Let $f$ be a continuous function on $[0,infty)$ such that $lim_{xto infty}(f(x)+int_0^x f(t)dt)$ exists. Find...












1














Let $f$ be a continuous function on $[0,infty)$ such that $lim_{xto infty}(f(x)+int_0^x f(t)dt)$ exists. Find $lim_{xto infty}f(x)$.



Useful hints will work. Please give some nice hint to solve the problem.






integration limits functions continuity







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  • Try taking $g=f+K$ with $K$ some constant. What happends? Can you guess the value of the limit?
    – xavierm02
    Feb 2 '17 at 14:36










  • @xavierm02 taking $g$=$f+K$ we will get $lim_{xto infty} f(x)+K$=$lim_{xto infty}g(x)$.Is it correct ? If yes whats next?
    – Piyush Raut
    Feb 2 '17 at 14:39












  • But what is $lim_{xto infty}(g(x)+int_0^x g(t)dt)$?
    – xavierm02
    Feb 2 '17 at 14:42












  • I didn't get that.Did you mean $lim_{xto infty}(f(x)+int_0^x f(t)dt)$?
    – Piyush Raut
    Feb 2 '17 at 14:45










  • No I meant with $g$, under the assumption that the limit exists with $f$.
    – xavierm02
    Feb 2 '17 at 14:45
















1














Let $f$ be a continuous function on $[0,infty)$ such that $lim_{xto infty}(f(x)+int_0^x f(t)dt)$ exists. Find $lim_{xto infty}f(x)$.



Useful hints will work. Please give some nice hint to solve the problem.






integration limits functions continuity







share|cite|improve this question
























  • Try taking $g=f+K$ with $K$ some constant. What happends? Can you guess the value of the limit?
    – xavierm02
    Feb 2 '17 at 14:36










  • @xavierm02 taking $g$=$f+K$ we will get $lim_{xto infty} f(x)+K$=$lim_{xto infty}g(x)$.Is it correct ? If yes whats next?
    – Piyush Raut
    Feb 2 '17 at 14:39












  • But what is $lim_{xto infty}(g(x)+int_0^x g(t)dt)$?
    – xavierm02
    Feb 2 '17 at 14:42












  • I didn't get that.Did you mean $lim_{xto infty}(f(x)+int_0^x f(t)dt)$?
    – Piyush Raut
    Feb 2 '17 at 14:45










  • No I meant with $g$, under the assumption that the limit exists with $f$.
    – xavierm02
    Feb 2 '17 at 14:45














1












1








1


1





Let $f$ be a continuous function on $[0,infty)$ such that $lim_{xto infty}(f(x)+int_0^x f(t)dt)$ exists. Find $lim_{xto infty}f(x)$.



Useful hints will work. Please give some nice hint to solve the problem.






integration limits functions continuity







share|cite|improve this question















Let $f$ be a continuous function on $[0,infty)$ such that $lim_{xto infty}(f(x)+int_0^x f(t)dt)$ exists. Find $lim_{xto infty}f(x)$.



Useful hints will work. Please give some nice hint to solve the problem.






integration limits functions continuity




integration limits functions continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 21:00









Shaun

8,820113681




8,820113681










asked Feb 2 '17 at 14:32









Piyush RautPiyush Raut

577




577












  • Try taking $g=f+K$ with $K$ some constant. What happends? Can you guess the value of the limit?
    – xavierm02
    Feb 2 '17 at 14:36










  • @xavierm02 taking $g$=$f+K$ we will get $lim_{xto infty} f(x)+K$=$lim_{xto infty}g(x)$.Is it correct ? If yes whats next?
    – Piyush Raut
    Feb 2 '17 at 14:39












  • But what is $lim_{xto infty}(g(x)+int_0^x g(t)dt)$?
    – xavierm02
    Feb 2 '17 at 14:42












  • I didn't get that.Did you mean $lim_{xto infty}(f(x)+int_0^x f(t)dt)$?
    – Piyush Raut
    Feb 2 '17 at 14:45










  • No I meant with $g$, under the assumption that the limit exists with $f$.
    – xavierm02
    Feb 2 '17 at 14:45


















