Mixing
The limit of $u(x,y)$ when $x^2+y^2toinfty$
$begingroup$
I cannot understand what the following does it mean:$limlimits{u(x,y)}=0$ , when $x^2+y^2toinfty$? Does it mean that at least one of the variables goes to $infty$ or both, or something else?
real-analysis analysis
asked Jan 30 at 18:09
dmtridmtri
1,7712521
$endgroup$
add a comment |
$begingroup$
I cannot understand what the following does it mean:$limlimits{u(x,y)}=0$ , when $x^2+y^2toinfty$? Does it mean that at least one of the variables goes to $infty$ or both, or something else?
real-analysis analysis
asked Jan 30 at 18:09
dmtridmtri
1,7712521
$endgroup$
$begingroup$
It really means that $x^2+y^2 to infty$. Note that $x^2+y^2$ is a number for every pair $(x,y)$. In order for this scalar quantity to tend to $infty$, at least one of the variables must tend to $infty$.
$endgroup$
– PierreCarre
Jan 30 at 18:13
add a comment |
$begingroup$
I cannot understand what the following does it mean:$limlimits{u(x,y)}=0$ , when $x^2+y^2toinfty$? Does it mean that at least one of the variables goes to $infty$ or both, or something else?
real-analysis analysis
asked Jan 30 at 18:09
dmtridmtri
1,7712521
$endgroup$
I cannot understand what the following does it mean:$limlimits{u(x,y)}=0$ , when $x^2+y^2toinfty$? Does it mean that at least one of the variables goes to $infty$ or both, or something else?
real-analysis analysis
real-analysis analysis
asked Jan 30 at 18:09
dmtridmtri
1,7712521
asked Jan 30 at 18:09
dmtridmtri
1,7712521
asked Jan 30 at 18:09
dmtridmtri
1,7712521
asked Jan 30 at 18:09
dmtridmtri
1,7712521
asked Jan 30 at 18:09
dmtridmtri
1,7712521
1,7712521
$begingroup$
It really means that $x^2+y^2 to infty$. Note that $x^2+y^2$ is a number for every pair $(x,y)$. In order for this scalar quantity to tend to $infty$, at least one of the variables must tend to $infty$.
$endgroup$
– PierreCarre
Jan 30 at 18:13
add a comment |
$begingroup$
It really means that $x^2+y^2 to infty$. Note that $x^2+y^2$ is a number for every pair $(x,y)$. In order for this scalar quantity to tend to $infty$, at least one of the variables must tend to $infty$.
$endgroup$
– PierreCarre
Jan 30 at 18:13
$begingroup$
It really means that $x^2+y^2 to infty$. Note that $x^2+y^2$ is a number for every pair $(x,y)$. In order for this scalar quantity to tend to $infty$, at least one of the variables must tend to $infty$.
$endgroup$
– PierreCarre
Jan 30 at 18:13
$begingroup$
It really means that $x^2+y^2 to infty$. Note that $x^2+y^2$ is a number for every pair $(x,y)$. In order for this scalar quantity to tend to $infty$, at least one of the variables must tend to $infty$.
$endgroup$
– PierreCarre
Jan 30 at 18:13
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I would not focus on the meaning of $x^2 + y^2 to infty$ in isolation.
The statement $lim_{x^2 + y^2 to infty}u(x,y) = L$ means for any $epsilon > 0$ there exists $K > 0$ such that $|u(x,y) - L| < epsilon $ for all $(x,y)$ where $x^2 + y^2 > K$.
answered Jan 30 at 18:29
RRLRRL
53.4k52574
$endgroup$
$begingroup$
I did write an answer once that allows picking apart the elements of the definition, in terms of filters - though I can't seem to find that right now. So, $x^2 + y^2 to infty$ is interpreted as the filter $mathscr{F}$ with basis ${ { (x,y) mid x^2 + y^2 > K } mid K in mathbb{R} }$, and then $u(x,y) to 0$ as $x^2 + y^2 to infty$ is interpreted as the filter $u_* mathscr{F} = { Ssubseteqmathbb{R} mid u^{-1}(S) in mathscr{F} }$ having limit 0 (i.e. having all neighborhoods of 0 as elements).
