Limits at infinity for sequences and funtions












0














First recall the definitions




Definition I: Let $(a_n)$ be a sequence of real or complex numbers. We say that $a_nto infty$ as $nto infty$, if
$$
forall R>0 exists Ninmathbb{N} forall ninmathbb{N}:[(ngeq N)implies (a_n>R) ].
$$



Definition II: Let $f:Dto mathbb{C}$, where $Dsubseteq mathbb{R}$ is not bounded above. We say that $f(x)to infty$ as $xto infty$, if
$$
forall R>0, exists Kinmathbb{R}, forall xin D: [(xgeq K)implies (f(x)>R)].
$$




Question: Let $f:[1,infty)to mathbb{C}$, and let $a_n=f(n)$ for $ngeq 1$. It is clear that if $f(x)toinfty$ as $xtoinfty$, then $a_ntoinfty$ as $ntoinfty$ (i.e. Definition II implies Definition I). Does the converse direction hold in general? If we assume that $a_ntoinfty$ as $ntoinfty$, then there are a lot of "gaps" (i.e. the elements of $[1,infty)setminus mathbb{N}$) that should be checked in Definition II.






real-analysis







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  • Yes, I think you are right. We can easily imagine a function that goes to hell and back between integers and yet always is "nice" on the integers. For example $f(x) = xcos(2pi*x)$ for $a_n= f(n) = n$ it's clear $a_nto infty$ but between integers will go way up and way down and it's clear definition 2 fails. .. Oooh.... also if $b_n = f(n + frac 12)=-n-frac 12to -infty$. We can't conclude both $f(x)to pm infty$. And just to be really annoying $c_n = f(n+frac 14)=(n+frac 14)(2npi + frac pi 2) =0$. So $c_nto 0$. ... we're.... boned if we try to conclude that direction.
    – fleablood
    Nov 27 '18 at 18:43
















0














First recall the definitions




Definition I: Let $(a_n)$ be a sequence of real or complex numbers. We say that $a_nto infty$ as $nto infty$, if
$$
forall R>0 exists Ninmathbb{N} forall ninmathbb{N}:[(ngeq N)implies (a_n>R) ].
$$



Definition II: Let $f:Dto mathbb{C}$, where $Dsubseteq mathbb{R}$ is not bounded above. We say that $f(x)to infty$ as $xto infty$, if
$$
forall R>0, exists Kinmathbb{R}, forall xin D: [(xgeq K)implies (f(x)>R)].
$$




Question: Let $f:[1,infty)to mathbb{C}$, and let $a_n=f(n)$ for $ngeq 1$. It is clear that if $f(x)toinfty$ as $xtoinfty$, then $a_ntoinfty$ as $ntoinfty$ (i.e. Definition II implies Definition I). Does the converse direction hold in general? If we assume that $a_ntoinfty$ as $ntoinfty$, then there are a lot of "gaps" (i.e. the elements of $[1,infty)setminus mathbb{N}$) that should be checked in Definition II.






real-analysis







share|cite|improve this question






















  • Yes, I think you are right. We can easily imagine a function that goes to hell and back between integers and yet always is "nice" on the integers. For example $f(x) = xcos(2pi*x)$ for $a_n= f(n) = n$ it's clear $a_nto infty$ but between integers will go way up and way down and it's clear definition 2 fails. .. Oooh.... also if $b_n = f(n + frac 12)=-n-frac 12to -infty$. We can't conclude both $f(x)to pm infty$. And just to be really annoying $c_n = f(n+frac 14)=(n+frac 14)(2npi + frac pi 2) =0$. So $c_nto 0$. ... we're.... boned if we try to conclude that direction.
    – fleablood
    Nov 27 '18 at 18:43














0












0








0







First recall the definitions




Definition I: Let $(a_n)$ be a sequence of real or complex numbers. We say that $a_nto infty$ as $nto infty$, if
$$
forall R>0 exists Ninmathbb{N} forall ninmathbb{N}:[(ngeq N)implies (a_n>R) ].
$$



