Evaluate $limlimits_(x,y)to(0,0)frac(x+y)^2x^2+y^2$
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Evaluate $displaystylelim_(x,y)to(0,0)dfrac(x+y)^2x^2+y^2$
Using polar, we have $x=rcos(theta),y=rsin(theta)$
Our limit becomes:
$$lim_rto 0dfrac(rcos(theta)+rsin(theta))^2r^2sin^2(theta)+r^2cos(theta)=lim_rto 0dfracr^2cos^2(theta)+2r^2cos(theta)sin(theta)+r^2sin^2(theta)r^2$$
Factoring and dividing removes the $r^2$ in the denominator, and we get $1$ as the limit.
However this is not right.
If we consider along the $x-axis$, our limit becomes $1$. If we consider along the line $y=x$, our limit becomes $1/2$, and are clearly not equal.
This means that the limit does not exist but my polar said it does and it equals $1$. Where did I mess up in my polar coordinates?
limits multivariable-calculus polar-coordinates
edited Aug 31 at 10:43
Martin Sleziak
43.6k6113260
asked Oct 17 '17 at 21:23
K Split X
3,847827
add a comment |Â
up vote
4
down vote
favorite
Evaluate $displaystylelim_(x,y)to(0,0)dfrac(x+y)^2x^2+y^2$
Using polar, we have $x=rcos(theta),y=rsin(theta)$
Our limit becomes:
$$lim_rto 0dfrac(rcos(theta)+rsin(theta))^2r^2sin^2(theta)+r^2cos(theta)=lim_rto 0dfracr^2cos^2(theta)+2r^2cos(theta)sin(theta)+r^2sin^2(theta)r^2$$
Factoring and dividing removes the $r^2$ in the denominator, and we get $1$ as the limit.
However this is not right.
If we consider along the $x-axis$, our limit becomes $1$. If we consider along the line $y=x$, our limit becomes $1/2$, and are clearly not equal.
This means that the limit does not exist but my polar said it does and it equals $1$. Where did I mess up in my polar coordinates?
limits multivariable-calculus polar-coordinates
edited Aug 31 at 10:43
Martin Sleziak
43.6k6113260
asked Oct 17 '17 at 21:23
K Split X
3,847827
1
It's a quite common mistake to think limits get easier when using polar coordinates, but often (and indeed here) people end up only considering what happens for constant $theta$, corresponding to straight lines. If you look closely at your calculations you'll probably also find that they are not valid along one of the along one of the axes, as you get get different results.
– Henrik
Oct 17 '17 at 21:27
10
you get actually $lim _ rrightarrow 0 frac r ^ 2 cos ^ 2 theta +2 r ^ 2 sin theta cos theta + r ^ 2 sin ^ 2 theta r ^ 2 cos ^ 2 theta + r ^ 2 sin ^ 2 theta =lim _ rrightarrow 0 frac r ^ 2 +2 r ^ 2 sin theta cos theta r ^ 2 =1+sin 2theta $ not $1$
– haqnatural
Oct 17 '17 at 21:29
3
Note that you're sending $r$ to zero, not $theta, $ which can vary arbitrarily with $r$ (and how it varies defines the various paths to approach zero). If you fixed $theta$, (which is taking the limit along a straight line), you'll find, after factoring the $r$, that you get that the limit is $(cos theta + sin theta)^2,$ not $1$. This clearly depends on $theta$, and hence the limit does not exist.
– stochasticboy321
Oct 17 '17 at 21:30
1
Also on a less important note, your limit along the line $y=x$ is $2$, not $1/2$. I mention only so it's clear that this accords with the straight line limit $1+sin(2theta)$ that somebody derived above. But as Henrik alluded to, even if the limit along all straight lines agreed it would not follow that the limit exists.
– spaceisdarkgreen
Oct 17 '17 at 22:33
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Evaluate $displaystylelim_(x,y)to(0,0)dfrac(x+y)^2x^2+y^2$
Using polar, we have $x=rcos(theta),y=rsin(theta)$
Our limit becomes:
$$lim_rto 0dfrac(rcos(theta)+rsin(theta))^2r^2sin^2(theta)+r^2cos(theta)=lim_rto 0dfracr^2cos^2(theta)+2r^2cos(theta)sin(theta)+r^2sin^2(theta)r^2$$
Factoring and dividing removes the $r^2$ in the denominator, and we get $1$ as the limit.
However this is not right.
If we consider along the $x-axis$, our limit becomes $1$. If we consider along the line $y=x$, our limit becomes $1/2$, and are clearly not equal.
