Mixing
Does this limit exist? $lim_{ntoinfty}frac{φ(f(n))}{ψ(g(n))}$
$begingroup$
EDİTED:
Is this statement true?
Suppose $left{f,g,varphi, psi right}:mathbb{R^{+}} rightarrow mathbb{R^{+}}$ and for $ntoinfty$
$left{f(n),g(n),varphi(n), psi(n) right}rightarrow infty$
If $lim_{ntoinfty}frac{φ(n)}{ψ(n)}$ doesn't exist, then $lim_{ntoinfty}frac{φ(f(n))}{ψ(g(n))}$ doesn't exist too.
I want to know that, if the assumption is wrong, under what conditions is true?
calculus limits examples-counterexamples
edited Jan 19 at 11:42
Maths Barry
438
asked Jan 19 at 9:00
ElementaryElementary
361111
$endgroup$
|
show 1 more comment
$begingroup$
EDİTED:
Is this statement true?
Suppose $left{f,g,varphi, psi right}:mathbb{R^{+}} rightarrow mathbb{R^{+}}$ and for $ntoinfty$
$left{f(n),g(n),varphi(n), psi(n) right}rightarrow infty$
If $lim_{ntoinfty}frac{φ(n)}{ψ(n)}$ doesn't exist, then $lim_{ntoinfty}frac{φ(f(n))}{ψ(g(n))}$ doesn't exist too.
I want to know that, if the assumption is wrong, under what conditions is true?
calculus limits examples-counterexamples
edited Jan 19 at 11:42
Maths Barry
438
asked Jan 19 at 9:00
ElementaryElementary
361111
$endgroup$
7
$begingroup$
That depends heavily on $f$ and $g$ ...
$endgroup$
– Martin R
Jan 19 at 9:01
$begingroup$
As an example: For constant $f=c_1$ and $g=c_2$ the limit will always exist.
$endgroup$
– maxmilgram
Jan 19 at 10:01
$begingroup$
I suppose that the limit is taking along the natural numbers, is that true?
$endgroup$
– Shashi
Jan 19 at 11:04
$begingroup$
The second statement looks like a statement about subsequences. I hope you know that a non convergent sequence might have a convergent subsequence... So from this you can already see that the statement can not be true.
$endgroup$
– Shashi
Jan 19 at 11:08
$begingroup$
@Shashi yes for positive integers., No, I did not know. The last thing you said, I learned now.
$endgroup$
– Elementary
Jan 19 at 11:18
|
show 1 more comment
$begingroup$
EDİTED:
Is this statement true?
Suppose $left{f,g,varphi, psi right}:mathbb{R^{+}} rightarrow mathbb{R^{+}}$ and for $ntoinfty$
$left{f(n),g(n),varphi(n), psi(n) right}rightarrow infty$
If $lim_{ntoinfty}frac{φ(n)}{ψ(n)}$ doesn't exist, then $lim_{ntoinfty}frac{φ(f(n))}{ψ(g(n))}$ doesn't exist too.
I want to know that, if the assumption is wrong, under what conditions is true?
calculus limits examples-counterexamples
edited Jan 19 at 11:42
Maths Barry
438
asked Jan 19 at 9:00
ElementaryElementary
361111
$endgroup$
EDİTED:
Is this statement true?
Suppose $left{f,g,varphi, psi right}:mathbb{R^{+}} rightarrow mathbb{R^{+}}$ and for $ntoinfty$
$left{f(n),g(n),varphi(n), psi(n) right}rightarrow infty$
If $lim_{ntoinfty}frac{φ(n)}{ψ(n)}$ doesn't exist, then $lim_{ntoinfty}frac{φ(f(n))}{ψ(g(n))}$ doesn't exist too.
I want to know that, if the assumption is wrong, under what conditions is true?
calculus limits examples-counterexamples
calculus limits examples-counterexamples
edited Jan 19 at 11:42
Maths Barry
438
asked Jan 19 at 9:00
ElementaryElementary
361111
edited Jan 19 at 11:42
Maths Barry
438
asked Jan 19 at 9:00
ElementaryElementary
361111
edited Jan 19 at 11:42
Maths Barry
438
edited Jan 19 at 11:42
Maths Barry
438
edited Jan 19 at 11:42
Maths Barry
438
438
asked Jan 19 at 9:00
ElementaryElementary
361111
asked Jan 19 at 9:00
ElementaryElementary
361111
asked Jan 19 at 9:00
ElementaryElementary
361111
361111
7
$begingroup$
That depends heavily on $f$ and $g$ ...