Definition II: Let $f:Dto mathbb{C}$, where $Dsubseteq mathbb{R}$ is not bounded above. We say that $f(x)to infty$ as $xto infty$, if
$$
forall R>0, exists Kinmathbb{R}, forall xin D: [(xgeq K)implies (f(x)>R)].
$$




Question: Let $f:[1,infty)to mathbb{C}$, and let $a_n=f(n)$ for $ngeq 1$. It is clear that if $f(x)toinfty$ as $xtoinfty$, then $a_ntoinfty$ as $ntoinfty$ (i.e. Definition II implies Definition I). Does the converse direction hold in general? If we assume that $a_ntoinfty$ as $ntoinfty$, then there are a lot of "gaps" (i.e. the elements of $[1,infty)setminus mathbb{N}$) that should be checked in Definition II.






real-analysis







share|cite|improve this question













First recall the definitions




Definition I: Let $(a_n)$ be a sequence of real or complex numbers. We say that $a_nto infty$ as $nto infty$, if
$$
forall R>0 exists Ninmathbb{N} forall ninmathbb{N}:[(ngeq N)implies (a_n>R) ].
$$



Definition II: Let $f:Dto mathbb{C}$, where $Dsubseteq mathbb{R}$ is not bounded above. We say that $f(x)to infty$ as $xto infty$, if
$$
forall R>0, exists Kinmathbb{R}, forall xin D: [(xgeq K)implies (f(x)>R)].
$$




Question: Let $f:[1,infty)to mathbb{C}$, and let $a_n=f(n)$ for $ngeq 1$. It is clear that if $f(x)toinfty$ as $xtoinfty$, then $a_ntoinfty$ as $ntoinfty$ (i.e. Definition II implies Definition I). Does the converse direction hold in general? If we assume that $a_ntoinfty$ as $ntoinfty$, then there are a lot of "gaps" (i.e. the elements of $[1,infty)setminus mathbb{N}$) that should be checked in Definition II.






real-analysis




real-analysis






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asked Nov 27 '18 at 17:54









UnknownW

953822




953822












  • Yes, I think you are right. We can easily imagine a function that goes to hell and back between integers and yet always is "nice" on the integers. For example $f(x) = xcos(2pi*x)$ for $a_n= f(n) = n$ it's clear $a_nto infty$ but between integers will go way up and way down and it's clear definition 2 fails. .. Oooh.... also if $b_n = f(n + frac 12)=-n-frac 12to -infty$. We can't conclude both $f(x)to pm infty$. And just to be really annoying $c_n = f(n+frac 14)=(n+frac 14)(2npi + frac pi 2) =0$. So $c_nto 0$. ... we're.... boned if we try to conclude that direction.
    – fleablood
    Nov 27 '18 at 18:43


















  • Yes, I think you are right. We can easily imagine a function that goes to hell and back between integers and yet always is "nice" on the integers. For example $f(x) = xcos(2pi*x)$ for $a_n= f(n) = n$ it's clear $a_nto infty$ but between integers will go way up and way down and it's clear definition 2 fails. .. Oooh.... also if $b_n = f(n + frac 12)=-n-frac 12to -infty$. We can't conclude both $f(x)to pm infty$. And just to be really annoying $c_n = f(n+frac 14)=(n+frac 14)(2npi + frac pi 2) =0$. So $c_nto 0$. ... we're.... boned if we try to conclude that direction.
    – fleablood
    Nov 27 '18 at 18:43
















Yes, I think you are right. We can easily imagine a function that goes to hell and back between integers and yet always is "nice" on the integers. For example $f(x) = xcos(2pi*x)$ for $a_n= f(n) = n$ it's clear $a_nto infty$ but between integers will go way up and way down and it's clear definition 2 fails. .. Oooh.... also if $b_n = f(n + frac 12)=-n-frac 12to -infty$. We can't conclude both $f(x)to pm infty$. And just to be really annoying $c_n = f(n+frac 14)=(n+frac 14)(2npi + frac pi 2) =0$. So $c_nto 0$. ... we're.... boned if we try to conclude that direction.
– fleablood
Nov 27 '18 at 18:43




