limit of and integral depending on n












2












$begingroup$


Can somebody give me some tips about how can find the next limit, please?
$$ lim_{nto infty} int_{0}^4 sqrt[n]{x^n+(4-x)^n} dx $$I found that $$int_{0}^4 sqrt[n]{x^n+(4-x)^n} =2int_{0}^2 sqrt[n]{x^n+(4-x)^n} $$ but I do not know how to determine some inequalities to find the limit with the squeeze theorem.






calculus limits definite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    I guess it should be $ntoinfty$, not $x$.
    $endgroup$
    – Mark
    Jan 7 at 14:34






  • 1




    $begingroup$
    Maybe you can change integral and limit? Just a suggestion I didn't check if dominated convergence or monotone convergence or something similar applies.
    $endgroup$
    – Math_QED
    Jan 7 at 14:56










  • $begingroup$
    @Math_QED, maybe OP doesn't know about these theorems. After all you need to know the theory of Lebesgue integration in order to prove them.
    $endgroup$
    – Mark
    Jan 7 at 15:18
















2












$begingroup$


Can somebody give me some tips about how can find the next limit, please?
$$ lim_{nto infty} int_{0}^4 sqrt[n]{x^n+(4-x)^n} dx $$I found that $$int_{0}^4 sqrt[n]{x^n+(4-x)^n} =2int_{0}^2 sqrt[n]{x^n+(4-x)^n} $$ but I do not know how to determine some inequalities to find the limit with the squeeze theorem.






calculus limits definite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    I guess it should be $ntoinfty$, not $x$.
    $endgroup$
    – Mark
    Jan 7 at 14:34






  • 1




    $begingroup$
    Maybe you can change integral and limit? Just a suggestion I didn't check if dominated convergence or monotone convergence or something similar applies.
    $endgroup$
    – Math_QED
    Jan 7 at 14:56










  • $begingroup$
    @Math_QED, maybe OP doesn't know about these theorems. After all you need to know the theory of Lebesgue integration in order to prove them.
    $endgroup$
    – Mark
    Jan 7 at 15:18














2












2








2





$begingroup$


Can somebody give me some tips about how can find the next limit, please?
$$ lim_{nto infty} int_{0}^4 sqrt[n]{x^n+(4-x)^n} dx $$I found that $$int_{0}^4 sqrt[n]{x^n+(4-x)^n} =2int_{0}^2 sqrt[n]{x^n+(4-x)^n} $$ but I do not know how to determine some inequalities to find the limit with the squeeze theorem.






calculus limits definite-integrals







share|cite|improve this question











$endgroup$




Can somebody give me some tips about how can find the next limit, please?
$$ lim_{nto infty} int_{0}^4 sqrt[n]{x^n+(4-x)^n} dx $$I found that $$int_{0}^4 sqrt[n]{x^n+(4-x)^n} =2int_{0}^2 sqrt[n]{x^n+(4-x)^n} $$ but I do not know how to determine some inequalities to find the limit with the squeeze theorem.






calculus limits definite-integrals




calculus limits definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 14:45







Gaboru

















asked Jan 7 at 14:33









GaboruGaboru

897




897












  • $begingroup$
    I guess it should be $ntoinfty$, not $x$.
    $endgroup$
    – Mark
    Jan 7 at 14:34






  • 1




    $begingroup$
    Maybe you can change integral and limit? Just a suggestion I didn't check if dominated convergence or monotone convergence or something similar applies.
    $endgroup$
    – Math_QED
    Jan 7 at 14:56










  • $begingroup$
    @Math_QED, maybe OP doesn't know about these theorems. After all you need to know the theory of Lebesgue integration in order to prove them.
    $endgroup$
    – Mark
    Jan 7 at 15:18


















  • $begingroup$
    I guess it should be $ntoinfty$, not $x$.
    $endgroup$
    – Mark
    Jan 7 at 14:34