$endgroup$
– Martin R
Jan 19 at 9:01
$begingroup$
As an example: For constant $f=c_1$ and $g=c_2$ the limit will always exist.
$endgroup$
– maxmilgram
Jan 19 at 10:01
$begingroup$
I suppose that the limit is taking along the natural numbers, is that true?
$endgroup$
– Shashi
Jan 19 at 11:04
$begingroup$
The second statement looks like a statement about subsequences. I hope you know that a non convergent sequence might have a convergent subsequence... So from this you can already see that the statement can not be true.
$endgroup$
– Shashi
Jan 19 at 11:08
$begingroup$
@Shashi yes for positive integers., No, I did not know. The last thing you said, I learned now.
$endgroup$
– Elementary
Jan 19 at 11:18
|
show 1 more comment
7
$begingroup$
That depends heavily on $f$ and $g$ ...
$endgroup$
– Martin R
Jan 19 at 9:01
$begingroup$
As an example: For constant $f=c_1$ and $g=c_2$ the limit will always exist.
$endgroup$
– maxmilgram
Jan 19 at 10:01
$begingroup$
I suppose that the limit is taking along the natural numbers, is that true?
$endgroup$
– Shashi
Jan 19 at 11:04
$begingroup$
The second statement looks like a statement about subsequences. I hope you know that a non convergent sequence might have a convergent subsequence... So from this you can already see that the statement can not be true.
$endgroup$
– Shashi
Jan 19 at 11:08
$begingroup$
@Shashi yes for positive integers., No, I did not know. The last thing you said, I learned now.
$endgroup$
– Elementary
Jan 19 at 11:18
7
7
$begingroup$
That depends heavily on $f$ and $g$ ...
$endgroup$
– Martin R
Jan 19 at 9:01
$begingroup$
That depends heavily on $f$ and $g$ ...
$endgroup$
– Martin R
Jan 19 at 9:01
$begingroup$
As an example: For constant $f=c_1$ and $g=c_2$ the limit will always exist.
$endgroup$
– maxmilgram
Jan 19 at 10:01
$begingroup$
As an example: For constant $f=c_1$ and $g=c_2$ the limit will always exist.
$endgroup$
– maxmilgram
Jan 19 at 10:01
$begingroup$
I suppose that the limit is taking along the natural numbers, is that true?
$endgroup$
– Shashi
Jan 19 at 11:04
$begingroup$
I suppose that the limit is taking along the natural numbers, is that true?
$endgroup$
– Shashi
Jan 19 at 11:04
$begingroup$
The second statement looks like a statement about subsequences. I hope you know that a non convergent sequence might have a convergent subsequence... So from this you can already see that the statement can not be true.
$endgroup$
– Shashi
Jan 19 at 11:08
$begingroup$
The second statement looks like a statement about subsequences. I hope you know that a non convergent sequence might have a convergent subsequence... So from this you can already see that the statement can not be true.
$endgroup$
– Shashi
Jan 19 at 11:08
$begingroup$
@Shashi yes for positive integers., No, I did not know. The last thing you said, I learned now.
$endgroup$
– Elementary
Jan 19 at 11:18
$begingroup$
@Shashi yes for positive integers., No, I did not know. The last thing you said, I learned now.
$endgroup$
– Elementary
Jan 19 at 11:18
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
The statement depends on the choice of $f$ and $g$ as mentioned in the comments. So as it stands now, the claim is not true.
Just consider sequences that satisfy
$$frac{phi(n) } {psi(n) } = 2+ (-1)^n$$
The limit of that sequence is clearly non-existent. How can we get that? Well just set $psi(n) =n$ and
$$phi(n) =psi(n) (2+ (-1)^n) $$
Now set $f(n) =g(n) =2n$ and see that they all satisfy the conditions mentioned by you. Let's see
$$lim_{ntoinfty} frac{phi(f(n) ) } {psi(g(n) ) }=lim_{ntoinfty} 2+1=3$$
That proves your claim wrong.
Honestly, I can't think of assumptions that make your statement true.
answered Jan 19 at 11:27
ShashiShashi
7,2551628
$endgroup$
$begingroup$
Thank you very much
$endgroup$
– Elementary
Jan 19 at 11:51
add a comment |
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$begingroup$
The statement depends on the choice of $f$ and $g$ as mentioned in the comments. So as it stands now, the claim is not true.
Just consider sequences that satisfy
$$frac{phi(n) } {psi(n) } = 2+ (-1)^n$$
The limit of that sequence is clearly non-existent. How can we get that? Well just set $psi(n) =n$ and
$$phi(n) =psi(n) (2+ (-1)^n) $$
Now set $f(n) =g(n) =2n$ and see that they all satisfy the conditions mentioned by you. Let's see
$$lim_{ntoinfty} frac{phi(f(n) ) } {psi(g(n) ) }=lim_{ntoinfty} 2+1=3$$
That proves your claim wrong.
