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?

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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










3












$begingroup$


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
    7 hours ago










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















3












$begingroup$


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
    7 hours ago










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













3












3








3


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 7 hours 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 8 hours ago









radooradoo

184




184




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
    7 hours ago










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
















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










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















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




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












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




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










3 Answers
3






active

oldest

votes


















5












$begingroup$

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









        5












        $begingroup$

        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









        $endgroup$

















          5












          $begingroup$

          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









          $endgroup$















            5












            5








            5





            $begingroup$

            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 7 hours ago









            DashiDashi

            746311




            746311





















                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 7 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 6 hours ago









                            Barry CipraBarry Cipra

                            60.5k655128




                            60.5k655128




















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                                Чепеларе Съдържание География | История | Население | Спортни и природни забележителности | Културни и исторически обекти | Религии | Обществени институции | Известни личности | Редовни събития | Галерия | Източници | Литература | Външни препратки | Навигация41°43′23.99″ с. ш. 24°41′09.99″ и. д. / 41.723333° с. ш. 24.686111° и. д.*ЧепелареЧепеларски Linux fest 2002Начало на Зимен сезон 2005/06Национални хайдушки празници „Капитан Петко Войвода“Град ЧепелареЧепеларе – народният ски курортbgrod.orgwww.terranatura.hit.bgСправка за населението на гр. Исперих, общ. Исперих, обл. РазградМузей на родопския карстМузей на спорта и скитеЧепеларебългарскибългарскианглийскитукИстория на градаСки писти в ЧепелареВремето в ЧепелареРадио и телевизия в ЧепелареЧепеларе мами с родопски чар и добри пистиЕвтин туризъм и снежни атракции в ЧепелареМестоположениеИнформация и снимки от музея на родопския карст3D панорами от ЧепелареЧепелареррр

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                                ЧепелареСки курорти в България градБългарияобласт СмолянПловдивСмолянПампоровообщина ЧепелареРодопитеБългарияДоспатрека ЧепеларсканекрополинеолитакърджалиитеПетко войводаМакедонската организацияВълчо СарафовКонстантин АнтоновПетър КузмановПрогледШирока лъкаМанастирКрънкаОсманската империяпощенска станциятелеграфтуристическопещеренскиорскаМечи чалалпийски скиЯгодинската пещераЕкатерина Дафовскабългари1964град Чепеларе от Уикипедия, свободната енциклопедия Направо към навигацията Направо към търсенето Тази статия е за града . За върха в Антарктика вижте Чепеларе (връх). Чепеларе Чепеларе Координати: 41°43′23.99″ с. ш. 24°41′09.99″ и. д.  /  41.723333° с. ш. 24.686111° и. д. 41.723333 , 24.686111 Чепеларе Чепеларе Общи данни Население 5 310  (ГРАО, 2015-03-15) * 5020 (НСИ) Землище 73,861 km² Надм. височина 1232±1 m Пощ. код 4850 Тел. код 03051 МПС код СМ ЕКАТТЕ 80371 Администрация Държава България Област Смолян Община     - кмет Чепеларе Слав
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