Yes, I think you are right. We can easily imagine a function that goes to hell and back between integers and yet always is "nice" on the integers. For example $f(x) = xcos(2pi*x)$ for $a_n= f(n) = n$ it's clear $a_nto infty$ but between integers will go way up and way down and it's clear definition 2 fails. .. Oooh.... also if $b_n = f(n + frac 12)=-n-frac 12to -infty$. We can't conclude both $f(x)to pm infty$. And just to be really annoying $c_n = f(n+frac 14)=(n+frac 14)(2npi + frac pi 2) =0$. So $c_nto 0$. ... we're.... boned if we try to conclude that direction.
– fleablood
Nov 27 '18 at 18:43










2 Answers
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You are right that $f(n) to infty$ as $n to infty$ does not imply $f(x) to infty$ as $x to infty$. But I would not say things like "Definition II implies Definition I" because the two definitions intrinsically have nothing to do with each other. You've forced a connection with the definition $a_n = f(n)$, but as you pointed out, the behavior of $f$ at natural numbers $n$ does not govern the behavior of $f$ anywhere else.




Concrete counterexample:



Let $f(x) = begin{cases} x & text{if }x in mathbb{N} \ 0 & text{otherwise}end{cases}$.
If $a_n = f(n)$ then $a_n to infty$ as $n to infty$. But $f(x) notto infty$ as $x to infty$.






share|cite|improve this answer





























    0














    You are correct.



    A good counter example is $f(x) = xcos(2pi x)$ and



    $a_n = f(n); b_n = f(n+frac 12);$ $c_n = f(n + frac 14)$



    $a_n = ncos (2pi n) = n to infty$



    $b_n = (n+frac 12)cos ((2n + 1)pi)=-n -frac 12 to -infty$



    $c_n =(n+frac 1r) cos(2npi + frac pi 2) = (n+frac 14)cdot 0 = 0to 0$.



    Clearly we can't conclude anything about the limits of $f(x)$ by the limits of $a_n, b_n$ and $c_n$.



    (Have you had the theorem that if a sequence converges then all subsequence converge to the same limit? If so this should be clear. If not this should give you a really good idea we the theorem holds.)



    And if we actually try to find the limit if $f(x)$.



    For Any $R >0$ and all $K in Bbb R$ we can find $x > max (|K|+1, R); x in mathbb N$ so $f(x+frac 12) = -x-frac 12 < -|K|-1 < K$ so $lim f(x) ne infty$.



    And $f(x) = x > |K|+1 > K$ so $lim f(x) ne -infty$



    and for $0< epsilon < 1$ $|f(x)-K| = ||K|+1 - K| ge 1 > epsilon$ so $lim f(x) ne K$.



    So $f(x)$ does not converge to any value nor does it converge to infinity or negative infinity.



    SO... Yes, you are correct.






    share|cite|improve this answer























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      2 Answers
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      2 Answers
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      1














      You are right that $f(n) to infty$ as $n to infty$ does not imply $f(x) to infty$ as $x to infty$. But I would not say things like "Definition II implies Definition I" because the two definitions intrinsically have nothing to do with each other. You've forced a connection with the definition $a_n = f(n)$, but as you pointed out, the behavior of $f$ at natural numbers $n$ does not govern the behavior of $f$ anywhere else.




      Concrete counterexample:



      Let $f(x) = begin{cases} x & text{if }x in mathbb{N} \ 0 & text{otherwise}end{cases}$.
      If $a_n = f(n)$ then $a_n to infty$ as $n to infty$. But $f(x) notto infty$ as $x to infty$.






      share|cite|improve this answer


























        1














        You are right that $f(n) to infty$ as $n to infty$ does not imply $f(x) to infty$ as $x to infty$. But I would not say things like "Definition II implies Definition I" because the two definitions intrinsically have nothing to do with each other. You've forced a connection with the definition $a_n = f(n)$, but as you pointed out, the behavior of $f$ at natural numbers $n$ does not govern the behavior of $f$ anywhere else.