  • Try taking $g=f+K$ with $K$ some constant. What happends? Can you guess the value of the limit?
    – xavierm02
    Feb 2 '17 at 14:36










  • @xavierm02 taking $g$=$f+K$ we will get $lim_{xto infty} f(x)+K$=$lim_{xto infty}g(x)$.Is it correct ? If yes whats next?
    – Piyush Raut
    Feb 2 '17 at 14:39












  • But what is $lim_{xto infty}(g(x)+int_0^x g(t)dt)$?
    – xavierm02
    Feb 2 '17 at 14:42












  • I didn't get that.Did you mean $lim_{xto infty}(f(x)+int_0^x f(t)dt)$?
    – Piyush Raut
    Feb 2 '17 at 14:45










  • No I meant with $g$, under the assumption that the limit exists with $f$.
    – xavierm02
    Feb 2 '17 at 14:45
















Try taking $g=f+K$ with $K$ some constant. What happends? Can you guess the value of the limit?
– xavierm02
Feb 2 '17 at 14:36




Try taking $g=f+K$ with $K$ some constant. What happends? Can you guess the value of the limit?
– xavierm02
Feb 2 '17 at 14:36












@xavierm02 taking $g$=$f+K$ we will get $lim_{xto infty} f(x)+K$=$lim_{xto infty}g(x)$.Is it correct ? If yes whats next?
– Piyush Raut
Feb 2 '17 at 14:39






@xavierm02 taking $g$=$f+K$ we will get $lim_{xto infty} f(x)+K$=$lim_{xto infty}g(x)$.Is it correct ? If yes whats next?
– Piyush Raut
Feb 2 '17 at 14:39














But what is $lim_{xto infty}(g(x)+int_0^x g(t)dt)$?
– xavierm02
Feb 2 '17 at 14:42






But what is $lim_{xto infty}(g(x)+int_0^x g(t)dt)$?
– xavierm02
Feb 2 '17 at 14:42














I didn't get that.Did you mean $lim_{xto infty}(f(x)+int_0^x f(t)dt)$?
– Piyush Raut
Feb 2 '17 at 14:45




I didn't get that.Did you mean $lim_{xto infty}(f(x)+int_0^x f(t)dt)$?
– Piyush Raut
Feb 2 '17 at 14:45












No I meant with $g$, under the assumption that the limit exists with $f$.
– xavierm02
Feb 2 '17 at 14:45




No I meant with $g$, under the assumption that the limit exists with $f$.
– xavierm02
Feb 2 '17 at 14:45










2 Answers
2






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oldest

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4














Let $F(x) = int_0^x f(t)dt$. Then, we have $lim_{xto infty} e^{-x}[e^x F(x)]' = c$ exists. From this, given $epsilon>0$, there is $N>0$ such that for all $x>N$,
$$
(c-epsilon)e^x leq[e^x F(x)]' leq (c+epsilon)e^x.
$$
Now, integrating from $N$ to $t$ gives:
$$(c-epsilon)(e^t-e^N)leq e^tF(t) -e^NF(N) leq (c+epsilon)(e^t-e^N),quadforall t>N
$$and hence
$$
(c-epsilon)(1-e^{N-t})leq F(t) -e^{N-t}F(N) leq (c+epsilon)(1-e^{N-t}), quad forall t>N.
$$Taking $ttoinfty$ yields
$$c-epsilon leq liminf_{tto infty}F(t)leq limsup_{tto infty}F(t)leq c+epsilon.
$$Since $epsilon>0$ was arbitrary, we have $lim_{ttoinfty} F(t) =c$ and thus $lim_{ttoinfty} f(t) = 0$.






share|cite|improve this answer





























    3





    +50









    One of the easiest approaches is the use of L'Hospital's Rule here. Let $F(x) =int_{0}^{x}f(t),dt$ so that $F'(x) =f(x) $ via Fundamental Theorem of Calculus. We are thus given that $F(x) +F'(x) to L$ as $xtoinfty$.



