Prove or disprove that $lim_{nto infty} frac{x_{n+1}-l}{x_n-l}=lim_{ntoinfty}frac{x_{n+1}}{x_n}$, $l=lim_{nto...












2












$begingroup$



Prove or disprove that $$lim_{nto infty} left(frac{x_{n+1}-l}{x_n-l}right)=lim_{ntoinfty}left(frac{x_{n+1}}{x_n}right)$$ where $l=lim_{nto infty} x_n$




I think that the above result is true,but I am not really sure how to prove it.If anyone has a counterexample I am looking forward to it.

EDIT1 : $x_n$ is any real sequence which is not constant.

EDIT2: What if we add the additional constraint that $l in mathbb{R}$?






real-analysis sequences-and-series







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  • $begingroup$
    math.stackexchange.com/questions/3047760/… I reasked the question with some changes
    $endgroup$
    – Love Invariants
    Dec 20 '18 at 17:10


















2












$begingroup$



Prove or disprove that $$lim_{nto infty} left(frac{x_{n+1}-l}{x_n-l}right)=lim_{ntoinfty}left(frac{x_{n+1}}{x_n}right)$$ where $l=lim_{nto infty} x_n$




I think that the above result is true,but I am not really sure how to prove it.If anyone has a counterexample I am looking forward to it.

EDIT1 : $x_n$ is any real sequence which is not constant.

EDIT2: What if we add the additional constraint that $l in mathbb{R}$?






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    math.stackexchange.com/questions/3047760/… I reasked the question with some changes
    $endgroup$
    – Love Invariants
    Dec 20 '18 at 17:10
















2












2








2





$begingroup$



Prove or disprove that $$lim_{nto infty} left(frac{x_{n+1}-l}{x_n-l}right)=lim_{ntoinfty}left(frac{x_{n+1}}{x_n}right)$$ where $l=lim_{nto infty} x_n$




I think that the above result is true,but I am not really sure how to prove it.If anyone has a counterexample I am looking forward to it.

EDIT1 : $x_n$ is any real sequence which is not constant.

EDIT2: What if we add the additional constraint that $l in mathbb{R}$?






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$





Prove or disprove that $$lim_{nto infty} left(frac{x_{n+1}-l}{x_n-l}right)=lim_{ntoinfty}left(frac{x_{n+1}}{x_n}right)$$ where $l=lim_{nto infty} x_n$




I think that the above result is true,but I am not really sure how to prove it.If anyone has a counterexample I am looking forward to it.

EDIT1 : $x_n$ is any real sequence which is not constant.

EDIT2: What if we add the additional constraint that $l in mathbb{R}$?






real-analysis sequences-and-series




real-analysis sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 16:03







JustAnAmateur

















asked Dec 20 '18 at 15:47









JustAnAmateurJustAnAmateur

1096




1096












  • $begingroup$
    math.stackexchange.com/questions/3047760/… I reasked the question with some changes
    $endgroup$
    – Love Invariants
    Dec 20 '18 at 17:10




















  • $begingroup$
    math.stackexchange.com/questions/3047760/… I reasked the question with some changes
    $endgroup$
    – Love Invariants
    Dec 20 '18 at 17:10


















$begingroup$
math.stackexchange.com/questions/3047760/… I reasked the question with some changes
$endgroup$
– Love Invariants
Dec 20 '18 at 17:10






$begingroup$
math.stackexchange.com/questions/3047760/… I reasked the question with some changes
$endgroup$
– Love Invariants
Dec 20 '18 at 17:10












2 Answers
2






active

oldest

votes


















4












$begingroup$

Let $x_n=frac{2^{n+1}-1}{2^n}$, for which $l=2$ and $$limfrac{x_{n+1}}{x_n}=1$$ and $$limfrac{x_{n+1}-2}{x_n-2}=frac12.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Question means, whether for any real value of $l$ there exist a sequence $x_n$ which satisfies the given condition.
    $endgroup$
    – Love Invariants
    Dec 20 '18 at 16:09












  • $begingroup$
    @LoveInvariants The OP also seems to be happy with a counter-example :-)
    $endgroup$
    – Math-fun
    Dec 20 '18 at 16:12










  • $begingroup$
    Actually a counter-example is what I want,thank you,excellent job.
    $endgroup$
    – JustAnAmateur
    Dec 20 '18 at 16:13