$endgroup$
– Daniel Schepler
Jan 30 at 19:22
add a comment |
$begingroup$
That sounds like it means at least one of the variables goes to $infty$. $x^2 + y^2 = r^2$ is often times used to describe a circle (centered at the origin) with radius r. If $x^2 + y^2 to infty$, then you are considering a circle (centered at the origin) with infinite radius. Do you have any information on $u(x,y)$?
answered Jan 30 at 18:15
Akash PatelAkash Patel
168110
$endgroup$
add a comment |
$begingroup$
Consider that $x^2+y^2=|(x,y)|_2$.
So the meaning of
$$
x^2+y^2toinfty
$$
is, that you consider something for points with large euclidean norm. If the euclidean norm goes to infinity, your points are getting more and more far distant to the orgin.
Even if the point is far from the orgin, the coordinates don't have to be both large. You can even conclude $x^2toinfty$ or $y^2toinfty$.
Consider
$$
x_n=(1-(-1)^n)ntext{ and }y_n=(1-(-1)^{n-1})n.
$$
You see $x_nnottoinfty$ and $y_nnottoinfty$ but
$$
x_n^2+y_n^2=4n^2toinfty.
$$
answered Jan 30 at 18:17
Mundron SchmidtMundron Schmidt
7,4942729
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I would not focus on the meaning of $x^2 + y^2 to infty$ in isolation.
The statement $lim_{x^2 + y^2 to infty}u(x,y) = L$ means for any $epsilon > 0$ there exists $K > 0$ such that $|u(x,y) - L| < epsilon $ for all $(x,y)$ where $x^2 + y^2 > K$.
answered Jan 30 at 18:29
RRLRRL
53.4k52574
$endgroup$
$begingroup$
I did write an answer once that allows picking apart the elements of the definition, in terms of filters - though I can't seem to find that right now. So, $x^2 + y^2 to infty$ is interpreted as the filter $mathscr{F}$ with basis ${ { (x,y) mid x^2 + y^2 > K } mid K in mathbb{R} }$, and then $u(x,y) to 0$ as $x^2 + y^2 to infty$ is interpreted as the filter $u_* mathscr{F} = { Ssubseteqmathbb{R} mid u^{-1}(S) in mathscr{F} }$ having limit 0 (i.e. having all neighborhoods of 0 as elements).
$endgroup$
– Daniel Schepler
Jan 30 at 19:22
add a comment |
$begingroup$
I would not focus on the meaning of $x^2 + y^2 to infty$ in isolation.
The statement $lim_{x^2 + y^2 to infty}u(x,y) = L$ means for any $epsilon > 0$ there exists $K > 0$ such that $|u(x,y) - L| < epsilon $ for all $(x,y)$ where $x^2 + y^2 > K$.
answered Jan 30 at 18:29
RRLRRL
53.4k52574
$endgroup$
$begingroup$
I did write an answer once that allows picking apart the elements of the definition, in terms of filters - though I can't seem to find that right now. So, $x^2 + y^2 to infty$ is interpreted as the filter $mathscr{F}$ with basis ${ { (x,y) mid x^2 + y^2 > K } mid K in mathbb{R} }$, and then $u(x,y) to 0$ as $x^2 + y^2 to infty$ is interpreted as the filter $u_* mathscr{F} = { Ssubseteqmathbb{R} mid u^{-1}(S) in mathscr{F} }$ having limit 0 (i.e. having all neighborhoods of 0 as elements).
$endgroup$
– Daniel Schepler
Jan 30 at 19:22
add a comment |
$begingroup$
I would not focus on the meaning of $x^2 + y^2 to infty$ in isolation.
The statement $lim_{x^2 + y^2 to infty}u(x,y) = L$ means for any $epsilon > 0$ there exists $K > 0$ such that $|u(x,y) - L| < epsilon $ for all $(x,y)$ where $x^2 + y^2 > K$.
answered Jan 30 at 18:29
RRLRRL
53.4k52574
$endgroup$
I would not focus on the meaning of $x^2 + y^2 to infty$ in isolation.
