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Does the limit $limlimits_(x,y)to (0,0)fracx^3-y^33x+2y$ exist?
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Does this limit exist? $$limlimits_(x,y)to (0,0)fracx^3-y^33x+2y$$
From what I understand, the line $3x+2y=0$ goes through $(0,0)$, so the limit doesn't exist.
But shouldn't I be able to find two paths with different limits in that case? I can't seem to find them.
limits multivariable-calculus
edited Jul 23 at 8:34
Lorenzo B.
1,5402418
asked Jul 23 at 7:38
Manuel
6
add a comment |Â
up vote
-2
down vote
favorite
Does this limit exist? $$limlimits_(x,y)to (0,0)fracx^3-y^33x+2y$$
From what I understand, the line $3x+2y=0$ goes through $(0,0)$, so the limit doesn't exist.
But shouldn't I be able to find two paths with different limits in that case? I can't seem to find them.
limits multivariable-calculus
edited Jul 23 at 8:34
Lorenzo B.
1,5402418
asked Jul 23 at 7:38
Manuel
6
The trick in these cases, when we want to show that the limit doesn't exist, is to approach along a path such that the main part of $3x+2y$ is equal to zero but we keep some remainder which determine the value for the denominator in such way that the limit takes a different values. In this case we can choose for $x=t to 0$and then $y=-frac32 t+t^n$. We need to choose $n$ in such way to obtain a different limit from the trivail one obtained for the path with $x=y$.
– gimusi
Jul 23 at 8:05
Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 10 at 23:26
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
Does this limit exist? $$limlimits_(x,y)to (0,0)fracx^3-y^33x+2y$$
From what I understand, the line $3x+2y=0$ goes through $(0,0)$, so the limit doesn't exist.
But shouldn't I be able to find two paths with different limits in that case? I can't seem to find them.
limits multivariable-calculus
edited Jul 23 at 8:34
Lorenzo B.
1,5402418
asked Jul 23 at 7:38
Manuel
6
Does this limit exist? $$limlimits_(x,y)to (0,0)fracx^3-y^33x+2y$$
From what I understand, the line $3x+2y=0$ goes through $(0,0)$, so the limit doesn't exist.
But shouldn't I be able to find two paths with different limits in that case? I can't seem to find them.
limits multivariable-calculus
edited Jul 23 at 8:34
Lorenzo B.
1,5402418
asked Jul 23 at 7:38
Manuel
6
edited Jul 23 at 8:34
Lorenzo B.
1,5402418
edited Jul 23 at 8:34
Lorenzo B.
1,5402418
edited Jul 23 at 8:34
Lorenzo B.
1,5402418
1,5402418
asked Jul 23 at 7:38
Manuel
6
asked Jul 23 at 7:38
Manuel
6
asked Jul 23 at 7:38
Manuel
6
6
The trick in these cases, when we want to show that the limit doesn't exist, is to approach along a path such that the main part of $3x+2y$ is equal to zero but we keep some remainder which determine the value for the denominator in such way that the limit takes a different values. In this case we can choose for $x=t to 0$and then $y=-frac32 t+t^n$. We need to choose $n$ in such way to obtain a different limit from the trivail one obtained for the path with $x=y$.
– gimusi
Jul 23 at 8:05
Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 10 at 23:26
add a comment |Â
The trick in these cases, when we want to show that the limit doesn't exist, is to approach along a path such that the main part of $3x+2y$ is equal to zero but we keep some remainder which determine the value for the denominator in such way that the limit takes a different values. In this case we can choose for $x=t to 0$and then $y=-frac32 t+t^n$. We need to choose $n$ in such way to obtain a different limit from the trivail one obtained for the path with $x=y$.
– gimusi
Jul 23 at 8:05
Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 10 at 23:26
The trick in these cases, when we want to show that the limit doesn't exist, is to approach along a path such that the main part of $3x+2y$ is equal to zero but we keep some remainder which determine the value for the denominator in such way that the limit takes a different values. In this case we can choose for $x=t to 0$and then $y=-frac32 t+t^n$. We need to choose $n$ in such way to obtain a different limit from the trivail one obtained for the path with $x=y$.
– gimusi
Jul 23 at 8:05
The trick in these cases, when we want to show that the limit doesn't exist, is to approach along a path such that the main part of $3x+2y$ is equal to zero but we keep some remainder which determine the value for the denominator in such way that the limit takes a different values. In this case we can choose for $x=t to 0$and then $y=-frac32 t+t^n$. We need to choose $n$ in such way to obtain a different limit from the trivail one obtained for the path with $x=y$.
– gimusi
Jul 23 at 8:05
Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 10 at 23:26
Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 10 at 23:26
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
Go along curves $y=ax+bx^3$ for suitable $a$ and $b$.
answered Jul 23 at 7:45
Empy2
31.8k12059
add a comment |Â
up vote
1
down vote
Note that the points such that $3x+2y=0$ are simply excluded from the domain and thus we can't conclude that the limit doesn't exist from here.
