Let $C_n =int^{1/n}_{1/(n+1)}frac{tan^{-1}(nx)}{sin^{-1}(nx)}dx $












5












$begingroup$


Let



$$C_n =int^{1/n}_{1/(n+1)}frac{tan^{-1}(nx)}{sin^{-1}(nx)}dxtextrm{.} $$



Then $lim_{n to infty} n ^2C_n$ equals :



a) $1$



b) $0$



c) $-1$



d) $frac12$



I am not getting any clue on this. How to proceed? Please help on this. I will be grateful to you.






calculus integration definite-integrals







share|cite|improve this question











$endgroup$

















    5












    $begingroup$


    Let



    $$C_n =int^{1/n}_{1/(n+1)}frac{tan^{-1}(nx)}{sin^{-1}(nx)}dxtextrm{.} $$



    Then $lim_{n to infty} n ^2C_n$ equals :



    a) $1$



    b) $0$



    c) $-1$



    d) $frac12$



    I am not getting any clue on this. How to proceed? Please help on this. I will be grateful to you.






    calculus integration definite-integrals







    share|cite|improve this question











    $endgroup$















      5












      5








      5


      2



      $begingroup$


      Let



      $$C_n =int^{1/n}_{1/(n+1)}frac{tan^{-1}(nx)}{sin^{-1}(nx)}dxtextrm{.} $$



      Then $lim_{n to infty} n ^2C_n$ equals :



      a) $1$



      b) $0$



      c) $-1$



      d) $frac12$



      I am not getting any clue on this. How to proceed? Please help on this. I will be grateful to you.






      calculus integration definite-integrals







      share|cite|improve this question











      $endgroup$




      Let



      $$C_n =int^{1/n}_{1/(n+1)}frac{tan^{-1}(nx)}{sin^{-1}(nx)}dxtextrm{.} $$



      Then $lim_{n to infty} n ^2C_n$ equals :



      a) $1$



      b) $0$



      c) $-1$



      d) $frac12$



      I am not getting any clue on this. How to proceed? Please help on this. I will be grateful to you.






      calculus integration definite-integrals




      calculus integration definite-integrals






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jun 5 '15 at 18:17









      Ken

      3,63151728




      3,63151728










      asked Aug 4 '14 at 16:53









      user108258user108258

      547819




      547819






















          2 Answers
          2






          active

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          4












          $begingroup$

          the thing here is not to be put off by the rather ferocious-looking integrand. as you can see, the limits of integration get closer together as $n$ increases. this means you can use an approximation based on a constant value of the integrand within the limits of integration.



          it is helpful to make a minor substitution of $z=nx$ so you will have changed limits of integration, and you must use that $dx=frac{dz}{n}$. this gives altogether:
          $$
          C_n = frac1n int_{frac{n}{n+1}}^1 frac {tan^{-1} z}{sin^{-1}z} dz
          $$
          you should be able to proceed from here, using the asymptotic equality of the ratio of $tan z$ and $sin z$ as $z to 0$. i don't think the question requires you to prove this, but you may wish to make an attempt






          share|cite|improve this answer









          $endgroup$





















            4












            $begingroup$

            You can use the Mean Value Theorem for integrals to get the limit. In fact, since
            $$ int_{frac{1}{n+1}}^{frac1n}frac{arctan nx}{arcsin nx}dx=frac{arctan nxi_n}{arcsin nxi_n}frac{1}{n(n+1)}$$
            where
            $$frac{1}{n+1}<xi_n<{frac1n},$$
            we have
            begin{eqnarray}
            lim_{ntoinfty}n^2C_n&=&lim_{ntoinfty}n^2int_{frac{1}{n+1}}^{frac1n}frac{arctan nx}{arcsin nx}dx\
            &=&lim_{ntoinfty}n^2frac{arctan nxi_n}{arcsin nxi_n}frac{1}{n(n+1)}\
            &=&lim_{ntoinfty}frac{arctan nxi_n}{arcsin nxi_n}.
            end{eqnarray}
            Since
            $$frac{1}{n+1}<xi_n<{frac1n}$$
            we have
            $$frac{n}{n+1}<nxi_n<1$$
            or $lim_{ntoinfty}nxi_n=1$. So
            $$ lim_{ntoinfty}n^2C_n=frac{arctan 1}{arcsin 1}=frac{frac{pi}{4}}{frac{pi}{2}}=frac12.$$
            So you have to choose (d).






