How to solve the limit $limlimits_{xto infty} (x arctan x - frac{xpi}{2})$












0












$begingroup$


Next week I have a math exam. While I was doing some exercises I came across this interesting limit:



$limlimits_{xto infty} (x arctan x - frac{xpi}{2})$



After struggling a lot, I decided to calculate this limit using my calculator. The answer turns out to be $-1$. The problem is that I don't know how to calculate this limit without a calculator. I tried using L'Hôpital's rule after converting the expression to a fraction. My steps:



$limlimits_{xto infty} (x arctan x - frac{xpi}{2}) = limlimits_{xto infty} frac{2x^2arctan x - x^2pi}{2x} stackrel{(H)}{=} limlimits_{xto infty} frac{4xarctan x - frac{2}{x^2+1}-2pi x+2}{2} = limlimits_{xto infty} frac{4x^2arctan x - frac{2x}{x^2+1}-2pi x^2+2x}{2x} stackrel{(H)}{=} limlimits_{xto infty} frac{8xarctan x - frac{2x^2+6}{(x^2+1)^2}-4pi x+6}{2} = dots$



This keeps going on without an end, I also don't see where I can simplify the expression when using L'Hôpital's rule. Am I missing a step or am I using the wrong method? What method can be used instead?






calculus limits trigonometry infinity







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    0












    $begingroup$


    Next week I have a math exam. While I was doing some exercises I came across this interesting limit:



    $limlimits_{xto infty} (x arctan x - frac{xpi}{2})$



    After struggling a lot, I decided to calculate this limit using my calculator. The answer turns out to be $-1$. The problem is that I don't know how to calculate this limit without a calculator. I tried using L'Hôpital's rule after converting the expression to a fraction. My steps:



    $limlimits_{xto infty} (x arctan x - frac{xpi}{2}) = limlimits_{xto infty} frac{2x^2arctan x - x^2pi}{2x} stackrel{(H)}{=} limlimits_{xto infty} frac{4xarctan x - frac{2}{x^2+1}-2pi x+2}{2} = limlimits_{xto infty} frac{4x^2arctan x - frac{2x}{x^2+1}-2pi x^2+2x}{2x} stackrel{(H)}{=} limlimits_{xto infty} frac{8xarctan x - frac{2x^2+6}{(x^2+1)^2}-4pi x+6}{2} = dots$



    This keeps going on without an end, I also don't see where I can simplify the expression when using L'Hôpital's rule. Am I missing a step or am I using the wrong method? What method can be used instead?






    calculus limits trigonometry infinity







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Next week I have a math exam. While I was doing some exercises I came across this interesting limit:



      $limlimits_{xto infty} (x arctan x - frac{xpi}{2})$



      After struggling a lot, I decided to calculate this limit using my calculator. The answer turns out to be $-1$. The problem is that I don't know how to calculate this limit without a calculator. I tried using L'Hôpital's rule after converting the expression to a fraction. My steps:



      $limlimits_{xto infty} (x arctan x - frac{xpi}{2}) = limlimits_{xto infty} frac{2x^2arctan x - x^2pi}{2x} stackrel{(H)}{=} limlimits_{xto infty} frac{4xarctan x - frac{2}{x^2+1}-2pi x+2}{2} = limlimits_{xto infty} frac{4x^2arctan x - frac{2x}{x^2+1}-2pi x^2+2x}{2x} stackrel{(H)}{=} limlimits_{xto infty} frac{8xarctan x - frac{2x^2+6}{(x^2+1)^2}-4pi x+6}{2} = dots$



      This keeps going on without an end, I also don't see where I can simplify the expression when using L'Hôpital's rule. Am I missing a step or am I using the wrong method? What method can be used instead?






      calculus limits trigonometry infinity







      share|cite|improve this question









      $endgroup$




      Next week I have a math exam. While I was doing some exercises I came across this interesting limit:



      $limlimits_{xto infty} (x arctan x - frac{xpi}{2})$



      After struggling a lot, I decided to calculate this limit using my calculator. The answer turns out to be $-1$. The problem is that I don't know how to calculate this limit without a calculator. I tried using L'Hôpital's rule after converting the expression to a fraction. My steps:



      $limlimits_{xto infty} (x arctan x - frac{xpi}{2}) = limlimits_{xto infty} frac{2x^2arctan x - x^2pi}{2x} stackrel{(H)}{=} limlimits_{xto infty} frac{4xarctan x - frac{2}{x^2+1}-2pi x+2}{2} = limlimits_{xto infty} frac{4x^2arctan x - frac{2x}{x^2+1}-2pi x^2+2x}{2x} stackrel{(H)}{=} limlimits_{xto infty} frac{8xarctan x - frac{2x^2+6}{(x^2+1)^2}-4pi x+6}{2} = dots$



      This keeps going on without an end, I also don't see where I can simplify the expression when using L'Hôpital's rule. Am I missing a step or am I using the wrong method? What method can be used instead?