The statement $lim_{x^2 + y^2 to infty}u(x,y) = L$ means for any $epsilon > 0$ there exists $K > 0$ such that $|u(x,y) - L| < epsilon $ for all $(x,y)$ where $x^2 + y^2 > K$.
answered Jan 30 at 18:29
RRLRRL
53.4k52574
answered Jan 30 at 18:29
RRLRRL
53.4k52574
answered Jan 30 at 18:29
RRLRRL
53.4k52574
answered Jan 30 at 18:29
RRLRRL
53.4k52574
53.4k52574
$begingroup$
I did write an answer once that allows picking apart the elements of the definition, in terms of filters - though I can't seem to find that right now. So, $x^2 + y^2 to infty$ is interpreted as the filter $mathscr{F}$ with basis ${ { (x,y) mid x^2 + y^2 > K } mid K in mathbb{R} }$, and then $u(x,y) to 0$ as $x^2 + y^2 to infty$ is interpreted as the filter $u_* mathscr{F} = { Ssubseteqmathbb{R} mid u^{-1}(S) in mathscr{F} }$ having limit 0 (i.e. having all neighborhoods of 0 as elements).
$endgroup$
– Daniel Schepler
Jan 30 at 19:22
add a comment |
$begingroup$
I did write an answer once that allows picking apart the elements of the definition, in terms of filters - though I can't seem to find that right now. So, $x^2 + y^2 to infty$ is interpreted as the filter $mathscr{F}$ with basis ${ { (x,y) mid x^2 + y^2 > K } mid K in mathbb{R} }$, and then $u(x,y) to 0$ as $x^2 + y^2 to infty$ is interpreted as the filter $u_* mathscr{F} = { Ssubseteqmathbb{R} mid u^{-1}(S) in mathscr{F} }$ having limit 0 (i.e. having all neighborhoods of 0 as elements).
$endgroup$
– Daniel Schepler
Jan 30 at 19:22
$begingroup$
I did write an answer once that allows picking apart the elements of the definition, in terms of filters - though I can't seem to find that right now. So, $x^2 + y^2 to infty$ is interpreted as the filter $mathscr{F}$ with basis ${ { (x,y) mid x^2 + y^2 > K } mid K in mathbb{R} }$, and then $u(x,y) to 0$ as $x^2 + y^2 to infty$ is interpreted as the filter $u_* mathscr{F} = { Ssubseteqmathbb{R} mid u^{-1}(S) in mathscr{F} }$ having limit 0 (i.e. having all neighborhoods of 0 as elements).
$endgroup$
– Daniel Schepler
Jan 30 at 19:22
$begingroup$
I did write an answer once that allows picking apart the elements of the definition, in terms of filters - though I can't seem to find that right now. So, $x^2 + y^2 to infty$ is interpreted as the filter $mathscr{F}$ with basis ${ { (x,y) mid x^2 + y^2 > K } mid K in mathbb{R} }$, and then $u(x,y) to 0$ as $x^2 + y^2 to infty$ is interpreted as the filter $u_* mathscr{F} = { Ssubseteqmathbb{R} mid u^{-1}(S) in mathscr{F} }$ having limit 0 (i.e. having all neighborhoods of 0 as elements).
$endgroup$
– Daniel Schepler
Jan 30 at 19:22
add a comment |
$begingroup$
That sounds like it means at least one of the variables goes to $infty$. $x^2 + y^2 = r^2$ is often times used to describe a circle (centered at the origin) with radius r. If $x^2 + y^2 to infty$, then you are considering a circle (centered at the origin) with infinite radius. Do you have any information on $u(x,y)$?
answered Jan 30 at 18:15
Akash PatelAkash Patel
168110
$endgroup$
add a comment |
$begingroup$
That sounds like it means at least one of the variables goes to $infty$. $x^2 + y^2 = r^2$ is often times used to describe a circle (centered at the origin) with radius r. If $x^2 + y^2 to infty$, then you are considering a circle (centered at the origin) with infinite radius. Do you have any information on $u(x,y)$?
answered Jan 30 at 18:15
Akash PatelAkash Patel
168110
$endgroup$
add a comment |
$begingroup$
That sounds like it means at least one of the variables goes to $infty$. $x^2 + y^2 = r^2$ is often times used to describe a circle (centered at the origin) with radius r. If $x^2 + y^2 to infty$, then you are considering a circle (centered at the origin) with infinite radius. Do you have any information on $u(x,y)$?
answered Jan 30 at 18:15
Akash PatelAkash Patel
168110
$endgroup$
That sounds like it means at least one of the variables goes to $infty$. $x^2 + y^2 = r^2$ is often times used to describe a circle (centered at the origin) with radius r. If $x^2 + y^2 to infty$, then you are considering a circle (centered at the origin) with infinite radius. Do you have any information on $u(x,y)$?
answered Jan 30 at 18:15
Akash PatelAkash Patel
168110
answered Jan 30 at 18:15
Akash PatelAkash Patel
168110
answered Jan 30 at 18:15
Akash PatelAkash Patel
168110
answered Jan 30 at 18:15
Akash PatelAkash Patel
168110
168110
add a comment |
add a comment |
$begingroup$
Consider that $x^2+y^2=|(x,y)|_2$.