  • 1




    $begingroup$
    The counter-example here is perfectly fine: either it works for all $ell$ or it doesn't work for all $ell$ (and that's the question). Anyway one can easily just change $2to frac{1}{ell}$ to get a counter-example that works for all $0 < ell < 1$.
    $endgroup$
    – Winther
    Dec 20 '18 at 16:16






  • 1




    $begingroup$
    ...and just for completeness $x_n = ellfrac{2^{2^n}-1}{2^{2^n}}$ works for all $ellnot= 0$. As $frac{x_{n+1}}{x_n} = 1 + 2^{-2^n}$ and $frac{x_{n+1}-ell}{x_n - ell} = 2^{-2^n}$.
    $endgroup$
    – Winther
    Dec 20 '18 at 16:36



















1












$begingroup$

Take $x_n=n$

Then LHS=$1over0$ and RHS=$1+{1overinfty}=1$

Both aren't equal.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Note that the question implicitly assumes that $ell = lim_{ntoinfty} x_n$ exists
    $endgroup$
    – Winther
    Dec 20 '18 at 16:01










  • $begingroup$
    Do you mean $l$ is a constant? It is nowhere written though about that.
    $endgroup$
    – Love Invariants
    Dec 20 '18 at 16:05











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2 Answers
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2 Answers
2






active

oldest

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active

oldest

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active

oldest

votes









4












$begingroup$

Let $x_n=frac{2^{n+1}-1}{2^n}$, for which $l=2$ and $$limfrac{x_{n+1}}{x_n}=1$$ and $$limfrac{x_{n+1}-2}{x_n-2}=frac12.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Question means, whether for any real value of $l$ there exist a sequence $x_n$ which satisfies the given condition.
    $endgroup$
    – Love Invariants
    Dec 20 '18 at 16:09












  • $begingroup$
    @LoveInvariants The OP also seems to be happy with a counter-example :-)
    $endgroup$
    – Math-fun
    Dec 20 '18 at 16:12










  • $begingroup$
    Actually a counter-example is what I want,thank you,excellent job.
    $endgroup$
    – JustAnAmateur
    Dec 20 '18 at 16:13






  • 1




    $begingroup$
    The counter-example here is perfectly fine: either it works for all $ell$ or it doesn't work for all $ell$ (and that's the question). Anyway one can easily just change $2to frac{1}{ell}$ to get a counter-example that works for all $0 < ell < 1$.
    $endgroup$
    – Winther
    Dec 20 '18 at 16:16






  • 1




    $begingroup$
    ...and just for completeness $x_n = ellfrac{2^{2^n}-1}{2^{2^n}}$ works for all $ellnot= 0$. As $frac{x_{n+1}}{x_n} = 1 + 2^{-2^n}$ and $frac{x_{n+1}-ell}{x_n - ell} = 2^{-2^n}$.
    $endgroup$
    – Winther
    Dec 20 '18 at 16:36
















4












$begingroup$

Let $x_n=frac{2^{n+1}-1}{2^n}$, for which $l=2$ and $$limfrac{x_{n+1}}{x_n}=1$$ and $$limfrac{x_{n+1}-2}{x_n-2}=frac12.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Question means, whether for any real value of $l$ there exist a sequence $x_n$ which satisfies the given condition.
    $endgroup$
    – Love Invariants
    Dec 20 '18 at 16:09












  • $begingroup$
    @LoveInvariants The OP also seems to be happy with a counter-example :-)
    $endgroup$
    – Math-fun
    Dec 20 '18 at 16:12










  • $begingroup$
    Actually a counter-example is what I want,thank you,excellent job.
    $endgroup$
    – JustAnAmateur
    Dec 20 '18 at 16:13






  • 1




    $begingroup$
    The counter-example here is perfectly fine: either it works for all $ell$ or it doesn't work for all $ell$ (and that's the question). Anyway one can easily just change $2to frac{1}{ell}$ to get a counter-example that works for all $0 < ell < 1$.
    $endgroup$
    – Winther
    Dec 20 '18 at 16:16






  • 1




    $begingroup$
    ...and just for completeness $x_n = ellfrac{2^{2^n}-1}{2^{2^n}}$ works for all $ellnot= 0$. As $frac{x_{n+1}}{x_n} = 1 + 2^{-2^n}$ and $frac{x_{n+1}-ell}{x_n - ell} = 2^{-2^n}$.
    $endgroup$
    – Winther
    Dec 20 '18 at 16:36