    We have
    begin{align}
    lim_{xtoinfty} F(x) &=lim_{xtoinfty} frac{e^xF(x)} {e^x}notag\
    &=lim_{xto infty} dfrac{e^x{F(x) +F'(x) }} {e^x}text{ (via L'Hospital's Rule)} notag\
    &=lim_{xtoinfty} F(x) +F'(x)notag\
    &=Lnotag
    end{align}
    and hence $$lim_{xto infty} f(x) =lim_{xtoinfty} F'(x) =lim_{xtoinfty} {F(x) + F'(x) }-F(x) =L-L=0$$






    share|cite|improve this answer

















    • 1




      Nice solution, +1. But the flow is weird because you write $lim$ when it hasn't been shown the limit exists. Also, you don't know the limit of the numerator, $e^xF(x),$ is $infty.$ Now to use L'Hopital in this case, you don't need to know that. But many readers may not know this little fact.
      – zhw.
      Nov 30 '18 at 17:51












    • @zhw : I should have added more details. Thanks for putting these details in your comment.
      – Paramanand Singh
      Nov 30 '18 at 17:55











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    2 Answers
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    4














    Let $F(x) = int_0^x f(t)dt$. Then, we have $lim_{xto infty} e^{-x}[e^x F(x)]' = c$ exists. From this, given $epsilon>0$, there is $N>0$ such that for all $x>N$,
    $$
    (c-epsilon)e^x leq[e^x F(x)]' leq (c+epsilon)e^x.
    $$
    Now, integrating from $N$ to $t$ gives:
    $$(c-epsilon)(e^t-e^N)leq e^tF(t) -e^NF(N) leq (c+epsilon)(e^t-e^N),quadforall t>N
    $$    and hence
    $$
    (c-epsilon)(1-e^{N-t})leq F(t) -e^{N-t}F(N) leq (c+epsilon)(1-e^{N-t}), quad forall t>N.
    $$    Taking $ttoinfty$ yields
    $$c-epsilon leq liminf_{tto infty}F(t)leq limsup_{tto infty}F(t)leq c+epsilon.
    $$    Since $epsilon>0$ was arbitrary, we have $lim_{ttoinfty} F(t) =c$ and thus $lim_{ttoinfty} f(t) = 0$.






    share|cite|improve this answer


























      4














      Let $F(x) = int_0^x f(t)dt$. Then, we have $lim_{xto infty} e^{-x}[e^x F(x)]' = c$ exists. From this, given $epsilon>0$, there is $N>0$ such that for all $x>N$,
      $$
      (c-epsilon)e^x leq[e^x F(x)]' leq (c+epsilon)e^x.
      $$
      Now, integrating from $N$ to $t$ gives:
      $$(c-epsilon)(e^t-e^N)leq e^tF(t) -e^NF(N) leq (c+epsilon)(e^t-e^N),quadforall t>N
      $$      and hence
      $$
      (c-epsilon)(1-e^{N-t})leq F(t) -e^{N-t}F(N) leq (c+epsilon)(1-e^{N-t}), quad forall t>N.
      $$      Taking $ttoinfty$ yields
      $$c-epsilon leq liminf_{tto infty}F(t)leq limsup_{tto infty}F(t)leq c+epsilon.
      $$      Since $epsilon>0$ was arbitrary, we have $lim_{ttoinfty} F(t) =c$ and thus $lim_{ttoinfty} f(t) = 0$.






      share|cite|improve this answer
























        4












        4








        4






        Let $F(x) = int_0^x f(t)dt$. Then, we have $lim_{xto infty} e^{-x}[e^x F(x)]' = c$ exists. From this, given $epsilon>0$, there is $N>0$ such that for all $x>N$,
        $$
        (c-epsilon)e^x leq[e^x F(x)]' leq (c+epsilon)e^x.
        $$
        Now, integrating from $N$ to $t$ gives:
        $$(c-epsilon)(e^t-e^N)leq e^tF(t) -e^NF(N) leq (c+epsilon)(e^t-e^N),quadforall t>N
        $$        and hence
        $$
        (c-epsilon)(1-e^{N-t})leq F(t) -e^{N-t}F(N) leq (c+epsilon)(1-e^{N-t}), quad forall t>N.
        $$        Taking $ttoinfty$ yields
        $$c-epsilon leq liminf_{tto infty}F(t)leq limsup_{tto infty}F(t)leq c+epsilon.
        $$        Since $epsilon>0$ was arbitrary, we have $lim_{ttoinfty} F(t) =c$ and thus $lim_{ttoinfty} f(t) = 0$.






