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How do I solve this limit?



The Next CEO of Stack Overflowrational limit problemHow to evaluate the following limit? $limlimits_{xtoinfty}xleft(fracpi2-arctan xright).$Limit only using squeeze theorem $ lim_{(x,y)to(0,2)} x,arctanleft(frac{1}{y-2}right)$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:



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



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
    1 hour 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:



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



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
    1 hour 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:



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



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:



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



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 49 mins 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 1 hour 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
    1 hour 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
    1 hour 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
1 hour ago




$begingroup$
I accidentally put k instead of n, sorry.
$endgroup$
– radoo
1 hour 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:
$$
lim_{xto 0 } exp(log((1-xarctan(nx))^{1/x^2})) = lim_{xto 0 } expleft(frac{log((1-xarctan(nx))}{x^2})right)
$$



by limit rules:



$$
lim_{xto 0 } expleft(frac{log((1-xarctan(nx))}{x^2})right) =expleft[ lim_{xto 0 } left(frac{log((1-xarctan(nx))}{x^2})right)right]
$$



apply L'Hospital's rule and after some algebra you should get:
$$
expleft[- frac{n+lim_{xto 0 }frac{arctan(nx)}{x}+n^2lim_{xto 0} xarctan(nx)}{2}right]
$$



simplifying



$$
expleft(-frac{n+n+n^2}{2}right) = e^{-n}
$$






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



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



      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





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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:
        $$
        lim_{xto 0 } exp(log((1-xarctan(nx))^{1/x^2})) = lim_{xto 0 } expleft(frac{log((1-xarctan(nx))}{x^2})right)
        $$



        by limit rules:



        $$
        lim_{xto 0 } expleft(frac{log((1-xarctan(nx))}{x^2})right) =expleft[ lim_{xto 0 } left(frac{log((1-xarctan(nx))}{x^2})right)right]
        $$



        apply L'Hospital's rule and after some algebra you should get:
        $$
        expleft[- frac{n+lim_{xto 0 }frac{arctan(nx)}{x}+n^2lim_{xto 0} xarctan(nx)}{2}right]
        $$



        simplifying



        $$
        expleft(-frac{n+n+n^2}{2}right) = e^{-n}
        $$






        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:
          $$
          lim_{xto 0 } exp(log((1-xarctan(nx))^{1/x^2})) = lim_{xto 0 } expleft(frac{log((1-xarctan(nx))}{x^2})right)
          $$



          by limit rules:



          $$
          lim_{xto 0 } expleft(frac{log((1-xarctan(nx))}{x^2})right) =expleft[ lim_{xto 0 } left(frac{log((1-xarctan(nx))}{x^2})right)right]
          $$



          apply L'Hospital's rule and after some algebra you should get:
          $$
          expleft[- frac{n+lim_{xto 0 }frac{arctan(nx)}{x}+n^2lim_{xto 0} xarctan(nx)}{2}right]
          $$



          simplifying



          $$
          expleft(-frac{n+n+n^2}{2}right) = e^{-n}
          $$






          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:
            $$
            lim_{xto 0 } exp(log((1-xarctan(nx))^{1/x^2})) = lim_{xto 0 } expleft(frac{log((1-xarctan(nx))}{x^2})right)
            $$



            by limit rules:



            $$
            lim_{xto 0 } expleft(frac{log((1-xarctan(nx))}{x^2})right) =expleft[ lim_{xto 0 } left(frac{log((1-xarctan(nx))}{x^2})right)right]
            $$



            apply L'Hospital's rule and after some algebra you should get:
            $$
            expleft[- frac{n+lim_{xto 0 }frac{arctan(nx)}{x}+n^2lim_{xto 0} xarctan(nx)}{2}right]
            $$



            simplifying



            $$
            expleft(-frac{n+n+n^2}{2}right) = e^{-n}
            $$






            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:
            $$
            lim_{xto 0 } exp(log((1-xarctan(nx))^{1/x^2})) = lim_{xto 0 } expleft(frac{log((1-xarctan(nx))}{x^2})right)
            $$



            by limit rules:



            $$
            lim_{xto 0 } expleft(frac{log((1-xarctan(nx))}{x^2})right) =expleft[ lim_{xto 0 } left(frac{log((1-xarctan(nx))}{x^2})right)right]
            $$



            apply L'Hospital's rule and after some algebra you should get:
            $$
            expleft[- frac{n+lim_{xto 0 }frac{arctan(nx)}{x}+n^2lim_{xto 0} xarctan(nx)}{2}right]
            $$



            simplifying



            $$
            expleft(-frac{n+n+n^2}{2}right) = e^{-n}
            $$







            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



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



                        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



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



                          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



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



                            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



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



                            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 21 mins ago









                            Barry CipraBarry Cipra

                            60.5k655128




                            60.5k655128






















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