  • 1




    $begingroup$
    Maybe you can change integral and limit? Just a suggestion I didn't check if dominated convergence or monotone convergence or something similar applies.
    $endgroup$
    – Math_QED
    Jan 7 at 14:56










  • $begingroup$
    @Math_QED, maybe OP doesn't know about these theorems. After all you need to know the theory of Lebesgue integration in order to prove them.
    $endgroup$
    – Mark
    Jan 7 at 15:18
















$begingroup$
I guess it should be $ntoinfty$, not $x$.
$endgroup$
– Mark
Jan 7 at 14:34




$begingroup$
I guess it should be $ntoinfty$, not $x$.
$endgroup$
– Mark
Jan 7 at 14:34




1




1




$begingroup$
Maybe you can change integral and limit? Just a suggestion I didn't check if dominated convergence or monotone convergence or something similar applies.
$endgroup$
– Math_QED
Jan 7 at 14:56




$begingroup$
Maybe you can change integral and limit? Just a suggestion I didn't check if dominated convergence or monotone convergence or something similar applies.
$endgroup$
– Math_QED
Jan 7 at 14:56












$begingroup$
@Math_QED, maybe OP doesn't know about these theorems. After all you need to know the theory of Lebesgue integration in order to prove them.
$endgroup$
– Mark
Jan 7 at 15:18




$begingroup$
@Math_QED, maybe OP doesn't know about these theorems. After all you need to know the theory of Lebesgue integration in order to prove them.
$endgroup$
– Mark
Jan 7 at 15:18










3 Answers
3






active

oldest

votes


















3












$begingroup$

Note that the p-norm
of a 2D vector is defined as
$$
left| {bf x} right|_{,p} = left( {x_{,1} ^{,p} + x_{,2} ^{,p} } right)^{,1/p}
$$
and it is known that
$$
mathop {lim }limits_{p, to ,infty } left| {bf x} right|_{,p} = left| {bf x} right|_{,infty } = max left{ {left| {x_{,1} } right|,left| {x_{,2} } right|} right}
$$



Therefore
$$
eqalign{
& mathop {lim }limits_{n, to ,infty } ,;int_{x = 0}^{,4} {root n of {x^{,n} + left( {4 - x} right)^{,n} } dx} = cr
& = ,;int_{x = 0}^{,4} {left( {mathop {lim }limits_{n, to ,infty } root n of {x^{,n} + left( {4 - x} right)^{,n} } } right)dx} = cr
& = ,;int_{x = 0}^{,4} {left( {max left{ {left| x right|,left| {4 - x} right|} right}} right)dx} = cr
& = ,;2int_{x = 0}^{,2} {left( {4 - x} right)dx} = ,;2int_{x = 2}^{,4} {x,dx} = 12 cr}
$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Hint: assuming Mark’s comment is correct,
    $2^{1/n}max(x,4-x) geq (x^n+(4-x)^n)^{1/n} geq max(x,4-x)$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Yes,I have edited. But how did you find that? Something like that it's written on the solution and I dind't know how to reach that.
      $endgroup$
      – Gaboru
      Jan 7 at 14:47










    • $begingroup$
      Take everything to the $n$-th powe, maybe?
      $endgroup$
      – Mindlack
      Jan 7 at 15:16










    • $begingroup$
      I was wondering where this suggestion came from initially but just realised it is merely the infinity minkowski norm!
      $endgroup$
      – DavidG
      Jan 9 at 4:35



















    0












    $begingroup$

    NOT A SOLUTION:



    Not sure if this will be of help. I would start by repositioning the integrand:



    begin{equation}
    I = int_0^2 sqrt[n]{x^n + left(4 - xright)^n}:dx = int_0^2 (4 - x) cdot sqrt[n]{left(frac{x}{4 - x}right)^n + 1}:dx
    end{equation}



    Now let $t = dfrac{x}{4 - x}$ :



    begin{equation}
    I = int_0^1 frac{4}{t + 1}sqrt[n]{t^n + 1}frac{4}{left(t + 1right)^2}:dt = 16 int_0^1 frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}:dt
    end{equation}