        share|cite|improve this answer












        Let $F(x) = int_0^x f(t)dt$. Then, we have $lim_{xto infty} e^{-x}[e^x F(x)]' = c$ exists. From this, given $epsilon>0$, there is $N>0$ such that for all $x>N$,
        $$
        (c-epsilon)e^x leq[e^x F(x)]' leq (c+epsilon)e^x.
        $$
        Now, integrating from $N$ to $t$ gives:
        $$(c-epsilon)(e^t-e^N)leq e^tF(t) -e^NF(N) leq (c+epsilon)(e^t-e^N),quadforall t>N
        $$        and hence
        $$
        (c-epsilon)(1-e^{N-t})leq F(t) -e^{N-t}F(N) leq (c+epsilon)(1-e^{N-t}), quad forall t>N.
        $$        Taking $ttoinfty$ yields
        $$c-epsilon leq liminf_{tto infty}F(t)leq limsup_{tto infty}F(t)leq c+epsilon.
        $$        Since $epsilon>0$ was arbitrary, we have $lim_{ttoinfty} F(t) =c$ and thus $lim_{ttoinfty} f(t) = 0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 '18 at 7:45









        SongSong

        6,948421




        6,948421























            3





            +50









            One of the easiest approaches is the use of L'Hospital's Rule here. Let $F(x) =int_{0}^{x}f(t),dt$ so that $F'(x) =f(x) $ via Fundamental Theorem of Calculus. We are thus given that $F(x) +F'(x) to L$ as $xtoinfty$.



            We have
            begin{align}
            lim_{xtoinfty} F(x) &=lim_{xtoinfty} frac{e^xF(x)} {e^x}notag\
            &=lim_{xto infty} dfrac{e^x{F(x) +F'(x) }} {e^x}text{ (via L'Hospital's Rule)} notag\
            &=lim_{xtoinfty} F(x) +F'(x)notag\
            &=Lnotag
            end{align}
            and hence $$lim_{xto infty} f(x) =lim_{xtoinfty} F'(x) =lim_{xtoinfty} {F(x) + F'(x) }-F(x) =L-L=0$$






            share|cite|improve this answer

















            • 1




              Nice solution, +1. But the flow is weird because you write $lim$ when it hasn't been shown the limit exists. Also, you don't know the limit of the numerator, $e^xF(x),$ is $infty.$ Now to use L'Hopital in this case, you don't need to know that. But many readers may not know this little fact.
              – zhw.
              Nov 30 '18 at 17:51












            • @zhw : I should have added more details. Thanks for putting these details in your comment.
              – Paramanand Singh
              Nov 30 '18 at 17:55
















            3





            +50









            One of the easiest approaches is the use of L'Hospital's Rule here. Let $F(x) =int_{0}^{x}f(t),dt$ so that $F'(x) =f(x) $ via Fundamental Theorem of Calculus. We are thus given that $F(x) +F'(x) to L$ as $xtoinfty$.



            We have
            begin{align}
            lim_{xtoinfty} F(x) &=lim_{xtoinfty} frac{e^xF(x)} {e^x}notag\
            &=lim_{xto infty} dfrac{e^x{F(x) +F'(x) }} {e^x}text{ (via L'Hospital's Rule)} notag\
            &=lim_{xtoinfty} F(x) +F'(x)notag\
            &=Lnotag
            end{align}
            and hence $$lim_{xto infty} f(x) =lim_{xtoinfty} F'(x) =lim_{xtoinfty} {F(x) + F'(x) }-F(x) =L-L=0$$






            share|cite|improve this answer

















            • 1




              Nice solution, +1. But the flow is weird because you write $lim$ when it hasn't been shown the limit exists. Also, you don't know the limit of the numerator, $e^xF(x),$ is $infty.$ Now to use L'Hopital in this case, you don't need to know that. But many readers may not know this little fact.
              – zhw.
              Nov 30 '18 at 17:51












            • @zhw : I should have added more details. Thanks for putting these details in your comment.
              – Paramanand Singh
              Nov 30 '18 at 17:55














            3





            +50







            3





            +50



            3




            +50




            One of the easiest approaches is the use of L'Hospital's Rule here. Let $F(x) =int_{0}^{x}f(t),dt$ so that $F'(x) =f(x) $ via Fundamental Theorem of Calculus. We are thus given that $F(x) +F'(x) to L$ as $xtoinfty$.