This means that the limit does not exist but my polar said it does and it equals $1$. Where did I mess up in my polar coordinates?
limits multivariable-calculus polar-coordinates
edited Aug 31 at 10:43
Martin Sleziak
43.6k6113260
asked Oct 17 '17 at 21:23
K Split X
3,847827
Evaluate $displaystylelim_(x,y)to(0,0)dfrac(x+y)^2x^2+y^2$
Using polar, we have $x=rcos(theta),y=rsin(theta)$
Our limit becomes:
$$lim_rto 0dfrac(rcos(theta)+rsin(theta))^2r^2sin^2(theta)+r^2cos(theta)=lim_rto 0dfracr^2cos^2(theta)+2r^2cos(theta)sin(theta)+r^2sin^2(theta)r^2$$
Factoring and dividing removes the $r^2$ in the denominator, and we get $1$ as the limit.
However this is not right.
If we consider along the $x-axis$, our limit becomes $1$. If we consider along the line $y=x$, our limit becomes $1/2$, and are clearly not equal.
This means that the limit does not exist but my polar said it does and it equals $1$. Where did I mess up in my polar coordinates?
limits multivariable-calculus polar-coordinates
limits multivariable-calculus polar-coordinates
edited Aug 31 at 10:43
Martin Sleziak
43.6k6113260
asked Oct 17 '17 at 21:23
K Split X
3,847827
edited Aug 31 at 10:43
Martin Sleziak
43.6k6113260
asked Oct 17 '17 at 21:23
K Split X
3,847827
edited Aug 31 at 10:43
Martin Sleziak
43.6k6113260
edited Aug 31 at 10:43
Martin Sleziak
43.6k6113260
edited Aug 31 at 10:43
Martin Sleziak
43.6k6113260
43.6k6113260
asked Oct 17 '17 at 21:23
K Split X
3,847827
asked Oct 17 '17 at 21:23
K Split X
3,847827
asked Oct 17 '17 at 21:23
K Split X
3,847827
3,847827
1
It's a quite common mistake to think limits get easier when using polar coordinates, but often (and indeed here) people end up only considering what happens for constant $theta$, corresponding to straight lines. If you look closely at your calculations you'll probably also find that they are not valid along one of the along one of the axes, as you get get different results.
– Henrik
Oct 17 '17 at 21:27
10
you get actually $lim _ rrightarrow 0 frac r ^ 2 cos ^ 2 theta +2 r ^ 2 sin theta cos theta + r ^ 2 sin ^ 2 theta r ^ 2 cos ^ 2 theta + r ^ 2 sin ^ 2 theta =lim _ rrightarrow 0 frac r ^ 2 +2 r ^ 2 sin theta cos theta r ^ 2 =1+sin 2theta $ not $1$
– haqnatural
Oct 17 '17 at 21:29
3
Note that you're sending $r$ to zero, not $theta, $ which can vary arbitrarily with $r$ (and how it varies defines the various paths to approach zero). If you fixed $theta$, (which is taking the limit along a straight line), you'll find, after factoring the $r$, that you get that the limit is $(cos theta + sin theta)^2,$ not $1$. This clearly depends on $theta$, and hence the limit does not exist.
– stochasticboy321
Oct 17 '17 at 21:30
1
Also on a less important note, your limit along the line $y=x$ is $2$, not $1/2$. I mention only so it's clear that this accords with the straight line limit $1+sin(2theta)$ that somebody derived above. But as Henrik alluded to, even if the limit along all straight lines agreed it would not follow that the limit exists.
– spaceisdarkgreen
Oct 17 '17 at 22:33
add a comment |Â
1
It's a quite common mistake to think limits get easier when using polar coordinates, but often (and indeed here) people end up only considering what happens for constant $theta$, corresponding to straight lines. If you look closely at your calculations you'll probably also find that they are not valid along one of the along one of the axes, as you get get different results.
– Henrik
Oct 17 '17 at 21:27
10
you get actually $lim _ rrightarrow 0 frac r ^ 2 cos ^ 2 theta +2 r ^ 2 sin theta cos theta + r ^ 2 sin ^ 2 theta r ^ 2 cos ^ 2 theta + r ^ 2 sin ^ 2 theta =lim _ rrightarrow 0 frac r ^ 2 +2 r ^ 2 sin theta cos theta r ^ 2 =1+sin 2theta $ not $1$
– haqnatural
Oct 17 '17 at 21:29
3
Note that you're sending $r$ to zero, not $theta, $ which can vary arbitrarily with $r$ (and how it varies defines the various paths to approach zero). If you fixed $theta$, (which is taking the limit along a straight line), you'll find, after factoring the $r$, that you get that the limit is $(cos theta + sin theta)^2,$ not $1$. This clearly depends on $theta$, and hence the limit does not exist.
