Limit involving inverse functions












3












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When I am given the limit



$$limlimits_{x rightarrow infty}frac{xarctansqrt{x^2 +1}}{sqrt{x^2+1}}$$



would it be possible to evaluate it giving some substitution?



L'Hospital's rule seemed an option but I ended up going in circles.






calculus limits inverse-function







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    When I am given the limit



    $$limlimits_{x rightarrow infty}frac{xarctansqrt{x^2 +1}}{sqrt{x^2+1}}$$



    would it be possible to evaluate it giving some substitution?



    L'Hospital's rule seemed an option but I ended up going in circles.






    calculus limits inverse-function







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      When I am given the limit



      $$limlimits_{x rightarrow infty}frac{xarctansqrt{x^2 +1}}{sqrt{x^2+1}}$$



      would it be possible to evaluate it giving some substitution?



      L'Hospital's rule seemed an option but I ended up going in circles.






      calculus limits inverse-function







      share|cite|improve this question











      $endgroup$




      When I am given the limit



      $$limlimits_{x rightarrow infty}frac{xarctansqrt{x^2 +1}}{sqrt{x^2+1}}$$



      would it be possible to evaluate it giving some substitution?



      L'Hospital's rule seemed an option but I ended up going in circles.






      calculus limits inverse-function




      calculus limits inverse-function






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 19 at 14:36









      Michael Rybkin

      2,853416




      2,853416










      asked Feb 19 at 14:29









      MadCapMadCap

      433




      433






















          3 Answers
          3






          active

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          7












          $begingroup$

          When $x>0$, $|x|=x$ and obviously if $xrightarrowinfty$, then $sqrt{x^2+1}rightarrowinfty$. And the last thing that you will need is the fact that the inverse tangent function approaches a value of $pi/2$ as its argument goes to infinity:



          $$
          limlimits_{x rightarrow infty}frac{xarctansqrt{x^2+1}}{sqrt{x^2+1}}=
          limlimits_{x rightarrow infty}frac{xarctansqrt{x^2 +1}}{sqrt{x^2}sqrt{1+frac{1}{x^2}}}=
          limlimits_{x rightarrow infty}frac{xarctansqrt{x^2 +1}}{|x|sqrt{1+frac{1}{x^2}}}=\
          limlimits_{x rightarrow infty}frac{xarctansqrt{x^2 +1}}{xsqrt{1+frac{1}{x^2}}}=
          limlimits_{x rightarrow infty}frac{arctansqrt{x^2 +1}}{sqrt{1+frac{1}{x^2}}}=
          frac{pi/2}{sqrt{1+0}}=frac{pi}{2}.
          $$






          share|cite|improve this answer











          $endgroup$





















            4












            $begingroup$

            You may proceed as follows:




            • Set $tan y = sqrt{1+x^2}$ and consider $y to frac{pi}{2}^-$


            begin{eqnarray*}frac{xarctansqrt{x^2 +1}}{sqrt{x^2+1}}
            & = & sqrt{tan^2y -1}frac{y}{tan y} \
            & = & frac{sqrt{sin^2 y - cos^2 y}}{sin y}cdot y\
            &stackrel{y to frac{pi}{2}^-}{longrightarrow} & frac{sqrt{1 - 0}}{1}cdot frac{pi}{2} = frac{pi}{2}
            end{eqnarray*}






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              Hint: It is true in general that if $lim f$ and $lim g$ both exist and are finite and nonzero, then $lim (fg)$ exists and equals $(lim f)(lim g)$.



              Take $f(x)=x/sqrt{x^2+1}$, $g(x)=tan^{-1}(x^2+1)$ and note that $xtoinfty$ implies $sqrt{x^2+1}toinfty$.






              share|cite|improve this answer









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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                7












                $begingroup$

                When $x>0$, $|x|=x$ and obviously if $xrightarrowinfty$, then $sqrt{x^2+1}rightarrowinfty$. And the last thing that you will need is the fact that the inverse tangent function approaches a value of $pi/2$ as its argument goes to infinity:



