$L= lim_{n to infty}sqrt{n}int_0^1dfrac{dx}{(1+x^2)^n}$ [duplicate]











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  • Finding $lim_{nto infty}sqrt n int_0^1 frac{,dx}{(1+x^2)^n}$

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Suppose the limit $$L= lim_{n to infty}sqrt{n}int_0^1dfrac{dx}{(1+x^2)^n}$$

and is larger than $frac{1}{2}$. Then



(A) $0.5<L<2$



(B) $2<L<3$



(C) $3<L<4$



(D) $Lge 4$



I used $x=tantheta$ but not getting the value






limits







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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











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    For the actual value of the limit, see math.stackexchange.com/q/2984468/321264.
    – StubbornAtom
    Nov 14 at 17:20

















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This question already has an answer here:




  • Finding $lim_{nto infty}sqrt n int_0^1 frac{,dx}{(1+x^2)^n}$

    3 answers




Suppose the limit $$L= lim_{n to infty}sqrt{n}int_0^1dfrac{dx}{(1+x^2)^n}$$

and is larger than $frac{1}{2}$. Then



(A) $0.5<L<2$



(B) $2<L<3$



(C) $3<L<4$



(D) $Lge 4$



I used $x=tantheta$ but not getting the value






limits







share|cite|improve this question















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 Nov 14 at 18:52


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 1




    For the actual value of the limit, see math.stackexchange.com/q/2984468/321264.
    – StubbornAtom
    Nov 14 at 17:20















up vote
1
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favorite









up vote
1
down vote

favorite












This question already has an answer here:




  • Finding $lim_{nto infty}sqrt n int_0^1 frac{,dx}{(1+x^2)^n}$

    3 answers




Suppose the limit $$L= lim_{n to infty}sqrt{n}int_0^1dfrac{dx}{(1+x^2)^n}$$

and is larger than $frac{1}{2}$. Then



(A) $0.5<L<2$



(B) $2<L<3$



(C) $3<L<4$



(D) $Lge 4$



I used $x=tantheta$ but not getting the value






limits







share|cite|improve this question
















This question already has an answer here:




  • Finding $lim_{nto infty}sqrt n int_0^1 frac{,dx}{(1+x^2)^n}$

    3 answers




Suppose the limit $$L= lim_{n to infty}sqrt{n}int_0^1dfrac{dx}{(1+x^2)^n}$$

and is larger than $frac{1}{2}$. Then



(A) $0.5<L<2$



(B) $2<L<3$



(C) $3<L<4$



(D) $Lge 4$



I used $x=tantheta$ but not getting the value





This question already has an answer here:




  • Finding $lim_{nto infty}sqrt n int_0^1 frac{,dx}{(1+x^2)^n}$

    3 answers







limits




limits






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edited Nov 14 at 17:18









Tianlalu

2,599632




2,599632










asked Nov 14 at 17:16









Samar Imam Zaidi

1,427417




1,427417




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 Nov 14 at 18:52


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    For the actual value of the limit, see math.stackexchange.com/q/2984468/321264.
    – StubbornAtom
    Nov 14 at 17:20
















  • 1




    For the actual value of the limit, see math.stackexchange.com/q/2984468/321264.
    – StubbornAtom
    Nov 14 at 17:20










1




1




For the actual value of the limit, see math.stackexchange.com/q/2984468/321264.
– StubbornAtom
Nov 14 at 17:20






For the actual value of the limit, see math.stackexchange.com/q/2984468/321264.
– StubbornAtom
Nov 14 at 17:20












2 Answers
2






active

oldest

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up vote
2
down vote













For $x ge 0$ you have
$$(1+x^2)^n ge 1+nx^2$$ hence
$$frac{1}{(1+x^2)^n} le frac{1}{1+n x^2}$$ and



$$u_n = int_0^1 frac{dx}{(1+x^2)^n} le int_0^1 frac{dx}{1+nx^2} = frac{1}{sqrt{n}}int_0^sqrt{n} frac{du}{1+u^2} le frac{1}{sqrt{n}}int_0^infty frac{du}{1+u^2} = frac{pi}{2 sqrt{n}}$$ using integral by substitution and the classical result
$$int_0^infty frac{du}{1+u^2} = frac{pi}{2}.$$



Therefore if $L$ exists as supposed in the hypothesis, we have
$$L le frac{pi}{2} < 2$$
proving that the right answer is (A).






share|cite|improve this answer























  • Nice, that's simpler than my answer. +1
    – zhw.
    Nov 14 at 19:59


















up vote
0
down vote













For $xin [0,1],$



$$frac{1}{1+x^2} le 1-frac{x^2}{2}.$$



Thus our expression is bounded above by



$$sqrt nint_0^1 (1-x^2/2)^n, dx.$$



Let $x=y/sqrt n.$ The above turns into



$$int_0^sqrt n (1-y^2/(2n))^n, dyle int_0^infty e^{-y^2/2},dy = sqrt 2frac{ sqrt pi}{2} <2.$$






share|cite|improve this answer




























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote













    For $x ge 0$ you have
    $$(1+x^2)^n ge 1+nx^2$$ hence
    $$frac{1}{(1+x^2)^n} le frac{1}{1+n x^2}$$ and



    $$u_n = int_0^1 frac{dx}{(1+x^2)^n} le int_0^1 frac{dx}{1+nx^2} = frac{1}{sqrt{n}}int_0^sqrt{n} frac{du}{1+u^2} le frac{1}{sqrt{n}}int_0^infty frac{du}{1+u^2} = frac{pi}{2 sqrt{n}}$$ using integral by substitution and the classical result
    $$int_0^infty frac{du}{1+u^2} = frac{pi}{2}.$$



