1-xarctan(nx))x^2)right)right]
          $$



          apply L'Hospital's rule and after some algebra you should get:
          $$
          expleft[- fracn+lim_xto 0 fracarctan(nx)x+n^2lim_xto 0 xarctan(nx)2right]
          $$



          simplifying



          $$
          expleft(-fracn+n+n^22right) = e^-n
          $$






          share|cite|improve this answer









          $endgroup$















            3












            3








            3





            $begingroup$

            You can do it the following way:



            $$
            lim_xto 0 (1-xarctan(nx))^1/x^2 = lim_xto 0 e^log((1-xarctan(nx))^1/x^2)
            $$



            doing some algebra on the exponent:
            $$
            lim_xto 0 exp(log((1-xarctan(nx))^1/x^2)) = lim_xto 0 expleft(fraclog((1-xarctan(nx))x^2)right)
            $$



            by limit rules:



            $$
            lim_xto 0 expleft(fraclog((1-xarctan(nx))x^2)right) =expleft[ lim_xto 0 left(fraclog((1-xarctan(nx))x^2)right)right]
            $$



            apply L'Hospital's rule and after some algebra you should get:
            $$
            expleft[- fracn+lim_xto 0 fracarctan(nx)x+n^2lim_xto 0 xarctan(nx)2right]
            $$



            simplifying



            $$
            expleft(-fracn+n+n^22right) = e^-n
            $$






            share|cite|improve this answer









            $endgroup$



            You can do it the following way:



            $$
            lim_xto 0 (1-xarctan(nx))^1/x^2 = lim_xto 0 e^log((1-xarctan(nx))^1/x^2)
            $$



            doing some algebra on the exponent:
            $$
            lim_xto 0 exp(log((1-xarctan(nx))^1/x^2)) = lim_xto 0 expleft(fraclog((1-xarctan(nx))x^2)right)
            $$



            by limit rules:



            $$
            lim_xto 0 expleft(fraclog((1-xarctan(nx))x^2)right) =expleft[ lim_xto 0 left(fraclog((1-xarctan(nx))x^2)right)right]
            $$



            apply L'Hospital's rule and after some algebra you should get:
            $$
            expleft[- fracn+lim_xto 0 fracarctan(nx)x+n^2lim_xto 0 xarctan(nx)2right]
            $$



            simplifying



            $$
            expleft(-fracn+n+n^22right) = e^-n
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            DashiDashi

            726311




            726311





















                2












                $begingroup$

                Hint: $lim_xto 0 (1-nx)^(1/x)=e^-n$



                Also look at the Taylor expansion of the arctan






                share|cite|improve this answer









                $endgroup$

















                  2












                  $begingroup$

                  Hint: $lim_xto 0 (1-nx)^(1/x)=e^-n$



                  Also look at the Taylor expansion of the arctan






                  share|cite|improve this answer









                  $endgroup$















                    2












                    2








                    2





                    $begingroup$

                    Hint: $lim_xto 0 (1-nx)^(1/x)=e^-n$



                    Also look at the Taylor expansion of the arctan






                    share|cite|improve this answer









                    $endgroup$



                    Hint: $lim_xto 0 (1-nx)^(1/x)=e^-n$



                    Also look at the Taylor expansion of the arctan







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 hours ago









                    A. PA. P

                    1386




                    1386





















                        0












                        $begingroup$

                        Noting that $u=xarctan nxto0$ and $(arctan nx)/xto n$ as $xto 0$, we have



                        $$(1-xarctan nx)^1/x^2=((1-xarctan nx)^1/(xarctan nx))^(arctan nx)/x=((1-u)^1/u)^(arctan nx)/xto (e^-1)^n=e^-n$$



                        using the general limit property $lim f(x)^g(x)=(lim f(x))^lim g(x)$, provided $lim f(x)$ and $lim g(x)$ both exist (with $lim f(x)ge0$) and are not both $0$.






                        share|cite|improve this answer









                        $endgroup$

















                          0












                          $begingroup$

                          Noting that $u=xarctan nxto0$ and $(arctan nx)/xto n$ as $xto 0$, we have



                          $$(1-xarctan nx)^1/x^2=((1-xarctan nx)^1/(xarctan nx))^(arctan nx)/x=((1-u)^1/u)^(arctan nx)/xto (e^-1)^n=e^-n$$



                          using the general limit property $lim f(x)^g(x)=(lim f(x))^lim g(x)$, provided $lim f(x)$ and $lim g(x)$ both exist (with $lim f(x)ge0$) and are not both $0$.






                          share|cite|improve this answer









                          $endgroup$















                            0












                            0








                            0





                            $begingroup$

                            Noting that $u=xarctan nxto0$ and $(arctan nx)/xto n$ as $xto 0$, we have



                            $$(1-xarctan nx)^1/x^2=((1-xarctan nx)^1/(xarctan nx))^(arctan nx)/x=((1-u)^1/u)^(arctan nx)/xto (e^-1)^n=e^-n$$



                            using the general limit property $lim f(x)^g(x)=(lim f(x))^lim g(x)$, provided $lim f(x)$ and $lim g(x)$ both exist (with $lim f(x)ge0$) and are not both $0$.






