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












$begingroup$


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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$endgroup$












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










  • $begingroup$
    @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?
    $endgroup$
    – Piyush Raut
    Feb 2 '17 at 14:39












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












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










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
















1












$begingroup$


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











$endgroup$












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










  • $begingroup$
    @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?
    $endgroup$
    – Piyush Raut
    Feb 2 '17 at 14:39












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












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










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














1












1








1


1



$begingroup$


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











$endgroup$




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

8,888113681




8,888113681










asked Feb 2 '17 at 14:32









Piyush RautPiyush Raut

577




577












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










  • $begingroup$
    @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?
    $endgroup$
    – Piyush Raut
    Feb 2 '17 at 14:39












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












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










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


















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










  • $begingroup$
    @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?
    $endgroup$
    – Piyush Raut
    Feb 2 '17 at 14:39












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












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










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
















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




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












$begingroup$
@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?
$endgroup$
– Piyush Raut
Feb 2 '17 at 14:39






$begingroup$
@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?
$endgroup$
– Piyush Raut
Feb 2 '17 at 14:39














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






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














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




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












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




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










2 Answers
2






active

oldest

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4












$begingroup$

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









$endgroup$





















    3





    +50







    $begingroup$

    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









    $endgroup$









    • 1




      $begingroup$
      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.
      $endgroup$
      – zhw.
      Nov 30 '18 at 17:51












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











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    2 Answers
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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    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









    $endgroup$


















      4












      $begingroup$

      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









      $endgroup$
















        4












        4








        4





        $begingroup$

        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









        $endgroup$



        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

        8,311625




        8,311625























            3





            +50







            $begingroup$

            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









            $endgroup$









            • 1




              $begingroup$
              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.
              $endgroup$
              – zhw.
              Nov 30 '18 at 17:51












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
















            3





            +50







            $begingroup$

            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









            $endgroup$









            • 1




              $begingroup$
              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.
              $endgroup$
              – zhw.
              Nov 30 '18 at 17:51












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














            3





            +50







            3





            +50



            3




            +50



            $begingroup$

            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









            $endgroup$



            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

            49.4k555162




            49.4k555162








            • 1




              $begingroup$
              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.
              $endgroup$
              – zhw.
              Nov 30 '18 at 17:51












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














            • 1




              $begingroup$
              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.
              $endgroup$
              – zhw.
              Nov 30 '18 at 17:51












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








            1




            1




            $begingroup$
            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.
            $endgroup$
            – zhw.
            Nov 30 '18 at 17:51






            $begingroup$
            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.
            $endgroup$
            – zhw.
            Nov 30 '18 at 17:51














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




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


















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