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Nice explanation xpaul.
              $endgroup$
              – juantheron
              Mar 17 '16 at 6:40











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            2 Answers
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            4












            $begingroup$

            the thing here is not to be put off by the rather ferocious-looking integrand. as you can see, the limits of integration get closer together as $n$ increases. this means you can use an approximation based on a constant value of the integrand within the limits of integration.



            it is helpful to make a minor substitution of $z=nx$ so you will have changed limits of integration, and you must use that $dx=frac{dz}{n}$. this gives altogether:
            $$
            C_n = frac1n int_{frac{n}{n+1}}^1 frac {tan^{-1} z}{sin^{-1}z} dz
            $$
            you should be able to proceed from here, using the asymptotic equality of the ratio of $tan z$ and $sin z$ as $z to 0$. i don't think the question requires you to prove this, but you may wish to make an attempt






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              the thing here is not to be put off by the rather ferocious-looking integrand. as you can see, the limits of integration get closer together as $n$ increases. this means you can use an approximation based on a constant value of the integrand within the limits of integration.



              it is helpful to make a minor substitution of $z=nx$ so you will have changed limits of integration, and you must use that $dx=frac{dz}{n}$. this gives altogether:
              $$
              C_n = frac1n int_{frac{n}{n+1}}^1 frac {tan^{-1} z}{sin^{-1}z} dz
              $$
              you should be able to proceed from here, using the asymptotic equality of the ratio of $tan z$ and $sin z$ as $z to 0$. i don't think the question requires you to prove this, but you may wish to make an attempt






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                the thing here is not to be put off by the rather ferocious-looking integrand. as you can see, the limits of integration get closer together as $n$ increases. this means you can use an approximation based on a constant value of the integrand within the limits of integration.



                it is helpful to make a minor substitution of $z=nx$ so you will have changed limits of integration, and you must use that $dx=frac{dz}{n}$. this gives altogether:
                $$
                C_n = frac1n int_{frac{n}{n+1}}^1 frac {tan^{-1} z}{sin^{-1}z} dz
                $$
                you should be able to proceed from here, using the asymptotic equality of the ratio of $tan z$ and $sin z$ as $z to 0$. i don't think the question requires you to prove this, but you may wish to make an attempt






                share|cite|improve this answer









                $endgroup$



                the thing here is not to be put off by the rather ferocious-looking integrand. as you can see, the limits of integration get closer together as $n$ increases. this means you can use an approximation based on a constant value of the integrand within the limits of integration.



                it is helpful to make a minor substitution of $z=nx$ so you will have changed limits of integration, and you must use that $dx=frac{dz}{n}$. this gives altogether:
                $$
                C_n = frac1n int_{frac{n}{n+1}}^1 frac {tan^{-1} z}{sin^{-1}z} dz
                $$
                you should be able to proceed from here, using the asymptotic equality of the ratio of $tan z$ and $sin z$ as $z to 0$. i don't think the question requires you to prove this, but you may wish to make an attempt







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 4 '14 at 17:17









                David HoldenDavid Holden

                14.7k21224




                14.7k21224























                    4












                    $begingroup$

                    You can use the Mean Value Theorem for integrals to get the limit. In fact, since
                    $$ int_{frac{1}{n+1}}^{frac1n}frac{arctan nx}{arcsin nx}dx=frac{arctan nxi_n}{arcsin nxi_n}frac{1}{n(n+1)}$$
                    where
                    $$frac{1}{n+1}<xi_n<{frac1n},$$
                    we have
                    begin{eqnarray}
                    lim_{ntoinfty}n^2C_n&=&lim_{ntoinfty}n^2int_{frac{1}{n+1}}^{frac1n}frac{arctan nx}{arcsin nx}dx\
                    &=&lim_{ntoinfty}n^2frac{arctan nxi_n}{arcsin nxi_n}frac{1}{n(n+1)}\
                    &=&lim_{ntoinfty}frac{arctan nxi_n}{arcsin nxi_n}.
                    end{eqnarray}
                    Since
                    $$frac{1}{n+1}<xi_n<{frac1n}$$
                    we have
                    $$frac{n}{n+1}<nxi_n<1$$
                    or $lim_{ntoinfty}nxi_n=1$. So
                    $$ lim_{ntoinfty}n^2C_n=frac{arctan 1}{arcsin 1}=frac{frac{pi}{4}}{frac{pi}{2}}=frac12.$$
                    So you have to choose (d).