      calculus limits trigonometry infinity




      calculus limits trigonometry infinity






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      asked Jan 2 at 21:55









      MichielvkMichielvk

      1065




      1065






















          3 Answers
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          7












          $begingroup$

          Observe
          begin{align}
          lim_{xrightarrow infty} xarctan x-xfrac{pi}{2}=lim_{xrightarrowinfty}frac{arctan x-frac{pi}{2}}{x^{-1}} = lim_{xrightarrow infty} frac{frac{1}{1+x^2}}{-x^{-2}}=-1
          end{align}






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            If you know that $lim_{xto+infty}arctan x = pi/2$, then factoring out $x$ in your original limit you would get a $+infty cdot 0$. You want to apply L'Hospital's Rule. Motivated by this, we do $$lim_{xto +infty} left(xarctan x - frac{xpi}{2}right) = lim_{xto +infty} xleft(arctan x - frac{pi}{2}right) = lim_{x to infty} frac{arctan x - (pi/2)}{1/x}.$$Now apply L'Hospital to get $$lim_{x to +infty}left(xarctan x - frac{xpi}{2}right) = lim_{xto +infty} frac{1/(1+x^2)}{-1/x^2} = -lim_{x to +infty}frac{x^2}{1+x^2} = -1.$$






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              You can do the substitution $x=1/t$, recalling that, for $t>0$,
              $$
              arctanfrac{1}{t}=frac{pi}{2}-arctan t
              $$
              Thus the limit becomes
              $$
              lim_{tto0^+}left(frac{1}{t}frac{pi}{2}-frac{1}{t}arctan t-frac{1}{t}frac{pi}{2}right)=lim_{tto0^+}-frac{arctan t}{t}
              $$
              Alternatively, substitute $u=arctan x$, so $x=tan u$ and the limit becomes
              $$
              lim_{utopi/2^-}(u-pi/2)tan u=lim_{utopi/2^-}frac{u-pi/2}{cot u}
              $$
              This is the reciprocal of the derivative at $pi/2$ of the cotangent.






              share|cite|improve this answer









              $endgroup$













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                3 Answers
                3






                active

                oldest

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                3 Answers
                3






                active

                oldest

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                active

                oldest

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                active

                oldest

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                7












                $begingroup$

                Observe
                begin{align}
                lim_{xrightarrow infty} xarctan x-xfrac{pi}{2}=lim_{xrightarrowinfty}frac{arctan x-frac{pi}{2}}{x^{-1}} = lim_{xrightarrow infty} frac{frac{1}{1+x^2}}{-x^{-2}}=-1
                end{align}






                share|cite|improve this answer









                $endgroup$


















                  7












                  $begingroup$

                  Observe
                  begin{align}
                  lim_{xrightarrow infty} xarctan x-xfrac{pi}{2}=lim_{xrightarrowinfty}frac{arctan x-frac{pi}{2}}{x^{-1}} = lim_{xrightarrow infty} frac{frac{1}{1+x^2}}{-x^{-2}}=-1
                  end{align}






                  share|cite|improve this answer









                  $endgroup$
















                    7












                    7








                    7





                    $begingroup$

                    Observe
                    begin{align}
                    lim_{xrightarrow infty} xarctan x-xfrac{pi}{2}=lim_{xrightarrowinfty}frac{arctan x-frac{pi}{2}}{x^{-1}} = lim_{xrightarrow infty} frac{frac{1}{1+x^2}}{-x^{-2}}=-1
                    end{align}






                    share|cite|improve this answer









                    $endgroup$



                    Observe
                    begin{align}
                    lim_{xrightarrow infty} xarctan x-xfrac{pi}{2}=lim_{xrightarrowinfty}frac{arctan x-frac{pi}{2}}{x^{-1}} = lim_{xrightarrow infty} frac{frac{1}{1+x^2}}{-x^{-2}}=-1
                    end{align}







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 2 at 22:01









                    Jacky ChongJacky Chong

                    18.2k21128




                    18.2k21128























                        1












                        $begingroup$

                        If you know that $lim_{xto+infty}arctan x = pi/2$, then factoring out $x$ in your original limit you would get a $+infty cdot 0$. You want to apply L'Hospital's Rule. Motivated by this, we do $$lim_{xto +infty} left(xarctan x - frac{xpi}{2}right) = lim_{xto +infty} xleft(arctan x - frac{pi}{2}right) = lim_{x to infty} frac{arctan x - (pi/2)}{1/x}.$$Now apply L'Hospital to get $$lim_{x to +infty}left(xarctan x - frac{xpi}{2}right) = lim_{xto +infty} frac{1/(1+x^2)}{-1/x^2} = -lim_{x to +infty}frac{x^2}{1+x^2} = -1.$$






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          If you know that $lim_{xto+infty}arctan x = pi/2$, then factoring out $x$ in your original limit you would get a $+infty cdot 0$. You want to apply L'Hospital's Rule. Motivated by this, we do $$lim_{xto +infty} left(xarctan x - frac{xpi}{2}right) = lim_{xto +infty} xleft(arctan x - frac{pi}{2}right) = lim_{x to infty} frac{arctan x - (pi/2)}{1/x}.$$Now apply L'Hospital to get $$lim_{x to +infty}left(xarctan x - frac{xpi}{2}right) = lim_{xto +infty} frac{1/(1+x^2)}{-1/x^2} = -lim_{x to +infty}frac{x^2}{1+x^2} = -1.$$