    I will leave it to qualified minds to take it further. If for real continuous functions the limit of a definite integral (with bounds not featuring the variable under the limit) is equal to the integral of the limit of the integrand i.e.



    begin{equation}
    lim_{nrightarrow infty} 16 int_0^1 frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}:dt = 16 int_0^1 left[ lim_{nrightarrow infty} frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}right]:dt
    end{equation}



    Then,



    begin{align}
    I = 16 int_0^1 left[ lim_{nrightarrow infty} frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}right]:dt = 16 int_0^1 left[ lim_{nrightarrow infty} frac{t}{left(t + 1right)^3}right]:dt = frac{1}{8}
    end{align}



    But I am unsure if that is correct or not. As before, I will leave it to more qualified minds.






    share|cite|improve this answer











    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Note that the p-norm
      of a 2D vector is defined as
      $$
      left| {bf x} right|_{,p} = left( {x_{,1} ^{,p} + x_{,2} ^{,p} } right)^{,1/p}
      $$
      and it is known that
      $$
      mathop {lim }limits_{p, to ,infty } left| {bf x} right|_{,p} = left| {bf x} right|_{,infty } = max left{ {left| {x_{,1} } right|,left| {x_{,2} } right|} right}
      $$



      Therefore
      $$
      eqalign{
      & mathop {lim }limits_{n, to ,infty } ,;int_{x = 0}^{,4} {root n of {x^{,n} + left( {4 - x} right)^{,n} } dx} = cr
      & = ,;int_{x = 0}^{,4} {left( {mathop {lim }limits_{n, to ,infty } root n of {x^{,n} + left( {4 - x} right)^{,n} } } right)dx} = cr
      & = ,;int_{x = 0}^{,4} {left( {max left{ {left| x right|,left| {4 - x} right|} right}} right)dx} = cr
      & = ,;2int_{x = 0}^{,2} {left( {4 - x} right)dx} = ,;2int_{x = 2}^{,4} {x,dx} = 12 cr}
      $$






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        Note that the p-norm
        of a 2D vector is defined as
        $$
        left| {bf x} right|_{,p} = left( {x_{,1} ^{,p} + x_{,2} ^{,p} } right)^{,1/p}
        $$
        and it is known that
        $$
        mathop {lim }limits_{p, to ,infty } left| {bf x} right|_{,p} = left| {bf x} right|_{,infty } = max left{ {left| {x_{,1} } right|,left| {x_{,2} } right|} right}
        $$



        Therefore
        $$
        eqalign{
        & mathop {lim }limits_{n, to ,infty } ,;int_{x = 0}^{,4} {root n of {x^{,n} + left( {4 - x} right)^{,n} } dx} = cr
        & = ,;int_{x = 0}^{,4} {left( {mathop {lim }limits_{n, to ,infty } root n of {x^{,n} + left( {4 - x} right)^{,n} } } right)dx} = cr
        & = ,;int_{x = 0}^{,4} {left( {max left{ {left| x right|,left| {4 - x} right|} right}} right)dx} = cr
        & = ,;2int_{x = 0}^{,2} {left( {4 - x} right)dx} = ,;2int_{x = 2}^{,4} {x,dx} = 12 cr}
        $$






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          Note that the p-norm
          of a 2D vector is defined as
          $$
          left| {bf x} right|_{,p} = left( {x_{,1} ^{,p} + x_{,2} ^{,p} } right)^{,1/p}
          $$
          and it is known that
          $$
          mathop {lim }limits_{p, to ,infty } left| {bf x} right|_{,p} = left| {bf x} right|_{,infty } = max left{ {left| {x_{,1} } right|,left| {x_{,2} } right|} right}
          $$



