Mixing
This polynomial limit doesn't exist but I can't seem to prove it.
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
$$lim_(x,y)to(0,0)fracx^2+y^2+5xyx-y$$
I've tried approximating using parabolas and even use polar coordinates but it seems to always result in $0$.
The limit seems simple so I must be forgetting something basic.
Hints? Thank you.
limits
edited 2 hours ago
Robert Howard
1,417622
asked 2 hours ago
João Simões
83
New contributor
João Simões is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
favorite
$$lim_(x,y)to(0,0)fracx^2+y^2+5xyx-y$$
I've tried approximating using parabolas and even use polar coordinates but it seems to always result in $0$.
The limit seems simple so I must be forgetting something basic.
Hints? Thank you.
limits
edited 2 hours ago
Robert Howard
1,417622
asked 2 hours ago
João Simões
83
New contributor
João Simões is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$$lim_(x,y)to(0,0)fracx^2+y^2+5xyx-y$$
I've tried approximating using parabolas and even use polar coordinates but it seems to always result in $0$.
The limit seems simple so I must be forgetting something basic.
Hints? Thank you.
limits
edited 2 hours ago
Robert Howard
1,417622
asked 2 hours ago
João Simões
83
New contributor
João Simões is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$$lim_(x,y)to(0,0)fracx^2+y^2+5xyx-y$$
I've tried approximating using parabolas and even use polar coordinates but it seems to always result in $0$.
The limit seems simple so I must be forgetting something basic.
Hints? Thank you.
limits
limits
edited 2 hours ago
Robert Howard
1,417622
asked 2 hours ago
João Simões
83
New contributor
João Simões is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
Robert Howard
1,417622
asked 2 hours ago
João Simões
83
New contributor
João Simões is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
Robert Howard
1,417622
edited 2 hours ago
Robert Howard
1,417622
edited 2 hours ago
Robert Howard
1,417622
1,417622
asked 2 hours ago
João Simões
83
New contributor
João Simões is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
João Simões
83
asked 2 hours ago
João Simões
83
83
New contributor
João Simões is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
João Simões is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
João Simões is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
The quotient blows up near the line $x=y$ no matter how close to $(0,0)$ you get, so the limit does not exist.
answered 2 hours ago
Henning Makholm
234k16299531
Small point: It's not a polynomial.
– zhw.
1 hour ago
@zhw: Right, fixed.
– Henning Makholm
24 mins ago
add a comment |Â
up vote
3
down vote
Write
beginalign
dfracx^2+y^2+5xyx-y &= x-y +dfrac7xyx-y \
&=x-y+yleft(dfrac7x-7y+7yx-yright) \
&=x-y+7y+dfrac7y^2x-y=x+6y+dfrac7y^2x-y.
endalign
To see that the limit fails to exist, consider $2$ paths: the first one is $(x,y) = (2t,t)$,and the limit is $0$. And the second path is $(x,y) = (t^2+t,t)$, and the limit is $7$. Thus you have $2$ different values of the limit, hence it does not exist.
edited 1 hour ago
Rócherz
2,2962518
answered 1 hour ago
DeepSea
69.9k54386
You might want to use beginalign and endalign for readability. $(+1)$ by the way.
– Mattos
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The quotient blows up near the line $x=y$ no matter how close to $(0,0)$ you get, so the limit does not exist.
answered 2 hours ago
Henning Makholm
234k16299531
Small point: It's not a polynomial.
– zhw.
1 hour ago
@zhw: Right, fixed.
– Henning Makholm
24 mins ago
add a comment |Â
up vote
2
down vote
accepted
The quotient blows up near the line $x=y$ no matter how close to $(0,0)$ you get, so the limit does not exist.
answered 2 hours ago
Henning Makholm
234k16299531
Small point: It's not a polynomial.
– zhw.
1 hour ago
@zhw: Right, fixed.
– Henning Makholm
24 mins ago
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The quotient blows up near the line $x=y$ no matter how close to $(0,0)$ you get, so the limit does not exist.
answered 2 hours ago
Henning Makholm
234k16299531
The quotient blows up near the line $x=y$ no matter how close to $(0,0)$ you get, so the limit does not exist.
answered 2 hours ago
Henning Makholm
234k16299531
edited 24 mins ago
answered 2 hours ago
Henning Makholm
234k16299531
answered 2 hours ago
Henning Makholm
234k16299531
answered 2 hours ago
Henning Makholm
234k16299531
234k16299531
Small point: It's not a polynomial.
– zhw.
1 hour ago
@zhw: Right, fixed.