        Concrete counterexample:



        Let $f(x) = begin{cases} x & text{if }x in mathbb{N} \ 0 & text{otherwise}end{cases}$.
        If $a_n = f(n)$ then $a_n to infty$ as $n to infty$. But $f(x) notto infty$ as $x to infty$.






        share|cite|improve this answer
























          1












          1








          1






          You are right that $f(n) to infty$ as $n to infty$ does not imply $f(x) to infty$ as $x to infty$. But I would not say things like "Definition II implies Definition I" because the two definitions intrinsically have nothing to do with each other. You've forced a connection with the definition $a_n = f(n)$, but as you pointed out, the behavior of $f$ at natural numbers $n$ does not govern the behavior of $f$ anywhere else.




          Concrete counterexample:



          Let $f(x) = begin{cases} x & text{if }x in mathbb{N} \ 0 & text{otherwise}end{cases}$.
          If $a_n = f(n)$ then $a_n to infty$ as $n to infty$. But $f(x) notto infty$ as $x to infty$.






          share|cite|improve this answer












          You are right that $f(n) to infty$ as $n to infty$ does not imply $f(x) to infty$ as $x to infty$. But I would not say things like "Definition II implies Definition I" because the two definitions intrinsically have nothing to do with each other. You've forced a connection with the definition $a_n = f(n)$, but as you pointed out, the behavior of $f$ at natural numbers $n$ does not govern the behavior of $f$ anywhere else.




          Concrete counterexample:



          Let $f(x) = begin{cases} x & text{if }x in mathbb{N} \ 0 & text{otherwise}end{cases}$.
          If $a_n = f(n)$ then $a_n to infty$ as $n to infty$. But $f(x) notto infty$ as $x to infty$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 27 '18 at 18:02









          angryavian

          39k23180




          39k23180























              0














              You are correct.



              A good counter example is $f(x) = xcos(2pi x)$ and



              $a_n = f(n); b_n = f(n+frac 12);$ $c_n = f(n + frac 14)$



              $a_n = ncos (2pi n) = n to infty$



              $b_n = (n+frac 12)cos ((2n + 1)pi)=-n -frac 12 to -infty$



              $c_n =(n+frac 1r) cos(2npi + frac pi 2) = (n+frac 14)cdot 0 = 0to 0$.



              Clearly we can't conclude anything about the limits of $f(x)$ by the limits of $a_n, b_n$ and $c_n$.



              (Have you had the theorem that if a sequence converges then all subsequence converge to the same limit? If so this should be clear. If not this should give you a really good idea we the theorem holds.)



              And if we actually try to find the limit if $f(x)$.



              For Any $R >0$ and all $K in Bbb R$ we can find $x > max (|K|+1, R); x in mathbb N$ so $f(x+frac 12) = -x-frac 12 < -|K|-1 < K$ so $lim f(x) ne infty$.



              And $f(x) = x > |K|+1 > K$ so $lim f(x) ne -infty$



              and for $0< epsilon < 1$ $|f(x)-K| = ||K|+1 - K| ge 1 > epsilon$ so $lim f(x) ne K$.



              So $f(x)$ does not converge to any value nor does it converge to infinity or negative infinity.



              SO... Yes, you are correct.






              share|cite|improve this answer




























                0














                You are correct.



                A good counter example is $f(x) = xcos(2pi x)$ and



                $a_n = f(n); b_n = f(n+frac 12);$ $c_n = f(n + frac 14)$



                $a_n = ncos (2pi n) = n to infty$



                $b_n = (n+frac 12)cos ((2n + 1)pi)=-n -frac 12 to -infty$



                $c_n =(n+frac 1r) cos(2npi + frac pi 2) = (n+frac 14)cdot 0 = 0to 0$.



                Clearly we can't conclude anything about the limits of $f(x)$ by the limits of $a_n, b_n$ and $c_n$.