                $$
                limlimits_{x rightarrow infty}frac{xarctansqrt{x^2+1}}{sqrt{x^2+1}}=
                limlimits_{x rightarrow infty}frac{xarctansqrt{x^2 +1}}{sqrt{x^2}sqrt{1+frac{1}{x^2}}}=
                limlimits_{x rightarrow infty}frac{xarctansqrt{x^2 +1}}{|x|sqrt{1+frac{1}{x^2}}}=\
                limlimits_{x rightarrow infty}frac{xarctansqrt{x^2 +1}}{xsqrt{1+frac{1}{x^2}}}=
                limlimits_{x rightarrow infty}frac{arctansqrt{x^2 +1}}{sqrt{1+frac{1}{x^2}}}=
                frac{pi/2}{sqrt{1+0}}=frac{pi}{2}.
                $$






                share|cite|improve this answer











                $endgroup$


















                  7












                  $begingroup$

                  When $x>0$, $|x|=x$ and obviously if $xrightarrowinfty$, then $sqrt{x^2+1}rightarrowinfty$. And the last thing that you will need is the fact that the inverse tangent function approaches a value of $pi/2$ as its argument goes to infinity:



                  $$
                  limlimits_{x rightarrow infty}frac{xarctansqrt{x^2+1}}{sqrt{x^2+1}}=
                  limlimits_{x rightarrow infty}frac{xarctansqrt{x^2 +1}}{sqrt{x^2}sqrt{1+frac{1}{x^2}}}=
                  limlimits_{x rightarrow infty}frac{xarctansqrt{x^2 +1}}{|x|sqrt{1+frac{1}{x^2}}}=\
                  limlimits_{x rightarrow infty}frac{xarctansqrt{x^2 +1}}{xsqrt{1+frac{1}{x^2}}}=
                  limlimits_{x rightarrow infty}frac{arctansqrt{x^2 +1}}{sqrt{1+frac{1}{x^2}}}=
                  frac{pi/2}{sqrt{1+0}}=frac{pi}{2}.
                  $$






                  share|cite|improve this answer











                  $endgroup$
















                    7












                    7








                    7





                    $begingroup$

                    When $x>0$, $|x|=x$ and obviously if $xrightarrowinfty$, then $sqrt{x^2+1}rightarrowinfty$. And the last thing that you will need is the fact that the inverse tangent function approaches a value of $pi/2$ as its argument goes to infinity:



                    $$
                    limlimits_{x rightarrow infty}frac{xarctansqrt{x^2+1}}{sqrt{x^2+1}}=
                    limlimits_{x rightarrow infty}frac{xarctansqrt{x^2 +1}}{sqrt{x^2}sqrt{1+frac{1}{x^2}}}=
                    limlimits_{x rightarrow infty}frac{xarctansqrt{x^2 +1}}{|x|sqrt{1+frac{1}{x^2}}}=\
                    limlimits_{x rightarrow infty}frac{xarctansqrt{x^2 +1}}{xsqrt{1+frac{1}{x^2}}}=
                    limlimits_{x rightarrow infty}frac{arctansqrt{x^2 +1}}{sqrt{1+frac{1}{x^2}}}=
                    frac{pi/2}{sqrt{1+0}}=frac{pi}{2}.
                    $$






                    share|cite|improve this answer











                    $endgroup$



                    When $x>0$, $|x|=x$ and obviously if $xrightarrowinfty$, then $sqrt{x^2+1}rightarrowinfty$. And the last thing that you will need is the fact that the inverse tangent function approaches a value of $pi/2$ as its argument goes to infinity:



                    $$
                    limlimits_{x rightarrow infty}frac{xarctansqrt{x^2+1}}{sqrt{x^2+1}}=
                    limlimits_{x rightarrow infty}frac{xarctansqrt{x^2 +1}}{sqrt{x^2}sqrt{1+frac{1}{x^2}}}=
                    limlimits_{x rightarrow infty}frac{xarctansqrt{x^2 +1}}{|x|sqrt{1+frac{1}{x^2}}}=\
                    limlimits_{x rightarrow infty}frac{xarctansqrt{x^2 +1}}{xsqrt{1+frac{1}{x^2}}}=
                    limlimits_{x rightarrow infty}frac{arctansqrt{x^2 +1}}{sqrt{1+frac{1}{x^2}}}=
                    frac{pi/2}{sqrt{1+0}}=frac{pi}{2}.
                    $$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Feb 19 at 18:39

























                    answered Feb 19 at 14:36









                    Michael RybkinMichael Rybkin

                    2,853416




                    2,853416























                        4












                        $begingroup$

                        You may proceed as follows:




                        • Set $tan y = sqrt{1+x^2}$ and consider $y to frac{pi}{2}^-$


                        begin{eqnarray*}frac{xarctansqrt{x^2 +1}}{sqrt{x^2+1}}
                        & = & sqrt{tan^2y -1}frac{y}{tan y} \
                        & = & frac{sqrt{sin^2 y - cos^2 y}}{sin y}cdot y\
                        &stackrel{y to frac{pi}{2}^-}{longrightarrow} & frac{sqrt{1 - 0}}{1}cdot frac{pi}{2} = frac{pi}{2}
                        end{eqnarray*}






                        share|cite|improve this answer









                        $endgroup$


















                          4












                          $begingroup$

                          You may proceed as follows:




                          • Set $tan y = sqrt{1+x^2}$ and consider $y to frac{pi}{2}^-$


                          begin{eqnarray*}frac{xarctansqrt{x^2 +1}}{sqrt{x^2+1}}
                          & = & sqrt{tan^2y -1}frac{y}{tan y} \
                          & = & frac{sqrt{sin^2 y - cos^2 y}}{sin y}cdot y\
                          &stackrel{y to frac{pi}{2}^-}{longrightarrow} & frac{sqrt{1 - 0}}{1}cdot frac{pi}{2} = frac{pi}{2}
                          end{eqnarray*}






                          share|cite|improve this answer









                          $endgroup$
















                            4












                            4








                            4





                            $begingroup$

                            You may proceed as follows:




                            • Set $tan y = sqrt{1+x^2}$ and consider $y to frac{pi}{2}^-$


                            begin{eqnarray*}frac{xarctansqrt{x^2 +1}}{sqrt{x^2+1}}
                            & = & sqrt{tan^2y -1}frac{y}{tan y} \
                            & = & frac{sqrt{sin^2 y - cos^2 y}}{sin y}cdot y\
                            &stackrel{y to frac{pi}{2}^-}{longrightarrow} & frac{sqrt{1 - 0}}{1}cdot frac{pi}{2} = frac{pi}{2}
                            end{eqnarray*}






                            share|cite|improve this answer









                            $endgroup$



                            You may proceed as follows:




                            • Set $tan y = sqrt{1+x^2}$ and consider $y to frac{pi}{2}^-$


                            begin{eqnarray*}frac{xarctansqrt{x^2 +1}}{sqrt{x^2+1}}
                            & = & sqrt{tan^2y -1}frac{y}{tan y} \
                            & = & frac{sqrt{sin^2 y - cos^2 y}}{sin y}cdot y\
                            &stackrel{y to frac{pi}{2}^-}{longrightarrow} & frac{sqrt{1 - 0}}{1}cdot frac{pi}{2} = frac{pi}{2}
                            end{eqnarray*}







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Feb 19 at 14:58









                            trancelocationtrancelocation

                            12.5k1826




                            12.5k1826























                                1












                                $begingroup$

                                Hint: It is true in general that if $lim f$ and $lim g$ both exist and are finite and nonzero, then $lim (fg)$ exists and equals $(lim f)(lim g)$.



                                Take $f(x)=x/sqrt{x^2+1}$, $g(x)=tan^{-1}(x^2+1)$ and note that $xtoinfty$ implies $sqrt{x^2+1}toinfty$.






                                share|cite|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  Hint: It is true in general that if $lim f$ and $lim g$ both exist and are finite and nonzero, then $lim (fg)$ exists and equals $(lim f)(lim g)$.



                                  Take $f(x)=x/sqrt{x^2+1}$, $g(x)=tan^{-1}(x^2+1)$ and note that $xtoinfty$ implies $sqrt{x^2+1}toinfty$.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    Hint: It is true in general that if $lim f$ and $lim g$ both exist and are finite and nonzero, then $lim (fg)$ exists and equals $(lim f)(lim g)$.



                                    Take $f(x)=x/sqrt{x^2+1}$, $g(x)=tan^{-1}(x^2+1)$ and note that $xtoinfty$ implies $sqrt{x^2+1}toinfty$.






                                    share|cite|improve this answer









                                    $endgroup$



                                    Hint: It is true in general that if $lim f$ and $lim g$ both exist and are finite and nonzero, then $lim (fg)$ exists and equals $(lim f)(lim g)$.



                                    Take $f(x)=x/sqrt{x^2+1}$, $g(x)=tan^{-1}(x^2+1)$ and note that $xtoinfty$ implies $sqrt{x^2+1}toinfty$.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Feb 19 at 22:05









                                    MPWMPW

                                    30.4k12157




                                    30.4k12157






























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