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            If you know that $lim_{xto+infty}arctan x = pi/2$, then factoring out $x$ in your original limit you would get a $+infty cdot 0$. You want to apply L'Hospital's Rule. Motivated by this, we do $$lim_{xto +infty} left(xarctan x - frac{xpi}{2}right) = lim_{xto +infty} xleft(arctan x - frac{pi}{2}right) = lim_{x to infty} frac{arctan x - (pi/2)}{1/x}.$$Now apply L'Hospital to get $$lim_{x to +infty}left(xarctan x - frac{xpi}{2}right) = lim_{xto +infty} frac{1/(1+x^2)}{-1/x^2} = -lim_{x to +infty}frac{x^2}{1+x^2} = -1.$$






                            share|cite|improve this answer









                            $endgroup$



                            If you know that $lim_{xto+infty}arctan x = pi/2$, then factoring out $x$ in your original limit you would get a $+infty cdot 0$. You want to apply L'Hospital's Rule. Motivated by this, we do $$lim_{xto +infty} left(xarctan x - frac{xpi}{2}right) = lim_{xto +infty} xleft(arctan x - frac{pi}{2}right) = lim_{x to infty} frac{arctan x - (pi/2)}{1/x}.$$Now apply L'Hospital to get $$lim_{x to +infty}left(xarctan x - frac{xpi}{2}right) = lim_{xto +infty} frac{1/(1+x^2)}{-1/x^2} = -lim_{x to +infty}frac{x^2}{1+x^2} = -1.$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 2 at 22:03









                            Ivo TerekIvo Terek

                            45.8k953142




                            45.8k953142























                                0












                                $begingroup$

                                You can do the substitution $x=1/t$, recalling that, for $t>0$,
                                $$
                                arctanfrac{1}{t}=frac{pi}{2}-arctan t
                                $$
                                Thus the limit becomes
                                $$
                                lim_{tto0^+}left(frac{1}{t}frac{pi}{2}-frac{1}{t}arctan t-frac{1}{t}frac{pi}{2}right)=lim_{tto0^+}-frac{arctan t}{t}
                                $$
                                Alternatively, substitute $u=arctan x$, so $x=tan u$ and the limit becomes
                                $$
                                lim_{utopi/2^-}(u-pi/2)tan u=lim_{utopi/2^-}frac{u-pi/2}{cot u}
                                $$
                                This is the reciprocal of the derivative at $pi/2$ of the cotangent.






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  You can do the substitution $x=1/t$, recalling that, for $t>0$,
                                  $$
                                  arctanfrac{1}{t}=frac{pi}{2}-arctan t
                                  $$
                                  Thus the limit becomes
                                  $$
                                  lim_{tto0^+}left(frac{1}{t}frac{pi}{2}-frac{1}{t}arctan t-frac{1}{t}frac{pi}{2}right)=lim_{tto0^+}-frac{arctan t}{t}
                                  $$
                                  Alternatively, substitute $u=arctan x$, so $x=tan u$ and the limit becomes
                                  $$
                                  lim_{utopi/2^-}(u-pi/2)tan u=lim_{utopi/2^-}frac{u-pi/2}{cot u}
                                  $$
                                  This is the reciprocal of the derivative at $pi/2$ of the cotangent.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    You can do the substitution $x=1/t$, recalling that, for $t>0$,
                                    $$
                                    arctanfrac{1}{t}=frac{pi}{2}-arctan t
                                    $$
                                    Thus the limit becomes
                                    $$
                                    lim_{tto0^+}left(frac{1}{t}frac{pi}{2}-frac{1}{t}arctan t-frac{1}{t}frac{pi}{2}right)=lim_{tto0^+}-frac{arctan t}{t}
                                    $$
                                    Alternatively, substitute $u=arctan x$, so $x=tan u$ and the limit becomes
                                    $$
                                    lim_{utopi/2^-}(u-pi/2)tan u=lim_{utopi/2^-}frac{u-pi/2}{cot u}
                                    $$
                                    This is the reciprocal of the derivative at $pi/2$ of the cotangent.






                                    share|cite|improve this answer









                                    $endgroup$



                                    You can do the substitution $x=1/t$, recalling that, for $t>0$,
                                    $$
                                    arctanfrac{1}{t}=frac{pi}{2}-arctan t
                                    $$
                                    Thus the limit becomes
                                    $$
                                    lim_{tto0^+}left(frac{1}{t}frac{pi}{2}-frac{1}{t}arctan t-frac{1}{t}frac{pi}{2}right)=lim_{tto0^+}-frac{arctan t}{t}
                                    $$
                                    Alternatively, substitute $u=arctan x$, so $x=tan u$ and the limit becomes
                                    $$
                                    lim_{utopi/2^-}(u-pi/2)tan u=lim_{utopi/2^-}frac{u-pi/2}{cot u}
                                    $$
                                    This is the reciprocal of the derivative at $pi/2$ of the cotangent.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Jan 2 at 22:41









                                    egregegreg

                                    181k1485202




                                    181k1485202






























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