– Henning Makholm
24 mins ago
add a comment |Â
Small point: It's not a polynomial.
– zhw.
1 hour ago
@zhw: Right, fixed.
– Henning Makholm
24 mins ago
Small point: It's not a polynomial.
– zhw.
1 hour ago
Small point: It's not a polynomial.
– zhw.
1 hour ago
@zhw: Right, fixed.
– Henning Makholm
24 mins ago
@zhw: Right, fixed.
– Henning Makholm
24 mins ago
add a comment |Â
up vote
3
down vote
Write
beginalign
dfracx^2+y^2+5xyx-y &= x-y +dfrac7xyx-y \
&=x-y+yleft(dfrac7x-7y+7yx-yright) \
&=x-y+7y+dfrac7y^2x-y=x+6y+dfrac7y^2x-y.
endalign
To see that the limit fails to exist, consider $2$ paths: the first one is $(x,y) = (2t,t)$,and the limit is $0$. And the second path is $(x,y) = (t^2+t,t)$, and the limit is $7$. Thus you have $2$ different values of the limit, hence it does not exist.
edited 1 hour ago
Rócherz
2,2962518
answered 1 hour ago
DeepSea
69.9k54386
You might want to use beginalign and endalign for readability. $(+1)$ by the way.
– Mattos
1 hour ago
add a comment |Â
up vote
3
down vote
Write
beginalign
dfracx^2+y^2+5xyx-y &= x-y +dfrac7xyx-y \
&=x-y+yleft(dfrac7x-7y+7yx-yright) \
&=x-y+7y+dfrac7y^2x-y=x+6y+dfrac7y^2x-y.
endalign
To see that the limit fails to exist, consider $2$ paths: the first one is $(x,y) = (2t,t)$,and the limit is $0$. And the second path is $(x,y) = (t^2+t,t)$, and the limit is $7$. Thus you have $2$ different values of the limit, hence it does not exist.
edited 1 hour ago
Rócherz
2,2962518
answered 1 hour ago
DeepSea
69.9k54386
You might want to use beginalign and endalign for readability. $(+1)$ by the way.
– Mattos
1 hour ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Write
beginalign
dfracx^2+y^2+5xyx-y &= x-y +dfrac7xyx-y \
&=x-y+yleft(dfrac7x-7y+7yx-yright) \
&=x-y+7y+dfrac7y^2x-y=x+6y+dfrac7y^2x-y.
endalign
To see that the limit fails to exist, consider $2$ paths: the first one is $(x,y) = (2t,t)$,and the limit is $0$. And the second path is $(x,y) = (t^2+t,t)$, and the limit is $7$. Thus you have $2$ different values of the limit, hence it does not exist.
edited 1 hour ago
Rócherz
2,2962518
answered 1 hour ago
DeepSea
69.9k54386
Write
beginalign
dfracx^2+y^2+5xyx-y &= x-y +dfrac7xyx-y \
&=x-y+yleft(dfrac7x-7y+7yx-yright) \
&=x-y+7y+dfrac7y^2x-y=x+6y+dfrac7y^2x-y.
endalign
To see that the limit fails to exist, consider $2$ paths: the first one is $(x,y) = (2t,t)$,and the limit is $0$. And the second path is $(x,y) = (t^2+t,t)$, and the limit is $7$. Thus you have $2$ different values of the limit, hence it does not exist.
edited 1 hour ago
Rócherz
2,2962518
answered 1 hour ago
DeepSea
69.9k54386
edited 1 hour ago
Rócherz
2,2962518
edited 1 hour ago
Rócherz
2,2962518
edited 1 hour ago
Rócherz
2,2962518
2,2962518
answered 1 hour ago
DeepSea
69.9k54386
answered 1 hour ago
DeepSea
69.9k54386
answered 1 hour ago
DeepSea
69.9k54386
69.9k54386
You might want to use beginalign and endalign for readability. $(+1)$ by the way.
– Mattos
1 hour ago
add a comment |Â
You might want to use beginalign and endalign for readability. $(+1)$ by the way.
– Mattos
1 hour ago
You might want to use beginalign and endalign for readability. $(+1)$ by the way.
– Mattos
1 hour ago
You might want to use beginalign and endalign for readability. $(+1)$ by the way.
– Mattos
1 hour ago
add a comment |Â
João Simões is a new contributor. Be nice, and check out our Code of Conduct.
João Simões is a new contributor. Be nice, and check out our Code of Conduct.
João Simões is a new contributor. Be nice, and check out our Code of Conduct.
João Simões is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2982435%2fthis-polynomial-limit-doesnt-exist-but-i-cant-seem-to-prove-it%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