          Therefore
          $$
          eqalign{
          & mathop {lim }limits_{n, to ,infty } ,;int_{x = 0}^{,4} {root n of {x^{,n} + left( {4 - x} right)^{,n} } dx} = cr
          & = ,;int_{x = 0}^{,4} {left( {mathop {lim }limits_{n, to ,infty } root n of {x^{,n} + left( {4 - x} right)^{,n} } } right)dx} = cr
          & = ,;int_{x = 0}^{,4} {left( {max left{ {left| x right|,left| {4 - x} right|} right}} right)dx} = cr
          & = ,;2int_{x = 0}^{,2} {left( {4 - x} right)dx} = ,;2int_{x = 2}^{,4} {x,dx} = 12 cr}
          $$






          share|cite|improve this answer









          $endgroup$



          Note that the p-norm
          of a 2D vector is defined as
          $$
          left| {bf x} right|_{,p} = left( {x_{,1} ^{,p} + x_{,2} ^{,p} } right)^{,1/p}
          $$
          and it is known that
          $$
          mathop {lim }limits_{p, to ,infty } left| {bf x} right|_{,p} = left| {bf x} right|_{,infty } = max left{ {left| {x_{,1} } right|,left| {x_{,2} } right|} right}
          $$



          Therefore
          $$
          eqalign{
          & mathop {lim }limits_{n, to ,infty } ,;int_{x = 0}^{,4} {root n of {x^{,n} + left( {4 - x} right)^{,n} } dx} = cr
          & = ,;int_{x = 0}^{,4} {left( {mathop {lim }limits_{n, to ,infty } root n of {x^{,n} + left( {4 - x} right)^{,n} } } right)dx} = cr
          & = ,;int_{x = 0}^{,4} {left( {max left{ {left| x right|,left| {4 - x} right|} right}} right)dx} = cr
          & = ,;2int_{x = 0}^{,2} {left( {4 - x} right)dx} = ,;2int_{x = 2}^{,4} {x,dx} = 12 cr}
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 9 at 18:40









          G CabG Cab

          19.5k31238




          19.5k31238























              1












              $begingroup$

              Hint: assuming Mark’s comment is correct,
              $2^{1/n}max(x,4-x) geq (x^n+(4-x)^n)^{1/n} geq max(x,4-x)$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Yes,I have edited. But how did you find that? Something like that it's written on the solution and I dind't know how to reach that.
                $endgroup$
                – Gaboru
                Jan 7 at 14:47










              • $begingroup$
                Take everything to the $n$-th powe, maybe?
                $endgroup$
                – Mindlack
                Jan 7 at 15:16










              • $begingroup$
                I was wondering where this suggestion came from initially but just realised it is merely the infinity minkowski norm!
                $endgroup$
                – DavidG
                Jan 9 at 4:35
















              1












              $begingroup$

              Hint: assuming Mark’s comment is correct,
              $2^{1/n}max(x,4-x) geq (x^n+(4-x)^n)^{1/n} geq max(x,4-x)$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Yes,I have edited. But how did you find that? Something like that it's written on the solution and I dind't know how to reach that.
                $endgroup$
                – Gaboru
                Jan 7 at 14:47










              • $begingroup$
                Take everything to the $n$-th powe, maybe?
                $endgroup$
                – Mindlack
                Jan 7 at 15:16










              • $begingroup$
                I was wondering where this suggestion came from initially but just realised it is merely the infinity minkowski norm!
                $endgroup$
                – DavidG
                Jan 9 at 4:35














              1












              1








              1





              $begingroup$

              Hint: assuming Mark’s comment is correct,
              $2^{1/n}max(x,4-x) geq (x^n+(4-x)^n)^{1/n} geq max(x,4-x)$






              share|cite|improve this answer









              $endgroup$



              Hint: assuming Mark’s comment is correct,
              $2^{1/n}max(x,4-x) geq (x^n+(4-x)^n)^{1/n} geq max(x,4-x)$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 7 at 14:40









              MindlackMindlack

              4,570210




              4,570210












              • $begingroup$
                Yes,I have edited. But how did you find that? Something like that it's written on the solution and I dind't know how to reach that.
                $endgroup$
                – Gaboru
                Jan 7 at 14:47










