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

Trouble understanding the speech of overseas colleagues

Is it okay to store user locations?

How to use tikz in fbox?

% symbol leads to superlong (forever?) compilations

Why didn't Theresa May consult with Parliament before negotiating a deal with the EU?

Rotate a column

What's the point of interval inversion?

Return the Closest Prime Number

Can a single photon have an energy density?

If the heap is initialized for security, then why is the stack uninitialized?

I believe this to be a fraud - hired, then asked to cash check and send cash as Bitcoin

What size rim is OK?

Return of the Riley Riddles in Reverse

How do I construct this japanese bowl?

Opposite of a diet

When airplanes disconnect from a tanker during air to air refueling, why do they bank so sharply to the right?

How can I get through very long and very dry, but also very useful technical documents when learning a new tool?

How do scammers retract money, while you can’t?

Anatomically Correct Mesopelagic Aves

Anatomically Correct Strange Women In Ponds Distributing Swords

Why is there a PLL in CPU?

How do I get the green key off the shelf in the Dobby level of Lego Harry Potter 2?

How to start emacs in "nothing" mode (`fundamental-mode`)

How easy is it to start Magic from scratch?



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
    1 hour 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
    1 hour 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.
Check out our Code of Conduct.





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
    1 hour 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
    1 hour 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
1 hour ago




$begingroup$
You can take the logarithm of the function and use L'Hopitals rule
$endgroup$
– Nimish
1 hour 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









      $endgroup$













        Your Answer





        StackExchange.ifUsing("editor", function ()
        return StackExchange.using("mathjaxEditing", function ()
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        );
        );
        , "mathjax-editing");

        StackExchange.ready(function()
        var channelOptions =
        tags: "".split(" "),
        id: "69"
        ;
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function()
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled)
        StackExchange.using("snippets", function()
        createEditor();
        );

        else
        createEditor();

        );

        function createEditor()
        StackExchange.prepareEditor(
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader:
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        ,
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        );



        );






        radoo is a new contributor. Be nice, and check out our Code of Conduct.









        draft saved

        draft discarded


















        StackExchange.ready(
        function ()
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3166364%2fhow-do-i-solve-this-limit%23new-answer', 'question_page');

        );

        Post as a guest















        Required, but never shown

























        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 1 hour 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 41 mins ago









                            Barry CipraBarry Cipra

                            60.5k655128




                            60.5k655128




















                                radoo is a new contributor. Be nice, and check out our Code of Conduct.









                                draft saved

                                draft discarded


















                                radoo is a new contributor. Be nice, and check out our Code of Conduct.












                                radoo is a new contributor. Be nice, and check out our Code of Conduct.











                                radoo is a new contributor. Be nice, and check out our Code of Conduct.














                                Thanks for contributing an answer to Mathematics Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid …


                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.

                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function ()
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3166364%2fhow-do-i-solve-this-limit%23new-answer', 'question_page');

                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                -

                                Popular posts from this blog

                                Станькаў Зьмест Назва | Гісторыя | Насельніцтва | Турыстычная інфармацыя | Галерэя | Асобы | Заўвагі | Крыніцы | Літаратура | Вонкавыя спасылкі | Навігацыйнае мэню53°37′48″ пн. ш. 27°13′40″ у. д. / 53.63° пн. ш. 27.22778° у. д. / 53.63; 27.2277853°37′48″ пн. ш. 27°13′40″ у. д. / 53.63° пн. ш. 27.22778° у. д. / 53.63; 27.22778djvu"Бабёнка – это я понимаю!". Як русіфікавалі тапонімы.Святыя руіны ажываюць210Станькаўв. СтанькаваСтаньково: история и новости - сайт деревниСтанькава: сядзіба Гутэн-Чапскіхррр

                                Image
                                Станькаўскі сельсаветНаселеныя пункты Койданаўскага раёнуНаселеныя пункты, згаданыя ў XV стагодзьдзі вёскаБеларусіРапусесельсаветуКойданаўскага раёнуМенска...
                                Read more

                                J. J. Abrams Índice Traxectoria | Filmografía | Premios | Notas | Véxase tamén | Menú de navegacióne"J.J. Abrams: Biography"Arquivado"'Star Trek' sequel on track"Arquivado"J.J. Abrams Producing Samurai Jack Movie"Arquivado"EXCLUSIVE: J.J. Abrams Goes Into Warp Speed with Star Trek and Beyond"Arquivado"David Semel To Direct Jonah Nolan/J.J. Abrams' CBS Pilot 'Person Of Interest'"Arquivado"Fox orders J.J. Abrams pilot 'Alcatraz'"ArquivadoJ. J. AbramsJ. J. AbramsWorldCat81800131p24091041000XX116709414031616ma11226833654496ID052246713376222X511412nm00091900000 0001 1772 5428no98124254ID0000002883100650044xx0054597000141374297344064w64f2mjx14255303415344

                                Image
                                Creadores de LostDirectores de LostProdutores de LostActores de cine dos Estados Unidos de AméricaDirectores de cine dos Estados Unidos de AméricaGuionistas dos Estados Unidos de AméricaNados en Nova YorkNados en 1966 Nova York27 de xuño1966escritordirectoractorcompositorxudeuestadounidenseÁnxelesCalifornia (function()var node=document.getElementById("mw-dismissablenotice-anonplace");if(node)node.outerHTML="u003Cdiv class="mw-dismissable-notice"u003Eu003Cdiv class="mw-dismissable-notice-close"u003E[u003Ca tabindex="0" role="button"u003Eagocharu003C/au003E]u003C/divu003Eu003Cdiv class="mw-dismissable-notice-body"u003Eu003Cdiv id="localNotice" lang="gl" dir="ltr"u003Eu003Ctable class="plainlinks fmbox fmbox-system plainlinks" style="background:#fffdf5; border:1px solid #ffc5a6; margin:1em auto 1em auto; padding: 0.5ex; text-align: center;"u003Enu003Ctbodyu003Eu003Ctru0...
                                Read more

                                Першая сусьветная вайна Зьмест Чыньнікі | Асноўныя падзеі | Беларусь у гады вайны | Балянс вайны | Глядзіце таксама | Крыніцы | Вонкавыя спасылкі | Навігацыйнае мэнюПершая сусьветная вайна: памяць павінна жыць124 (27489)41990-763xЗападный мир отмечает 90 лет со дня окончания Первой мировой войныПершая сусьветная вайнаСпадчына Першай сусьветнай вайныГісторыя вайныАрхіў дакумэнтаў вайныМапы Эўропы ў ходзе вайныГісторыя пад знакам Пагоні/Першая сусветная вайна і Беларусь

                                Image
                                Першая сусьветная вайна вайна19141918АнтантайВялікабрытаніяФранцыяРасеяСэрбіяЯпоніяІталія1915ЗШАЦэнтральнымі дзяржаваміАўстра-ВугоршчынаНямеччынаТурэччынаБаў...
                                Read more