                            share|cite|improve this answer









                            $endgroup$



                            Noting that $u=xarctan nxto0$ and $(arctan nx)/xto n$ as $xto 0$, we have



                            $$(1-xarctan nx)^1/x^2=((1-xarctan nx)^1/(xarctan nx))^(arctan nx)/x=((1-u)^1/u)^(arctan nx)/xto (e^-1)^n=e^-n$$



                            using the general limit property $lim f(x)^g(x)=(lim f(x))^lim g(x)$, provided $lim f(x)$ and $lim g(x)$ both exist (with $lim f(x)ge0$) and are not both $0$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 53 mins ago









                            Barry CipraBarry Cipra

                            60.5k655128




                            60.5k655128




















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                                Objetos do disco dispersoObjetos transnetunianos objeto transnetunianodisco dispersoSistema Solarcorpo celesteressonância orbitalplanetaplanetamagnitude absolutadiâmetro9 de abril2013Outer Solar System Origins Surveyórbitaexcentricidadesemieixo maiorUAperiélioSolafélio 2013 GY136 Data da descoberta 9 de abril de 2013 Descoberto por Outer Solar System Origins Survey Categoria Transnetuniano Objeto do disco disperso Elementos orbitais Semieixo maior 55,961 UA Perélio 32,595 UA Afélio 79,327 UA Excentricidade 0,418 Inclinação 10,9° Características físicas Dimensões 133 km Magnitude absoluta 7,6 ver 2013 GY 136 , também escrito como 2013 GY136 , é um objeto transnetuniano que está localizado no disco disperso, uma região do Sistema Solar. Este corpo celeste está em uma ressonância orbital de 2:5 com o planeta planeta. Ele possui uma magnitude absoluta de 7,6 [ 1 ] e tem um diâmetro estimad...
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                                Mortes em março de 2019 Referências Menu de navegação«Zhores Alferov, Nobel de Física bielorrusso, morre aos 88 anos - Ciência»«Fallece Rafael Torija, o bispo emérito de Ciudad Real»«Peter Hurford dies at 88»«Keith Flint, vocalista do The Prodigy, morre aos 49 anos»«Luke Perry, ator de 'Barrados no baile' e 'Riverdale', morre aos 52 anos»«Former Rangers and Scotland captain Eric Caldow dies, aged 84»«Morreu, aos 61 anos, a antiga lenda do wrestling King Kong Bundy»«Fallece el actor y director teatral Abraham Stavans»«In Memoriam Guillaume Faye»«Sidney Sheinberg, a Force Behind Universal and Spielberg, Is Dead at 84»«Carmine Persico, Colombo Crime Family Boss, Is Dead at 85»«Dirigent Michael Gielen gestorben»«Ciclista tricampeã mundial e prata na Rio 2016 é encontrada morta em casa aos 23 anos»«Pagan Community Notes: Raven Grimassi dies, Indianapolis pop-up event cancelled, Circle Sanctuary announces new podcast, and more!»«Hal Blaine, Wrecking Crew Drummer, Dies at 90»«Morre Coutinho, que editou dupla lendária com Pelé no Santos»«Cantor Demétrius, ídolo da Jovem Guarda, morre em SP»«Ex-presidente do Vasco, Eurico Miranda morre no Rio de Janeiro»«Bronze no Mundial de basquete de 1971, Laís Elena morre aos 76 anos»«Diretor de Corridas da F1, Charlie Whiting morre aos 66 anos às vésperas do GP da Austrália»«Morreu o cardeal Danneels, da Bélgica»«Morreu o cartoonista Augusto Cid»«Morreu a atriz Maria Isabel de Lizandra, de "Vale Tudo" e novelas da Tupi»«WS Merwin, prize-winning poet of nature, dies at 91»«Atriz Márcia Real morre em São Paulo aos 88 anos»«Mauritanie: décès de l'ancien président Mohamed Mahmoud ould Louly»«Morreu Dick Dale, o rei da surf guitar e de "Pulp Fiction"»«Falleció Víctor Genes»«João Carlos Marinho, autor de 'O Gênio do Crime', morre em SP»«Legendary Horror Director and SFX Artist John Carl Buechler Dies at 66»«Morre em Salvador a religiosa Makota Valdina»«مرگ بازیکن‌ سابق نساجی بر اثر سقوط سنگ در مازندران»«Domingos Oliveira morre no Rio»«Morre Airton Ravagniani, ex-São Paulo, Fla, Vasco, Grêmio e Sport - Notícias»«Morre o escritor Flavio Moreira da Costa»«Larry Cohen, Writer-Director of 'It's Alive' and 'Hell Up in Harlem,' Dies at 77»«Scott Walker, experimental singer-songwriter, dead at 76»«Joseph Pilato, Day of the Dead Star and Horror Favorite, Dies at 70»«Sheffield United set to pay tribute to legendary goalkeeper Ted Burgin who has died at 91»«Morre Rafael Henzel, sobrevivente de acidente aéreo da Chapecoense»«Morre Valery Bykovsky, um dos primeiros cosmonautas da União Soviética»«Agnès Varda, cineasta da Nouvelle Vague, morre aos 90 anos»«Agnès Varda, cineasta francesa, morre aos 90 anos»«Tania Mallet, James Bond Actress and Helen Mirren's Cousin, Dies at 77»e

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                                Mortos em 2019Eventos de março de 2019 ←JaneiroFevereiroMarçoAbrilMaioJunhoJulhoAgostoSetembroOutubroNovembroDezembro→https://g1.globo.com/pop-arte/noticia/2019/04/01/rapper-nipsey-hussle-e-morto-a-tiros-em-los-angeles.ghtml Mortes em 2019 : ← Janeiro – Fevereiro – Março – Abril – Maio – Junho – Julho – Agosto – Setembro – Outubro – Novembro – Dezembro → Está é uma relação de pessoas notáveis que morreram durante o mês de março de 2019, listando nome, nacionalidade, ocupação e ano de nascimento. Keith Flint , músico inglês. Luke Perry , ator e dublador americano. Coutinho , futebolista brasileiro. Eurico Miranda , dirigente esportivo e político brasileiro. Charlie Whiting , diretor de provas britânico. Valery Bykovsky , cosmonauta russo. Agnès Varda , cineasta belgo-francesa. Dia Nome Profissão ou motivo de...
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