              • $begingroup$
                Take everything to the $n$-th powe, maybe?
                $endgroup$
                – Mindlack
                Jan 7 at 15:16










              • $begingroup$
                I was wondering where this suggestion came from initially but just realised it is merely the infinity minkowski norm!
                $endgroup$
                – DavidG
                Jan 9 at 4:35


















              • $begingroup$
                Yes,I have edited. But how did you find that? Something like that it's written on the solution and I dind't know how to reach that.
                $endgroup$
                – Gaboru
                Jan 7 at 14:47










              • $begingroup$
                Take everything to the $n$-th powe, maybe?
                $endgroup$
                – Mindlack
                Jan 7 at 15:16










              • $begingroup$
                I was wondering where this suggestion came from initially but just realised it is merely the infinity minkowski norm!
                $endgroup$
                – DavidG
                Jan 9 at 4:35
















              $begingroup$
              Yes,I have edited. But how did you find that? Something like that it's written on the solution and I dind't know how to reach that.
              $endgroup$
              – Gaboru
              Jan 7 at 14:47




              $begingroup$
              Yes,I have edited. But how did you find that? Something like that it's written on the solution and I dind't know how to reach that.
              $endgroup$
              – Gaboru
              Jan 7 at 14:47












              $begingroup$
              Take everything to the $n$-th powe, maybe?
              $endgroup$
              – Mindlack
              Jan 7 at 15:16




              $begingroup$
              Take everything to the $n$-th powe, maybe?
              $endgroup$
              – Mindlack
              Jan 7 at 15:16












              $begingroup$
              I was wondering where this suggestion came from initially but just realised it is merely the infinity minkowski norm!
              $endgroup$
              – DavidG
              Jan 9 at 4:35




              $begingroup$
              I was wondering where this suggestion came from initially but just realised it is merely the infinity minkowski norm!
              $endgroup$
              – DavidG
              Jan 9 at 4:35











              0












              $begingroup$

              NOT A SOLUTION:



              Not sure if this will be of help. I would start by repositioning the integrand:



              begin{equation}
              I = int_0^2 sqrt[n]{x^n + left(4 - xright)^n}:dx = int_0^2 (4 - x) cdot sqrt[n]{left(frac{x}{4 - x}right)^n + 1}:dx
              end{equation}



              Now let $t = dfrac{x}{4 - x}$ :



              begin{equation}
              I = int_0^1 frac{4}{t + 1}sqrt[n]{t^n + 1}frac{4}{left(t + 1right)^2}:dt = 16 int_0^1 frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}:dt
              end{equation}



              I will leave it to qualified minds to take it further. If for real continuous functions the limit of a definite integral (with bounds not featuring the variable under the limit) is equal to the integral of the limit of the integrand i.e.



              begin{equation}
              lim_{nrightarrow infty} 16 int_0^1 frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}:dt = 16 int_0^1 left[ lim_{nrightarrow infty} frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}right]:dt
              end{equation}



              Then,



              begin{align}
              I = 16 int_0^1 left[ lim_{nrightarrow infty} frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}right]:dt = 16 int_0^1 left[ lim_{nrightarrow infty} frac{t}{left(t + 1right)^3}right]:dt = frac{1}{8}
              end{align}



              But I am unsure if that is correct or not. As before, I will leave it to more qualified minds.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                NOT A SOLUTION:



                Not sure if this will be of help. I would start by repositioning the integrand:



                begin{equation}
                I = int_0^2 sqrt[n]{x^n + left(4 - xright)^n}:dx = int_0^2 (4 - x) cdot sqrt[n]{left(frac{x}{4 - x}right)^n + 1}:dx
                end{equation}



                Now let $t = dfrac{x}{4 - x}$ :



                begin{equation}
                I = int_0^1 frac{4}{t + 1}sqrt[n]{t^n + 1}frac{4}{left(t + 1right)^2}:dt = 16 int_0^1 frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}:dt
                end{equation}