– stochasticboy321
Oct 17 '17 at 21:30
1
Also on a less important note, your limit along the line $y=x$ is $2$, not $1/2$. I mention only so it's clear that this accords with the straight line limit $1+sin(2theta)$ that somebody derived above. But as Henrik alluded to, even if the limit along all straight lines agreed it would not follow that the limit exists.
– spaceisdarkgreen
Oct 17 '17 at 22:33
1
1
It's a quite common mistake to think limits get easier when using polar coordinates, but often (and indeed here) people end up only considering what happens for constant $theta$, corresponding to straight lines. If you look closely at your calculations you'll probably also find that they are not valid along one of the along one of the axes, as you get get different results.
– Henrik
Oct 17 '17 at 21:27
It's a quite common mistake to think limits get easier when using polar coordinates, but often (and indeed here) people end up only considering what happens for constant $theta$, corresponding to straight lines. If you look closely at your calculations you'll probably also find that they are not valid along one of the along one of the axes, as you get get different results.
– Henrik
Oct 17 '17 at 21:27
10
10
you get actually $lim _ rrightarrow 0 frac r ^ 2 cos ^ 2 theta +2 r ^ 2 sin theta cos theta + r ^ 2 sin ^ 2 theta r ^ 2 cos ^ 2 theta + r ^ 2 sin ^ 2 theta =lim _ rrightarrow 0 frac r ^ 2 +2 r ^ 2 sin theta cos theta r ^ 2 =1+sin 2theta $ not $1$
– haqnatural
Oct 17 '17 at 21:29
you get actually $lim _ rrightarrow 0 frac r ^ 2 cos ^ 2 theta +2 r ^ 2 sin theta cos theta + r ^ 2 sin ^ 2 theta r ^ 2 cos ^ 2 theta + r ^ 2 sin ^ 2 theta =lim _ rrightarrow 0 frac r ^ 2 +2 r ^ 2 sin theta cos theta r ^ 2 =1+sin 2theta $ not $1$
– haqnatural
Oct 17 '17 at 21:29
3
3
Note that you're sending $r$ to zero, not $theta, $ which can vary arbitrarily with $r$ (and how it varies defines the various paths to approach zero). If you fixed $theta$, (which is taking the limit along a straight line), you'll find, after factoring the $r$, that you get that the limit is $(cos theta + sin theta)^2,$ not $1$. This clearly depends on $theta$, and hence the limit does not exist.
– stochasticboy321
Oct 17 '17 at 21:30
Note that you're sending $r$ to zero, not $theta, $ which can vary arbitrarily with $r$ (and how it varies defines the various paths to approach zero). If you fixed $theta$, (which is taking the limit along a straight line), you'll find, after factoring the $r$, that you get that the limit is $(cos theta + sin theta)^2,$ not $1$. This clearly depends on $theta$, and hence the limit does not exist.
– stochasticboy321
Oct 17 '17 at 21:30
1
1
Also on a less important note, your limit along the line $y=x$ is $2$, not $1/2$. I mention only so it's clear that this accords with the straight line limit $1+sin(2theta)$ that somebody derived above. But as Henrik alluded to, even if the limit along all straight lines agreed it would not follow that the limit exists.
– spaceisdarkgreen
Oct 17 '17 at 22:33
Also on a less important note, your limit along the line $y=x$ is $2$, not $1/2$. I mention only so it's clear that this accords with the straight line limit $1+sin(2theta)$ that somebody derived above. But as Henrik alluded to, even if the limit along all straight lines agreed it would not follow that the limit exists.
– spaceisdarkgreen
Oct 17 '17 at 22:33
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Its been extensively discussed and OP is probably a bit exhausted by seeing many view points that some are clear and some are not. Let's give a "nice proof " that the limit does not exist. Just choose two paths: path $1$ is: $x = -y = t$, then the limit is $0$, and path $2$ is: $x = y = t$, then the limit is $2$. Different values of limits show there is no limit at $(0,0)$.
answered Oct 17 '17 at 23:23
DeepSea
69.1k54284
This doesn't address at all the actual question that the OP asked: "Where did I mess up in my polar coordinates?"
– zipirovich
Oct 17 '17 at 23:26
The problem and solution is clear. I messed up on taking the limit, I took it with respect to $theta$ instead of $r$. @DeepSea the second half of my question shows how the limit does not exist,
– K Split X
Oct 18 '17 at 1:48
@KSplitX: "some one" downvoted me..haha..you see...