            We have
            begin{align}
            lim_{xtoinfty} F(x) &=lim_{xtoinfty} frac{e^xF(x)} {e^x}notag\
            &=lim_{xto infty} dfrac{e^x{F(x) +F'(x) }} {e^x}text{ (via L'Hospital's Rule)} notag\
            &=lim_{xtoinfty} F(x) +F'(x)notag\
            &=Lnotag
            end{align}
            and hence $$lim_{xto infty} f(x) =lim_{xtoinfty} F'(x) =lim_{xtoinfty} {F(x) + F'(x) }-F(x) =L-L=0$$






            share|cite|improve this answer












            One of the easiest approaches is the use of L'Hospital's Rule here. Let $F(x) =int_{0}^{x}f(t),dt$ so that $F'(x) =f(x) $ via Fundamental Theorem of Calculus. We are thus given that $F(x) +F'(x) to L$ as $xtoinfty$.



            We have
            begin{align}
            lim_{xtoinfty} F(x) &=lim_{xtoinfty} frac{e^xF(x)} {e^x}notag\
            &=lim_{xto infty} dfrac{e^x{F(x) +F'(x) }} {e^x}text{ (via L'Hospital's Rule)} notag\
            &=lim_{xtoinfty} F(x) +F'(x)notag\
            &=Lnotag
            end{align}
            and hence $$lim_{xto infty} f(x) =lim_{xtoinfty} F'(x) =lim_{xtoinfty} {F(x) + F'(x) }-F(x) =L-L=0$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 30 '18 at 2:37









            Paramanand SinghParamanand Singh

            49.2k555161




            49.2k555161








            • 1




              Nice solution, +1. But the flow is weird because you write $lim$ when it hasn't been shown the limit exists. Also, you don't know the limit of the numerator, $e^xF(x),$ is $infty.$ Now to use L'Hopital in this case, you don't need to know that. But many readers may not know this little fact.
              – zhw.
              Nov 30 '18 at 17:51












            • @zhw : I should have added more details. Thanks for putting these details in your comment.
              – Paramanand Singh
              Nov 30 '18 at 17:55














            • 1




              Nice solution, +1. But the flow is weird because you write $lim$ when it hasn't been shown the limit exists. Also, you don't know the limit of the numerator, $e^xF(x),$ is $infty.$ Now to use L'Hopital in this case, you don't need to know that. But many readers may not know this little fact.
              – zhw.
              Nov 30 '18 at 17:51












            • @zhw : I should have added more details. Thanks for putting these details in your comment.
              – Paramanand Singh
              Nov 30 '18 at 17:55








            1




            1




            Nice solution, +1. But the flow is weird because you write $lim$ when it hasn't been shown the limit exists. Also, you don't know the limit of the numerator, $e^xF(x),$ is $infty.$ Now to use L'Hopital in this case, you don't need to know that. But many readers may not know this little fact.
            – zhw.
            Nov 30 '18 at 17:51






            Nice solution, +1. But the flow is weird because you write $lim$ when it hasn't been shown the limit exists. Also, you don't know the limit of the numerator, $e^xF(x),$ is $infty.$ Now to use L'Hopital in this case, you don't need to know that. But many readers may not know this little fact.
            – zhw.
            Nov 30 '18 at 17:51














            @zhw : I should have added more details. Thanks for putting these details in your comment.
            – Paramanand Singh
            Nov 30 '18 at 17:55




            @zhw : I should have added more details. Thanks for putting these details in your comment.
            – Paramanand Singh
            Nov 30 '18 at 17:55


















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            Quarter-circle Tiles

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            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
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            Mont Emei

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            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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