                I will leave it to qualified minds to take it further. If for real continuous functions the limit of a definite integral (with bounds not featuring the variable under the limit) is equal to the integral of the limit of the integrand i.e.



                begin{equation}
                lim_{nrightarrow infty} 16 int_0^1 frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}:dt = 16 int_0^1 left[ lim_{nrightarrow infty} frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}right]:dt
                end{equation}



                Then,



                begin{align}
                I = 16 int_0^1 left[ lim_{nrightarrow infty} frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}right]:dt = 16 int_0^1 left[ lim_{nrightarrow infty} frac{t}{left(t + 1right)^3}right]:dt = frac{1}{8}
                end{align}



                But I am unsure if that is correct or not. As before, I will leave it to more qualified minds.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  NOT A SOLUTION:



                  Not sure if this will be of help. I would start by repositioning the integrand:



                  begin{equation}
                  I = int_0^2 sqrt[n]{x^n + left(4 - xright)^n}:dx = int_0^2 (4 - x) cdot sqrt[n]{left(frac{x}{4 - x}right)^n + 1}:dx
                  end{equation}



                  Now let $t = dfrac{x}{4 - x}$ :



                  begin{equation}
                  I = int_0^1 frac{4}{t + 1}sqrt[n]{t^n + 1}frac{4}{left(t + 1right)^2}:dt = 16 int_0^1 frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}:dt
                  end{equation}



                  I will leave it to qualified minds to take it further. If for real continuous functions the limit of a definite integral (with bounds not featuring the variable under the limit) is equal to the integral of the limit of the integrand i.e.



                  begin{equation}
                  lim_{nrightarrow infty} 16 int_0^1 frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}:dt = 16 int_0^1 left[ lim_{nrightarrow infty} frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}right]:dt
                  end{equation}



                  Then,



                  begin{align}
                  I = 16 int_0^1 left[ lim_{nrightarrow infty} frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}right]:dt = 16 int_0^1 left[ lim_{nrightarrow infty} frac{t}{left(t + 1right)^3}right]:dt = frac{1}{8}
                  end{align}



                  But I am unsure if that is correct or not. As before, I will leave it to more qualified minds.






                  share|cite|improve this answer











                  $endgroup$



                  NOT A SOLUTION:



                  Not sure if this will be of help. I would start by repositioning the integrand:



                  begin{equation}
                  I = int_0^2 sqrt[n]{x^n + left(4 - xright)^n}:dx = int_0^2 (4 - x) cdot sqrt[n]{left(frac{x}{4 - x}right)^n + 1}:dx
                  end{equation}



                  Now let $t = dfrac{x}{4 - x}$ :



                  begin{equation}
                  I = int_0^1 frac{4}{t + 1}sqrt[n]{t^n + 1}frac{4}{left(t + 1right)^2}:dt = 16 int_0^1 frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}:dt
                  end{equation}



                  I will leave it to qualified minds to take it further. If for real continuous functions the limit of a definite integral (with bounds not featuring the variable under the limit) is equal to the integral of the limit of the integrand i.e.



                  begin{equation}
                  lim_{nrightarrow infty} 16 int_0^1 frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}:dt = 16 int_0^1 left[ lim_{nrightarrow infty} frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}right]:dt
                  end{equation}



                  Then,



                  begin{align}
                  I = 16 int_0^1 left[ lim_{nrightarrow infty} frac{sqrt[n]{t^n + 1}}{left(t + 1right)^3}right]:dt = 16 int_0^1 left[ lim_{nrightarrow infty} frac{t}{left(t + 1right)^3}right]:dt = frac{1}{8}
                  end{align}



                  But I am unsure if that is correct or not. As before, I will leave it to more qualified minds.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 8 at 13:01

























                  answered Jan 8 at 12:55









                  DavidGDavidG

                  2,1121723




                  2,1121723






























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