4












4








4





$begingroup$

Let $x_n=frac{2^{n+1}-1}{2^n}$, for which $l=2$ and $$limfrac{x_{n+1}}{x_n}=1$$ and $$limfrac{x_{n+1}-2}{x_n-2}=frac12.$$






share|cite|improve this answer









$endgroup$



Let $x_n=frac{2^{n+1}-1}{2^n}$, for which $l=2$ and $$limfrac{x_{n+1}}{x_n}=1$$ and $$limfrac{x_{n+1}-2}{x_n-2}=frac12.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 20 '18 at 16:07









Math-funMath-fun

7,1531527




7,1531527












  • $begingroup$
    Question means, whether for any real value of $l$ there exist a sequence $x_n$ which satisfies the given condition.
    $endgroup$
    – Love Invariants
    Dec 20 '18 at 16:09












  • $begingroup$
    @LoveInvariants The OP also seems to be happy with a counter-example :-)
    $endgroup$
    – Math-fun
    Dec 20 '18 at 16:12










  • $begingroup$
    Actually a counter-example is what I want,thank you,excellent job.
    $endgroup$
    – JustAnAmateur
    Dec 20 '18 at 16:13






  • 1




    $begingroup$
    The counter-example here is perfectly fine: either it works for all $ell$ or it doesn't work for all $ell$ (and that's the question). Anyway one can easily just change $2to frac{1}{ell}$ to get a counter-example that works for all $0 < ell < 1$.
    $endgroup$
    – Winther
    Dec 20 '18 at 16:16






  • 1




    $begingroup$
    ...and just for completeness $x_n = ellfrac{2^{2^n}-1}{2^{2^n}}$ works for all $ellnot= 0$. As $frac{x_{n+1}}{x_n} = 1 + 2^{-2^n}$ and $frac{x_{n+1}-ell}{x_n - ell} = 2^{-2^n}$.
    $endgroup$
    – Winther
    Dec 20 '18 at 16:36


















  • $begingroup$
    Question means, whether for any real value of $l$ there exist a sequence $x_n$ which satisfies the given condition.
    $endgroup$
    – Love Invariants
    Dec 20 '18 at 16:09












  • $begingroup$
    @LoveInvariants The OP also seems to be happy with a counter-example :-)
    $endgroup$
    – Math-fun
    Dec 20 '18 at 16:12










  • $begingroup$
    Actually a counter-example is what I want,thank you,excellent job.
    $endgroup$
    – JustAnAmateur
    Dec 20 '18 at 16:13






  • 1




    $begingroup$
    The counter-example here is perfectly fine: either it works for all $ell$ or it doesn't work for all $ell$ (and that's the question). Anyway one can easily just change $2to frac{1}{ell}$ to get a counter-example that works for all $0 < ell < 1$.
    $endgroup$
    – Winther
    Dec 20 '18 at 16:16






  • 1




    $begingroup$
    ...and just for completeness $x_n = ellfrac{2^{2^n}-1}{2^{2^n}}$ works for all $ellnot= 0$. As $frac{x_{n+1}}{x_n} = 1 + 2^{-2^n}$ and $frac{x_{n+1}-ell}{x_n - ell} = 2^{-2^n}$.
    $endgroup$
    – Winther
    Dec 20 '18 at 16:36
















$begingroup$
Question means, whether for any real value of $l$ there exist a sequence $x_n$ which satisfies the given condition.
$endgroup$
– Love Invariants
Dec 20 '18 at 16:09






$begingroup$
Question means, whether for any real value of $l$ there exist a sequence $x_n$ which satisfies the given condition.
$endgroup$
– Love Invariants
Dec 20 '18 at 16:09














$begingroup$
@LoveInvariants The OP also seems to be happy with a counter-example :-)
$endgroup$
– Math-fun
Dec 20 '18 at 16:12




$begingroup$
@LoveInvariants The OP also seems to be happy with a counter-example :-)
$endgroup$
– Math-fun
Dec 20 '18 at 16:12












$begingroup$
Actually a counter-example is what I want,thank you,excellent job.
$endgroup$
– JustAnAmateur
Dec 20 '18 at 16:13




$begingroup$
Actually a counter-example is what I want,thank you,excellent job.
$endgroup$
– JustAnAmateur
Dec 20 '18 at 16:13