    Therefore if $L$ exists as supposed in the hypothesis, we have
    $$L le frac{pi}{2} < 2$$
    proving that the right answer is (A).






    share|cite|improve this answer























    • Nice, that's simpler than my answer. +1
      – zhw.
      Nov 14 at 19:59















    up vote
    2
    down vote













    For $x ge 0$ you have
    $$(1+x^2)^n ge 1+nx^2$$ hence
    $$frac{1}{(1+x^2)^n} le frac{1}{1+n x^2}$$ and



    $$u_n = int_0^1 frac{dx}{(1+x^2)^n} le int_0^1 frac{dx}{1+nx^2} = frac{1}{sqrt{n}}int_0^sqrt{n} frac{du}{1+u^2} le frac{1}{sqrt{n}}int_0^infty frac{du}{1+u^2} = frac{pi}{2 sqrt{n}}$$ using integral by substitution and the classical result
    $$int_0^infty frac{du}{1+u^2} = frac{pi}{2}.$$



    Therefore if $L$ exists as supposed in the hypothesis, we have
    $$L le frac{pi}{2} < 2$$
    proving that the right answer is (A).






    share|cite|improve this answer























    • Nice, that's simpler than my answer. +1
      – zhw.
      Nov 14 at 19:59













    up vote
    2
    down vote










    up vote
    2
    down vote









    For $x ge 0$ you have
    $$(1+x^2)^n ge 1+nx^2$$ hence
    $$frac{1}{(1+x^2)^n} le frac{1}{1+n x^2}$$ and



    $$u_n = int_0^1 frac{dx}{(1+x^2)^n} le int_0^1 frac{dx}{1+nx^2} = frac{1}{sqrt{n}}int_0^sqrt{n} frac{du}{1+u^2} le frac{1}{sqrt{n}}int_0^infty frac{du}{1+u^2} = frac{pi}{2 sqrt{n}}$$ using integral by substitution and the classical result
    $$int_0^infty frac{du}{1+u^2} = frac{pi}{2}.$$



    Therefore if $L$ exists as supposed in the hypothesis, we have
    $$L le frac{pi}{2} < 2$$
    proving that the right answer is (A).






    share|cite|improve this answer














    For $x ge 0$ you have
    $$(1+x^2)^n ge 1+nx^2$$ hence
    $$frac{1}{(1+x^2)^n} le frac{1}{1+n x^2}$$ and



    $$u_n = int_0^1 frac{dx}{(1+x^2)^n} le int_0^1 frac{dx}{1+nx^2} = frac{1}{sqrt{n}}int_0^sqrt{n} frac{du}{1+u^2} le frac{1}{sqrt{n}}int_0^infty frac{du}{1+u^2} = frac{pi}{2 sqrt{n}}$$ using integral by substitution and the classical result
    $$int_0^infty frac{du}{1+u^2} = frac{pi}{2}.$$



    Therefore if $L$ exists as supposed in the hypothesis, we have
    $$L le frac{pi}{2} < 2$$
    proving that the right answer is (A).







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 14 at 18:11

























    answered Nov 14 at 18:05









    mathcounterexamples.net

    23.6k21753




    23.6k21753












    • Nice, that's simpler than my answer. +1
      – zhw.
      Nov 14 at 19:59


















    • Nice, that's simpler than my answer. +1
      – zhw.
      Nov 14 at 19:59
















    Nice, that's simpler than my answer. +1
    – zhw.
    Nov 14 at 19:59




    Nice, that's simpler than my answer. +1
    – zhw.
    Nov 14 at 19:59










    up vote
    0
    down vote













    For $xin [0,1],$



    $$frac{1}{1+x^2} le 1-frac{x^2}{2}.$$



    Thus our expression is bounded above by



    $$sqrt nint_0^1 (1-x^2/2)^n, dx.$$



    Let $x=y/sqrt n.$ The above turns into



    $$int_0^sqrt n (1-y^2/(2n))^n, dyle int_0^infty e^{-y^2/2},dy = sqrt 2frac{ sqrt pi}{2} <2.$$






    share|cite|improve this answer

























      up vote
      0
      down vote













      For $xin [0,1],$



      $$frac{1}{1+x^2} le 1-frac{x^2}{2}.$$



      Thus our expression is bounded above by



      $$sqrt nint_0^1 (1-x^2/2)^n, dx.$$



      Let $x=y/sqrt n.$ The above turns into



      $$int_0^sqrt n (1-y^2/(2n))^n, dyle int_0^infty e^{-y^2/2},dy = sqrt 2frac{ sqrt pi}{2} <2.$$






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        For $xin [0,1],$



        $$frac{1}{1+x^2} le 1-frac{x^2}{2}.$$



        Thus our expression is bounded above by



        $$sqrt nint_0^1 (1-x^2/2)^n, dx.$$



        Let $x=y/sqrt n.$ The above turns into



        $$int_0^sqrt n (1-y^2/(2n))^n, dyle int_0^infty e^{-y^2/2},dy = sqrt 2frac{ sqrt pi}{2} <2.$$






        share|cite|improve this answer












        For $xin [0,1],$



        $$frac{1}{1+x^2} le 1-frac{x^2}{2}.$$



        Thus our expression is bounded above by



        $$sqrt nint_0^1 (1-x^2/2)^n, dx.$$



        Let $x=y/sqrt n.$ The above turns into



        $$int_0^sqrt n (1-y^2/(2n))^n, dyle int_0^infty e^{-y^2/2},dy = sqrt 2frac{ sqrt pi}{2} <2.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 14 at 18:21









        zhw.

        70.4k42975




        70.4k42975















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