Honestly, I can't think of assumptions that make your statement true.
answered Jan 19 at 11:27
ShashiShashi
7,2551628
$endgroup$
$begingroup$
Thank you very much
$endgroup$
– Elementary
Jan 19 at 11:51
add a comment |
$begingroup$
The statement depends on the choice of $f$ and $g$ as mentioned in the comments. So as it stands now, the claim is not true.
Just consider sequences that satisfy
$$frac{phi(n) } {psi(n) } = 2+ (-1)^n$$
The limit of that sequence is clearly non-existent. How can we get that? Well just set $psi(n) =n$ and
$$phi(n) =psi(n) (2+ (-1)^n) $$
Now set $f(n) =g(n) =2n$ and see that they all satisfy the conditions mentioned by you. Let's see
$$lim_{ntoinfty} frac{phi(f(n) ) } {psi(g(n) ) }=lim_{ntoinfty} 2+1=3$$
That proves your claim wrong.
Honestly, I can't think of assumptions that make your statement true.
answered Jan 19 at 11:27
ShashiShashi
7,2551628
$endgroup$
$begingroup$
Thank you very much
$endgroup$
– Elementary
Jan 19 at 11:51
add a comment |
$begingroup$
The statement depends on the choice of $f$ and $g$ as mentioned in the comments. So as it stands now, the claim is not true.
Just consider sequences that satisfy
$$frac{phi(n) } {psi(n) } = 2+ (-1)^n$$
The limit of that sequence is clearly non-existent. How can we get that? Well just set $psi(n) =n$ and
$$phi(n) =psi(n) (2+ (-1)^n) $$
Now set $f(n) =g(n) =2n$ and see that they all satisfy the conditions mentioned by you. Let's see
$$lim_{ntoinfty} frac{phi(f(n) ) } {psi(g(n) ) }=lim_{ntoinfty} 2+1=3$$
That proves your claim wrong.
Honestly, I can't think of assumptions that make your statement true.
answered Jan 19 at 11:27
ShashiShashi
7,2551628
$endgroup$
The statement depends on the choice of $f$ and $g$ as mentioned in the comments. So as it stands now, the claim is not true.
Just consider sequences that satisfy
$$frac{phi(n) } {psi(n) } = 2+ (-1)^n$$
The limit of that sequence is clearly non-existent. How can we get that? Well just set $psi(n) =n$ and
$$phi(n) =psi(n) (2+ (-1)^n) $$
Now set $f(n) =g(n) =2n$ and see that they all satisfy the conditions mentioned by you. Let's see
$$lim_{ntoinfty} frac{phi(f(n) ) } {psi(g(n) ) }=lim_{ntoinfty} 2+1=3$$
That proves your claim wrong.
Honestly, I can't think of assumptions that make your statement true.
answered Jan 19 at 11:27
ShashiShashi
7,2551628
edited Jan 19 at 11:37
answered Jan 19 at 11:27
ShashiShashi
7,2551628
answered Jan 19 at 11:27
ShashiShashi
7,2551628
answered Jan 19 at 11:27
ShashiShashi
7,2551628
7,2551628
$begingroup$
Thank you very much
$endgroup$
– Elementary
Jan 19 at 11:51
add a comment |
$begingroup$
Thank you very much
$endgroup$
– Elementary
Jan 19 at 11:51
$begingroup$
Thank you very much
$endgroup$
– Elementary
Jan 19 at 11:51
$begingroup$
Thank you very much
$endgroup$
– Elementary
Jan 19 at 11:51
add a comment |
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7
$begingroup$
That depends heavily on $f$ and $g$ ...
$endgroup$
– Martin R
Jan 19 at 9:01
$begingroup$
As an example: For constant $f=c_1$ and $g=c_2$ the limit will always exist.
$endgroup$
– maxmilgram
Jan 19 at 10:01
$begingroup$
I suppose that the limit is taking along the natural numbers, is that true?
$endgroup$
– Shashi
Jan 19 at 11:04
$begingroup$
The second statement looks like a statement about subsequences. I hope you know that a non convergent sequence might have a convergent subsequence... So from this you can already see that the statement can not be true.
$endgroup$
– Shashi
Jan 19 at 11:08
$begingroup$
@Shashi yes for positive integers., No, I did not know. The last thing you said, I learned now.
$endgroup$
– Elementary
Jan 19 at 11:18