– DeepSea
Oct 18 '17 at 4:59
Wasn't me, but Ill make it even
– K Split X
Oct 18 '17 at 11:32
add a comment |Â
up vote
0
down vote
The above limit did not exists. Indeed for the path $y=-x$ one gets that the limit equals zero, while for the path $y=x$ one gets that
$$lim_xrightarrow 0 dfrac(x+x)^2x^2+x^2=2 neq 0.$$
answered Aug 30 at 3:57
Nelson Faustino
1376
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Its been extensively discussed and OP is probably a bit exhausted by seeing many view points that some are clear and some are not. Let's give a "nice proof " that the limit does not exist. Just choose two paths: path $1$ is: $x = -y = t$, then the limit is $0$, and path $2$ is: $x = y = t$, then the limit is $2$. Different values of limits show there is no limit at $(0,0)$.
answered Oct 17 '17 at 23:23
DeepSea
69.1k54284
This doesn't address at all the actual question that the OP asked: "Where did I mess up in my polar coordinates?"
– zipirovich
Oct 17 '17 at 23:26
The problem and solution is clear. I messed up on taking the limit, I took it with respect to $theta$ instead of $r$. @DeepSea the second half of my question shows how the limit does not exist,
– K Split X
Oct 18 '17 at 1:48
@KSplitX: "some one" downvoted me..haha..you see...
– DeepSea
Oct 18 '17 at 4:59
Wasn't me, but Ill make it even
– K Split X
Oct 18 '17 at 11:32
add a comment |Â
up vote
1
down vote
accepted
Its been extensively discussed and OP is probably a bit exhausted by seeing many view points that some are clear and some are not. Let's give a "nice proof " that the limit does not exist. Just choose two paths: path $1$ is: $x = -y = t$, then the limit is $0$, and path $2$ is: $x = y = t$, then the limit is $2$. Different values of limits show there is no limit at $(0,0)$.
answered Oct 17 '17 at 23:23
DeepSea
69.1k54284
This doesn't address at all the actual question that the OP asked: "Where did I mess up in my polar coordinates?"
– zipirovich
Oct 17 '17 at 23:26
The problem and solution is clear. I messed up on taking the limit, I took it with respect to $theta$ instead of $r$. @DeepSea the second half of my question shows how the limit does not exist,
– K Split X
Oct 18 '17 at 1:48
@KSplitX: "some one" downvoted me..haha..you see...
– DeepSea
Oct 18 '17 at 4:59
Wasn't me, but Ill make it even
– K Split X
Oct 18 '17 at 11:32
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Its been extensively discussed and OP is probably a bit exhausted by seeing many view points that some are clear and some are not. Let's give a "nice proof " that the limit does not exist. Just choose two paths: path $1$ is: $x = -y = t$, then the limit is $0$, and path $2$ is: $x = y = t$, then the limit is $2$. Different values of limits show there is no limit at $(0,0)$.
answered Oct 17 '17 at 23:23
DeepSea
69.1k54284
Its been extensively discussed and OP is probably a bit exhausted by seeing many view points that some are clear and some are not. Let's give a "nice proof " that the limit does not exist. Just choose two paths: path $1$ is: $x = -y = t$, then the limit is $0$, and path $2$ is: $x = y = t$, then the limit is $2$. Different values of limits show there is no limit at $(0,0)$.
answered Oct 17 '17 at 23:23
DeepSea
69.1k54284
answered Oct 17 '17 at 23:23
DeepSea
69.1k54284
answered Oct 17 '17 at 23:23
DeepSea
69.1k54284
answered Oct 17 '17 at 23:23
DeepSea
69.1k54284
69.1k54284
This doesn't address at all the actual question that the OP asked: "Where did I mess up in my polar coordinates?"
– zipirovich
Oct 17 '17 at 23:26
The problem and solution is clear. I messed up on taking the limit, I took it with respect to $theta$ instead of $r$. @DeepSea the second half of my question shows how the limit does not exist,
– K Split X
Oct 18 '17 at 1:48
@KSplitX: "some one" downvoted me..haha..you see...
– DeepSea
Oct 18 '17 at 4:59
Wasn't me, but Ill make it even
– K Split X
Oct 18 '17 at 11:32
add a comment |Â
This doesn't address at all the actual question that the OP asked: "Where did I mess up in my polar coordinates?"
– zipirovich
Oct 17 '17 at 23:26
The problem and solution is clear. I messed up on taking the limit, I took it with respect to $theta$ instead of $r$. @DeepSea the second half of my question shows how the limit does not exist,
– K Split X
Oct 18 '17 at 1:48
@KSplitX: "some one" downvoted me..haha..you see...