1




1




$begingroup$
The counter-example here is perfectly fine: either it works for all $ell$ or it doesn't work for all $ell$ (and that's the question). Anyway one can easily just change $2to frac{1}{ell}$ to get a counter-example that works for all $0 < ell < 1$.
$endgroup$
– Winther
Dec 20 '18 at 16:16




$begingroup$
The counter-example here is perfectly fine: either it works for all $ell$ or it doesn't work for all $ell$ (and that's the question). Anyway one can easily just change $2to frac{1}{ell}$ to get a counter-example that works for all $0 < ell < 1$.
$endgroup$
– Winther
Dec 20 '18 at 16:16




1




1




$begingroup$
...and just for completeness $x_n = ellfrac{2^{2^n}-1}{2^{2^n}}$ works for all $ellnot= 0$. As $frac{x_{n+1}}{x_n} = 1 + 2^{-2^n}$ and $frac{x_{n+1}-ell}{x_n - ell} = 2^{-2^n}$.
$endgroup$
– Winther
Dec 20 '18 at 16:36




$begingroup$
...and just for completeness $x_n = ellfrac{2^{2^n}-1}{2^{2^n}}$ works for all $ellnot= 0$. As $frac{x_{n+1}}{x_n} = 1 + 2^{-2^n}$ and $frac{x_{n+1}-ell}{x_n - ell} = 2^{-2^n}$.
$endgroup$
– Winther
Dec 20 '18 at 16:36











1












$begingroup$

Take $x_n=n$

Then LHS=$1over0$ and RHS=$1+{1overinfty}=1$

Both aren't equal.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Note that the question implicitly assumes that $ell = lim_{ntoinfty} x_n$ exists
    $endgroup$
    – Winther
    Dec 20 '18 at 16:01










  • $begingroup$
    Do you mean $l$ is a constant? It is nowhere written though about that.
    $endgroup$
    – Love Invariants
    Dec 20 '18 at 16:05
















1












$begingroup$

Take $x_n=n$

Then LHS=$1over0$ and RHS=$1+{1overinfty}=1$

Both aren't equal.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Note that the question implicitly assumes that $ell = lim_{ntoinfty} x_n$ exists
    $endgroup$
    – Winther
    Dec 20 '18 at 16:01










  • $begingroup$
    Do you mean $l$ is a constant? It is nowhere written though about that.
    $endgroup$
    – Love Invariants
    Dec 20 '18 at 16:05














1












1








1





$begingroup$

Take $x_n=n$

Then LHS=$1over0$ and RHS=$1+{1overinfty}=1$

Both aren't equal.






share|cite|improve this answer









$endgroup$



Take $x_n=n$

Then LHS=$1over0$ and RHS=$1+{1overinfty}=1$

Both aren't equal.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 20 '18 at 16:00









Love InvariantsLove Invariants

86115




86115








  • 1




    $begingroup$
    Note that the question implicitly assumes that $ell = lim_{ntoinfty} x_n$ exists
    $endgroup$
    – Winther
    Dec 20 '18 at 16:01










  • $begingroup$
    Do you mean $l$ is a constant? It is nowhere written though about that.
    $endgroup$
    – Love Invariants
    Dec 20 '18 at 16:05














  • 1




    $begingroup$
    Note that the question implicitly assumes that $ell = lim_{ntoinfty} x_n$ exists
    $endgroup$
    – Winther
    Dec 20 '18 at 16:01










  • $begingroup$
    Do you mean $l$ is a constant? It is nowhere written though about that.
    $endgroup$
    – Love Invariants
    Dec 20 '18 at 16:05








1




1




$begingroup$
Note that the question implicitly assumes that $ell = lim_{ntoinfty} x_n$ exists
$endgroup$
– Winther
Dec 20 '18 at 16:01




$begingroup$
Note that the question implicitly assumes that $ell = lim_{ntoinfty} x_n$ exists
$endgroup$
– Winther
Dec 20 '18 at 16:01












$begingroup$
Do you mean $l$ is a constant? It is nowhere written though about that.
$endgroup$
– Love Invariants
Dec 20 '18 at 16:05




$begingroup$
Do you mean $l$ is a constant? It is nowhere written though about that.
$endgroup$
– Love Invariants
Dec 20 '18 at 16:05


















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