Is my use of Fatou's Lemma correct in this case?












2












$begingroup$


I am asked to determine:



$$lim_{nto infty}int_{-infty}^{infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}dlambda{x}$$



My idea:



$$lim_{nto infty}int_{-infty}^{infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}dlambda{x}=limsup_{nto infty}int_{-infty}^{infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}dlambda{x}leqint_{-infty}^{infty}limsup_{nto infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}dlambda{x}$$



and $lim_{nto infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}=0$



Therefore, $int_{-infty}^{infty}limsup_{nto infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}dlambda{x}=0$



Now note that $frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}geq0,$ for any $x in ]-infty,infty[$



This then implies that: $lim_{nto infty}int_{-infty}^{infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}dlambda{x}=0$






real-analysis integration measure-theory convergence lebesgue-measure







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$endgroup$

















    2












    $begingroup$


    I am asked to determine:



    $$lim_{nto infty}int_{-infty}^{infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}dlambda{x}$$



    My idea:



    $$lim_{nto infty}int_{-infty}^{infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}dlambda{x}=limsup_{nto infty}int_{-infty}^{infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}dlambda{x}leqint_{-infty}^{infty}limsup_{nto infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}dlambda{x}$$



    and $lim_{nto infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}=0$



    Therefore, $int_{-infty}^{infty}limsup_{nto infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}dlambda{x}=0$



    Now note that $frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}geq0,$ for any $x in ]-infty,infty[$



    This then implies that: $lim_{nto infty}int_{-infty}^{infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}dlambda{x}=0$






    real-analysis integration measure-theory convergence lebesgue-measure







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I am asked to determine:



      $$lim_{nto infty}int_{-infty}^{infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}dlambda{x}$$



      My idea:



      $$lim_{nto infty}int_{-infty}^{infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}dlambda{x}=limsup_{nto infty}int_{-infty}^{infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}dlambda{x}leqint_{-infty}^{infty}limsup_{nto infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}dlambda{x}$$



      and $lim_{nto infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}=0$



      Therefore, $int_{-infty}^{infty}limsup_{nto infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}dlambda{x}=0$



      Now note that $frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}geq0,$ for any $x in ]-infty,infty[$



      This then implies that: $lim_{nto infty}int_{-infty}^{infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}dlambda{x}=0$






      real-analysis integration measure-theory convergence lebesgue-measure







      share|cite|improve this question











      $endgroup$




      I am asked to determine:



      $$lim_{nto infty}int_{-infty}^{infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}dlambda{x}$$



      My idea:



      $$lim_{nto infty}int_{-infty}^{infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}dlambda{x}=limsup_{nto infty}int_{-infty}^{infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}dlambda{x}leqint_{-infty}^{infty}limsup_{nto infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}dlambda{x}$$



      and $lim_{nto infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}=0$



      Therefore, $int_{-infty}^{infty}limsup_{nto infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}dlambda{x}=0$



      Now note that $frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}geq0,$ for any $x in ]-infty,infty[$



      This then implies that: $lim_{nto infty}int_{-infty}^{infty}frac{1}{1+e^{nx}}e^{frac{-x^2}{n}}dlambda{x}=0$






      real-analysis integration measure-theory convergence lebesgue-measure




      real-analysis integration measure-theory convergence lebesgue-measure






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 27 at 8:34









      Aweygan

      14.7k21442




      14.7k21442










      asked Jan 27 at 8:29









      MinaThumaMinaThuma

      1968




      1968






















          1 Answer
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          $begingroup$

          Something is wrong. The pointwise limit should be
          $$lim_{nto infty}frac{e^{frac{-x^2}{n}}}{1+e^{nx}}=
          begin{cases}
          0& text{if $x>0$,}\
          1/2 &text{if $x=0$,}\
          1 &text{if $x< 0$.}
          end{cases}$$






          share|cite|improve this answer











          $endgroup$













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            1 Answer
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            1 Answer
            1






            active

            oldest

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            active

            oldest

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            active

            oldest

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            3












            $begingroup$

            Something is wrong. The pointwise limit should be
            $$lim_{nto infty}frac{e^{frac{-x^2}{n}}}{1+e^{nx}}=
            begin{cases}
            0& text{if $x>0$,}\
            1/2 &text{if $x=0$,}\
            1 &text{if $x< 0$.}
            end{cases}$$






            share|cite|improve this answer











            $endgroup$


















              3












              $begingroup$

              Something is wrong. The pointwise limit should be
              $$lim_{nto infty}frac{e^{frac{-x^2}{n}}}{1+e^{nx}}=
              begin{cases}
              0& text{if $x>0$,}\
              1/2 &text{if $x=0$,}\
              1 &text{if $x< 0$.}
              end{cases}$$






              share|cite|improve this answer











              $endgroup$
















                3












                3








                3





                $begingroup$

                Something is wrong. The pointwise limit should be
                $$lim_{nto infty}frac{e^{frac{-x^2}{n}}}{1+e^{nx}}=
                begin{cases}
                0& text{if $x>0$,}\
                1/2 &text{if $x=0$,}\
                1 &text{if $x< 0$.}
                end{cases}$$






                share|cite|improve this answer











                $endgroup$



                Something is wrong. The pointwise limit should be
                $$lim_{nto infty}frac{e^{frac{-x^2}{n}}}{1+e^{nx}}=
                begin{cases}
                0& text{if $x>0$,}\
                1/2 &text{if $x=0$,}\
                1 &text{if $x< 0$.}
                end{cases}$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 27 at 9:19

























                answered Jan 27 at 8:47









                Robert ZRobert Z

                101k1070143




                101k1070143






























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