– DeepSea
Oct 18 '17 at 4:59
Wasn't me, but Ill make it even
– K Split X
Oct 18 '17 at 11:32
This doesn't address at all the actual question that the OP asked: "Where did I mess up in my polar coordinates?"
– zipirovich
Oct 17 '17 at 23:26
This doesn't address at all the actual question that the OP asked: "Where did I mess up in my polar coordinates?"
– zipirovich
Oct 17 '17 at 23:26
The problem and solution is clear. I messed up on taking the limit, I took it with respect to $theta$ instead of $r$. @DeepSea the second half of my question shows how the limit does not exist,
– K Split X
Oct 18 '17 at 1:48
The problem and solution is clear. I messed up on taking the limit, I took it with respect to $theta$ instead of $r$. @DeepSea the second half of my question shows how the limit does not exist,
– K Split X
Oct 18 '17 at 1:48
@KSplitX: "some one" downvoted me..haha..you see...
– DeepSea
Oct 18 '17 at 4:59
@KSplitX: "some one" downvoted me..haha..you see...
– DeepSea
Oct 18 '17 at 4:59
Wasn't me, but Ill make it even
– K Split X
Oct 18 '17 at 11:32
Wasn't me, but Ill make it even
– K Split X
Oct 18 '17 at 11:32
add a comment |Â
up vote
0
down vote
The above limit did not exists. Indeed for the path $y=-x$ one gets that the limit equals zero, while for the path $y=x$ one gets that
$$lim_xrightarrow 0 dfrac(x+x)^2x^2+x^2=2 neq 0.$$
answered Aug 30 at 3:57
Nelson Faustino
1376
add a comment |Â
up vote
0
down vote
The above limit did not exists. Indeed for the path $y=-x$ one gets that the limit equals zero, while for the path $y=x$ one gets that
$$lim_xrightarrow 0 dfrac(x+x)^2x^2+x^2=2 neq 0.$$
answered Aug 30 at 3:57
Nelson Faustino
1376
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The above limit did not exists. Indeed for the path $y=-x$ one gets that the limit equals zero, while for the path $y=x$ one gets that
$$lim_xrightarrow 0 dfrac(x+x)^2x^2+x^2=2 neq 0.$$
answered Aug 30 at 3:57
Nelson Faustino
1376
The above limit did not exists. Indeed for the path $y=-x$ one gets that the limit equals zero, while for the path $y=x$ one gets that
$$lim_xrightarrow 0 dfrac(x+x)^2x^2+x^2=2 neq 0.$$
answered Aug 30 at 3:57
Nelson Faustino
1376
answered Aug 30 at 3:57
Nelson Faustino
1376
answered Aug 30 at 3:57
Nelson Faustino
1376
answered Aug 30 at 3:57
Nelson Faustino
1376
1376
add a comment |Â
add a comment |Â
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1
It's a quite common mistake to think limits get easier when using polar coordinates, but often (and indeed here) people end up only considering what happens for constant $theta$, corresponding to straight lines. If you look closely at your calculations you'll probably also find that they are not valid along one of the along one of the axes, as you get get different results.
– Henrik
Oct 17 '17 at 21:27
10
you get actually $lim _ rrightarrow 0 frac r ^ 2 cos ^ 2 theta +2 r ^ 2 sin theta cos theta + r ^ 2 sin ^ 2 theta r ^ 2 cos ^ 2 theta + r ^ 2 sin ^ 2 theta =lim _ rrightarrow 0 frac r ^ 2 +2 r ^ 2 sin theta cos theta r ^ 2 =1+sin 2theta $ not $1$
– haqnatural
Oct 17 '17 at 21:29
3
Note that you're sending $r$ to zero, not $theta, $ which can vary arbitrarily with $r$ (and how it varies defines the various paths to approach zero). If you fixed $theta$, (which is taking the limit along a straight line), you'll find, after factoring the $r$, that you get that the limit is $(cos theta + sin theta)^2,$ not $1$. This clearly depends on $theta$, and hence the limit does not exist.
– stochasticboy321
Oct 17 '17 at 21:30
1
Also on a less important note, your limit along the line $y=x$ is $2$, not $1/2$. I mention only so it's clear that this accords with the straight line limit $1+sin(2theta)$ that somebody derived above. But as Henrik alluded to, even if the limit along all straight lines agreed it would not follow that the limit exists.
– spaceisdarkgreen
Oct 17 '17 at 22:33