                (Have you had the theorem that if a sequence converges then all subsequence converge to the same limit? If so this should be clear. If not this should give you a really good idea we the theorem holds.)



                And if we actually try to find the limit if $f(x)$.



                For Any $R >0$ and all $K in Bbb R$ we can find $x > max (|K|+1, R); x in mathbb N$ so $f(x+frac 12) = -x-frac 12 < -|K|-1 < K$ so $lim f(x) ne infty$.



                And $f(x) = x > |K|+1 > K$ so $lim f(x) ne -infty$



                and for $0< epsilon < 1$ $|f(x)-K| = ||K|+1 - K| ge 1 > epsilon$ so $lim f(x) ne K$.



                So $f(x)$ does not converge to any value nor does it converge to infinity or negative infinity.



                SO... Yes, you are correct.






                share|cite|improve this answer


























                  0












                  0








                  0






                  You are correct.



                  A good counter example is $f(x) = xcos(2pi x)$ and



                  $a_n = f(n); b_n = f(n+frac 12);$ $c_n = f(n + frac 14)$



                  $a_n = ncos (2pi n) = n to infty$



                  $b_n = (n+frac 12)cos ((2n + 1)pi)=-n -frac 12 to -infty$



                  $c_n =(n+frac 1r) cos(2npi + frac pi 2) = (n+frac 14)cdot 0 = 0to 0$.



                  Clearly we can't conclude anything about the limits of $f(x)$ by the limits of $a_n, b_n$ and $c_n$.



                  (Have you had the theorem that if a sequence converges then all subsequence converge to the same limit? If so this should be clear. If not this should give you a really good idea we the theorem holds.)



                  And if we actually try to find the limit if $f(x)$.



                  For Any $R >0$ and all $K in Bbb R$ we can find $x > max (|K|+1, R); x in mathbb N$ so $f(x+frac 12) = -x-frac 12 < -|K|-1 < K$ so $lim f(x) ne infty$.



                  And $f(x) = x > |K|+1 > K$ so $lim f(x) ne -infty$



                  and for $0< epsilon < 1$ $|f(x)-K| = ||K|+1 - K| ge 1 > epsilon$ so $lim f(x) ne K$.



                  So $f(x)$ does not converge to any value nor does it converge to infinity or negative infinity.



                  SO... Yes, you are correct.






                  share|cite|improve this answer














                  You are correct.



                  A good counter example is $f(x) = xcos(2pi x)$ and



                  $a_n = f(n); b_n = f(n+frac 12);$ $c_n = f(n + frac 14)$



                  $a_n = ncos (2pi n) = n to infty$



                  $b_n = (n+frac 12)cos ((2n + 1)pi)=-n -frac 12 to -infty$



                  $c_n =(n+frac 1r) cos(2npi + frac pi 2) = (n+frac 14)cdot 0 = 0to 0$.



                  Clearly we can't conclude anything about the limits of $f(x)$ by the limits of $a_n, b_n$ and $c_n$.



                  (Have you had the theorem that if a sequence converges then all subsequence converge to the same limit? If so this should be clear. If not this should give you a really good idea we the theorem holds.)



                  And if we actually try to find the limit if $f(x)$.



                  For Any $R >0$ and all $K in Bbb R$ we can find $x > max (|K|+1, R); x in mathbb N$ so $f(x+frac 12) = -x-frac 12 < -|K|-1 < K$ so $lim f(x) ne infty$.



                  And $f(x) = x > |K|+1 > K$ so $lim f(x) ne -infty$



                  and for $0< epsilon < 1$ $|f(x)-K| = ||K|+1 - K| ge 1 > epsilon$ so $lim f(x) ne K$.



                  So $f(x)$ does not converge to any value nor does it converge to infinity or negative infinity.



                  SO... Yes, you are correct.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 27 '18 at 19:05

























                  answered Nov 27 '18 at 18:51









                  fleablood

                  68.2k22684




                  68.2k22684






























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