So the meaning of
$$
x^2+y^2toinfty
$$
is, that you consider something for points with large euclidean norm. If the euclidean norm goes to infinity, your points are getting more and more far distant to the orgin.
Even if the point is far from the orgin, the coordinates don't have to be both large. You can even conclude $x^2toinfty$ or $y^2toinfty$.
Consider
$$
x_n=(1-(-1)^n)ntext{ and }y_n=(1-(-1)^{n-1})n.
$$
You see $x_nnottoinfty$ and $y_nnottoinfty$ but
$$
x_n^2+y_n^2=4n^2toinfty.
$$
answered Jan 30 at 18:17
Mundron SchmidtMundron Schmidt
7,4942729
$endgroup$
add a comment |
$begingroup$
Consider that $x^2+y^2=|(x,y)|_2$.
So the meaning of
$$
x^2+y^2toinfty
$$
is, that you consider something for points with large euclidean norm. If the euclidean norm goes to infinity, your points are getting more and more far distant to the orgin.
Even if the point is far from the orgin, the coordinates don't have to be both large. You can even conclude $x^2toinfty$ or $y^2toinfty$.
Consider
$$
x_n=(1-(-1)^n)ntext{ and }y_n=(1-(-1)^{n-1})n.
$$
You see $x_nnottoinfty$ and $y_nnottoinfty$ but
$$
x_n^2+y_n^2=4n^2toinfty.
$$
answered Jan 30 at 18:17
Mundron SchmidtMundron Schmidt
7,4942729
$endgroup$
add a comment |
$begingroup$
Consider that $x^2+y^2=|(x,y)|_2$.
So the meaning of
$$
x^2+y^2toinfty
$$
is, that you consider something for points with large euclidean norm. If the euclidean norm goes to infinity, your points are getting more and more far distant to the orgin.
Even if the point is far from the orgin, the coordinates don't have to be both large. You can even conclude $x^2toinfty$ or $y^2toinfty$.
Consider
$$
x_n=(1-(-1)^n)ntext{ and }y_n=(1-(-1)^{n-1})n.
$$
You see $x_nnottoinfty$ and $y_nnottoinfty$ but
$$
x_n^2+y_n^2=4n^2toinfty.
$$
answered Jan 30 at 18:17
Mundron SchmidtMundron Schmidt
7,4942729
$endgroup$
Consider that $x^2+y^2=|(x,y)|_2$.
So the meaning of
$$
x^2+y^2toinfty
$$
is, that you consider something for points with large euclidean norm. If the euclidean norm goes to infinity, your points are getting more and more far distant to the orgin.
Even if the point is far from the orgin, the coordinates don't have to be both large. You can even conclude $x^2toinfty$ or $y^2toinfty$.
Consider
$$
x_n=(1-(-1)^n)ntext{ and }y_n=(1-(-1)^{n-1})n.
$$
You see $x_nnottoinfty$ and $y_nnottoinfty$ but
$$
x_n^2+y_n^2=4n^2toinfty.
$$
answered Jan 30 at 18:17
Mundron SchmidtMundron Schmidt
7,4942729
answered Jan 30 at 18:17
Mundron SchmidtMundron Schmidt
7,4942729
answered Jan 30 at 18:17
Mundron SchmidtMundron Schmidt
7,4942729
answered Jan 30 at 18:17
Mundron SchmidtMundron Schmidt
7,4942729
7,4942729
add a comment |
add a comment |
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$begingroup$
It really means that $x^2+y^2 to infty$. Note that $x^2+y^2$ is a number for every pair $(x,y)$. In order for this scalar quantity to tend to $infty$, at least one of the variables must tend to $infty$.
$endgroup$
– PierreCarre
Jan 30 at 18:13