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

How do spells that require an ability check vs. the caster's spell save DC work?

How to use tikz in fbox?

Is a stroke of luck acceptable after a series of unfavorable events?

What is meant by a M next to a roman numeral?

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

How to write papers efficiently when English isn't my first language?

How to make a software documentation "officially" citable?

What size rim is OK?

Why do professional authors make "consistency" mistakes? And how to avoid them?

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

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

Can a caster that cast Polymorph on themselves stop concentrating at any point even if their Int is low?

What is the purpose of the Evocation wizard's Potent Cantrip feature?

Why didn't Khan get resurrected in the Genesis Explosion?

Visit to the USA with ESTA approved before trip to Iran

Anatomically Correct Strange Women In Ponds Distributing Swords

Explicit solution of a Hamiltonian system

Why does standard notation not preserve intervals (visually)

Science fiction novels about a solar system spanning civilisation where people change their bodies at will

Why Were Madagascar and New Zealand Discovered So Late?

Return the Closest Prime Number

Rotate a column

Too much space between section and text in a twocolumn document

How do I construct this japanese bowl?



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
    3 hours ago










  • $begingroup$
    You can take the logarithm of the function and use L'Hopitals rule
    $endgroup$
    – Nimish
    3 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
    3 hours ago










  • $begingroup$
    You can take the logarithm of the function and use L'Hopitals rule
    $endgroup$
    – Nimish
    3 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 2 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 3 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
    3 hours ago










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
















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










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















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




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












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




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










3 Answers
3






active

oldest

votes


















4












$begingroup$

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









        4












        $begingroup$

        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









        $endgroup$

















          4












          $begingroup$

          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









          $endgroup$















            4












            4








            4





            $begingroup$

            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 3 hours ago









            DashiDashi

            736311




            736311





















                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 3 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 2 hours 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

                                Reverse int within the 32-bit signed integer range: [−2^31, 2^31 − 1]Combining two 32-bit integers into one 64-bit integerDetermine if an int is within rangeLossy packing 32 bit integer to 16 bitComputing the square root of a 64-bit integerKeeping integer addition within boundsSafe multiplication of two 64-bit signed integersLeetcode 10: Regular Expression MatchingSigned integer-to-ascii x86_64 assembler macroReverse the digits of an Integer“Add two numbers given in reverse order from a linked list”

                                Image
                                C++ debug of nlohmann json using GDB Is this toilet slogan correct usage of the English language? Symbol used to indicate indivisibility What if a revenant (monster) gains fire resistance? Is aluminum electrical wire used on aircraft? Closed-form expression for certain product Strong empirical falsification of quantum mechanics based on vacuum energy density How should I respond when I lied about my education and the company finds out through background check? Pre-mixing cryogenic fuels and using only one fuel tank How do you respond to a colleague from another team when they're wrongly expecting that you'll help them? How can I block email signup overlays or javascript popups in Safari? How to implement a feedback to keep the DC gain at zero for this conceptual passive filter? Count the occurrence of each unique word in the file Why electric field inside a cavity of a non-conducting sphere not zero? If infinitesimal transformations commute why dont
                                Read more

                                Category:Fedor von Bock Media in category "Fedor von Bock"Navigation menuUpload mediaISNI: 0000 0000 5511 3417VIAF ID: 24712551GND ID: 119294796Library of Congress authority ID: n96068363BnF ID: 12534305fSUDOC authorities ID: 034604189Open Library ID: OL338253ANKCR AUT ID: jn19990000869National Library of Israel ID: 000514068National Thesaurus for Author Names ID: 341574317ReasonatorScholiaStatistics

                                Image
                                1880 births1945 deathsFedor (given name)Recipients of the Pour le Mérite (military class)Recipients of the Military Merit Cross (Austria-Hungary)Knight's Cross of the Iron Cross recipientsRecipients of the Order of Michael the BraveRecipients of the Order of the Crown (Prussia)Recipients of the Order of the Crown (Württemberg)Recipients of the House Order of HohenzollernRecipients of the Order of the Zähringer LionRecipients of the Hanseatic CrossRecipients of the House Order of the Wendish CrownRecipients of the Order of the Iron CrownGrand Master of the Order of Military Merit (Bulgaria)Recipients of the Order of the Yugoslav CrownRecipients of the Order of Merit of the Republic of HungaryRecipients of the Cross of HonorKnights Grand Cross of the Order of the Crown of ItalyBock (surname)19th-century men of Germany20th-century men of GermanyField Marshals of the Heer (Wehrmacht)Generals of the ReichswehrHeer Knight's Cross of the Iron Cross recipientsMen of Germany by nameMili
                                Read more

                                Kiel Indholdsfortegnelse Historie | Transport og færgeforbindelser | Sejlsport og anden sport | Kultur | Kendte personer fra Kiel | Noter | Litteratur | Eksterne henvisninger | Navigationsmenuwww.kiel.de54°19′31″N 10°8′26″Ø / 54.32528°N 10.14056°Ø / 54.32528; 10.14056Oberbürgermeister Dr. Ulf Kämpferwww.statistik-nord.deDen danske Stats StatistikKiels hjemmesiderrrWorldCat312794080n790547494030481-4

                                Image
                                DitmarskenFlensborgLauenburgKielLübeckNeumünsterNordfrislandØstholstenPinnebergPlönRendsborg-EgernførdeSlesvig-FlensborgSegebergSteinburgStormarnStuttgartMannheimKarlsruheFreiburg im BreisgauHeidelbergHeilbronnUlmPforzheimReutlingenBerlinBremenBremerhavenFrankfurt am MainWiesbadenKasselDarmstadtOffenbach am MainHannoverBraunschweigOsnabrückOldenburgWolfsburgGöttingenHildesheimSalzgitterMainzLudwigshafen am RheinKoblenzTrierDresdenLeipzigChemnitzKielLübeckKaunasKlaipėdaPärnuRigaTallinnTartuViljandiPolotskElblagGdanskKrakówStargard SzczecinskiSzczecinTorunWroclawBerwick-upon-TweedBostonGreat YarmouthKing's LynnKingston upon HullIpswichLeithLondonNewcastle upon TyneYork Byer i Slesvig-HolstenKielHanse NordtysklandhovedstadendelstatenSlesvig-HolstenTysklandsKielerfjordenØstersøenLangelandÆrøHamborgEjderenVesterhavetdansktyskslavisk1865Kielerkanalen1918novemberrevolutionenoprør blandt matrosernekejserlige marineWeimarrepublikken2. verdenskrigværfter2. verdenskrig1200-talletKunsthalle zu
                                Read more