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Nice explanation xpaul.
                      $endgroup$
                      – juantheron
                      Mar 17 '16 at 6:40
















                    4












                    $begingroup$

                    You can use the Mean Value Theorem for integrals to get the limit. In fact, since
                    $$ int_{frac{1}{n+1}}^{frac1n}frac{arctan nx}{arcsin nx}dx=frac{arctan nxi_n}{arcsin nxi_n}frac{1}{n(n+1)}$$
                    where
                    $$frac{1}{n+1}<xi_n<{frac1n},$$
                    we have
                    begin{eqnarray}
                    lim_{ntoinfty}n^2C_n&=&lim_{ntoinfty}n^2int_{frac{1}{n+1}}^{frac1n}frac{arctan nx}{arcsin nx}dx\
                    &=&lim_{ntoinfty}n^2frac{arctan nxi_n}{arcsin nxi_n}frac{1}{n(n+1)}\
                    &=&lim_{ntoinfty}frac{arctan nxi_n}{arcsin nxi_n}.
                    end{eqnarray}
                    Since
                    $$frac{1}{n+1}<xi_n<{frac1n}$$
                    we have
                    $$frac{n}{n+1}<nxi_n<1$$
                    or $lim_{ntoinfty}nxi_n=1$. So
                    $$ lim_{ntoinfty}n^2C_n=frac{arctan 1}{arcsin 1}=frac{frac{pi}{4}}{frac{pi}{2}}=frac12.$$
                    So you have to choose (d).






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Nice explanation xpaul.
                      $endgroup$
                      – juantheron
                      Mar 17 '16 at 6:40














                    4












                    4








                    4





                    $begingroup$

                    You can use the Mean Value Theorem for integrals to get the limit. In fact, since
                    $$ int_{frac{1}{n+1}}^{frac1n}frac{arctan nx}{arcsin nx}dx=frac{arctan nxi_n}{arcsin nxi_n}frac{1}{n(n+1)}$$
                    where
                    $$frac{1}{n+1}<xi_n<{frac1n},$$
                    we have
                    begin{eqnarray}
                    lim_{ntoinfty}n^2C_n&=&lim_{ntoinfty}n^2int_{frac{1}{n+1}}^{frac1n}frac{arctan nx}{arcsin nx}dx\
                    &=&lim_{ntoinfty}n^2frac{arctan nxi_n}{arcsin nxi_n}frac{1}{n(n+1)}\
                    &=&lim_{ntoinfty}frac{arctan nxi_n}{arcsin nxi_n}.
                    end{eqnarray}
                    Since
                    $$frac{1}{n+1}<xi_n<{frac1n}$$
                    we have
                    $$frac{n}{n+1}<nxi_n<1$$
                    or $lim_{ntoinfty}nxi_n=1$. So
                    $$ lim_{ntoinfty}n^2C_n=frac{arctan 1}{arcsin 1}=frac{frac{pi}{4}}{frac{pi}{2}}=frac12.$$
                    So you have to choose (d).






                    share|cite|improve this answer











                    $endgroup$



                    You can use the Mean Value Theorem for integrals to get the limit. In fact, since
                    $$ int_{frac{1}{n+1}}^{frac1n}frac{arctan nx}{arcsin nx}dx=frac{arctan nxi_n}{arcsin nxi_n}frac{1}{n(n+1)}$$
                    where
                    $$frac{1}{n+1}<xi_n<{frac1n},$$
                    we have
                    begin{eqnarray}
                    lim_{ntoinfty}n^2C_n&=&lim_{ntoinfty}n^2int_{frac{1}{n+1}}^{frac1n}frac{arctan nx}{arcsin nx}dx\
                    &=&lim_{ntoinfty}n^2frac{arctan nxi_n}{arcsin nxi_n}frac{1}{n(n+1)}\
                    &=&lim_{ntoinfty}frac{arctan nxi_n}{arcsin nxi_n}.
                    end{eqnarray}
                    Since
                    $$frac{1}{n+1}<xi_n<{frac1n}$$
                    we have
                    $$frac{n}{n+1}<nxi_n<1$$
                    or $lim_{ntoinfty}nxi_n=1$. So
                    $$ lim_{ntoinfty}n^2C_n=frac{arctan 1}{arcsin 1}=frac{frac{pi}{4}}{frac{pi}{2}}=frac12.$$
                    So you have to choose (d).







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 12 at 14:47

























                    answered Aug 5 '14 at 1:30









                    xpaulxpaul

                    22.6k24455




                    22.6k24455












                    • $begingroup$
                      Nice explanation xpaul.
                      $endgroup$
                      – juantheron
                      Mar 17 '16 at 6:40


















                    • $begingroup$
                      Nice explanation xpaul.
                      $endgroup$
                      – juantheron
                      Mar 17 '16 at 6:40
















                    $begingroup$
                    Nice explanation xpaul.
                    $endgroup$
                    – juantheron
                    Mar 17 '16 at 6:40




                    $begingroup$
                    Nice explanation xpaul.
                    $endgroup$
                    – juantheron
                    Mar 17 '16 at 6:40


















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