To show that we need to find two different paths with different limits as $(x,y)to (0,0)$.
Notably in this case we have that
for $x=y$ we have $dfracx^3-y^33x+2y=0$
for $x=tto 0$ and $y=-frac32t+t^3$ we have $=dfracx^3-y^33x+2y=dfract^3-(-frac32t+t^3)^33t-3t+2t^3to Lin mathbbRsetminus0$
answered Jul 23 at 7:44
gimusi
65.2k73583
What I meant by 3x+2y=0 is that the function tends to + or -infinity near those points, so I assume the delta-epsilon definition would fail. Anyway, thank you very much. This helped a lot.
– Manuel
Jul 23 at 8:08
@Manuel It is not a reason to conclude that the limit doesn't exist. What about, for example, for $lim_(x,y)to (0,0) fracsin xyxy$?
– gimusi
Jul 23 at 8:10
@Manuel Note also that, usually, when we refer to the existence of the limit we mean that the limit is the same (finite or infinite) for any path. Therefore to show that the limit doesn't exist it suffices to find (at least) two different paths with different limits.
– gimusi
Jul 23 at 8:11
I'd say your example is a different situation, because at the points where the denominator = 0, your function approaches 1.
– Manuel
Jul 23 at 8:21
@Manuel And what about $lim_(x,y)to (0,0) frac1x^2y^2$?
– gimusi
Jul 23 at 8:26
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Go along curves $y=ax+bx^3$ for suitable $a$ and $b$.
answered Jul 23 at 7:45
Empy2
31.8k12059
add a comment |Â
up vote
1
down vote
Go along curves $y=ax+bx^3$ for suitable $a$ and $b$.
answered Jul 23 at 7:45
Empy2
31.8k12059
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Go along curves $y=ax+bx^3$ for suitable $a$ and $b$.
answered Jul 23 at 7:45
Empy2
31.8k12059
Go along curves $y=ax+bx^3$ for suitable $a$ and $b$.
answered Jul 23 at 7:45
Empy2
31.8k12059
answered Jul 23 at 7:45
Empy2
31.8k12059
answered Jul 23 at 7:45
Empy2
31.8k12059
answered Jul 23 at 7:45
Empy2
31.8k12059
31.8k12059
add a comment |Â
add a comment |Â
up vote
1
down vote
Note that the points such that $3x+2y=0$ are simply excluded from the domain and thus we can't conclude that the limit doesn't exist from here.
To show that we need to find two different paths with different limits as $(x,y)to (0,0)$.
Notably in this case we have that
for $x=y$ we have $dfracx^3-y^33x+2y=0$
for $x=tto 0$ and $y=-frac32t+t^3$ we have $=dfracx^3-y^33x+2y=dfract^3-(-frac32t+t^3)^33t-3t+2t^3to Lin mathbbRsetminus0$
answered Jul 23 at 7:44
gimusi
65.2k73583
What I meant by 3x+2y=0 is that the function tends to + or -infinity near those points, so I assume the delta-epsilon definition would fail. Anyway, thank you very much. This helped a lot.
– Manuel
Jul 23 at 8:08
@Manuel It is not a reason to conclude that the limit doesn't exist. What about, for example, for $lim_(x,y)to (0,0) fracsin xyxy$?
– gimusi
Jul 23 at 8:10
@Manuel Note also that, usually, when we refer to the existence of the limit we mean that the limit is the same (finite or infinite) for any path. Therefore to show that the limit doesn't exist it suffices to find (at least) two different paths with different limits.
– gimusi
Jul 23 at 8:11
I'd say your example is a different situation, because at the points where the denominator = 0, your function approaches 1.
– Manuel
Jul 23 at 8:21
@Manuel And what about $lim_(x,y)to (0,0) frac1x^2y^2$?
– gimusi
Jul 23 at 8:26
 |Â
show 2 more comments
up vote
1
down vote
Note that the points such that $3x+2y=0$ are simply excluded from the domain and thus we can't conclude that the limit doesn't exist from here.
To show that we need to find two different paths with different limits as $(x,y)to (0,0)$.
Notably in this case we have that
for $x=y$ we have $dfracx^3-y^33x+2y=0$
for $x=tto 0$ and $y=-frac32t+t^3$ we have $=dfracx^3-y^33x+2y=dfract^3-(-frac32t+t^3)^33t-3t+2t^3to Lin mathbbRsetminus0$
answered Jul 23 at 7:44
gimusi
65.2k73583
What I meant by 3x+2y=0 is that the function tends to + or -infinity near those points, so I assume the delta-epsilon definition would fail. Anyway, thank you very much. This helped a lot.
– Manuel
Jul 23 at 8:08
@Manuel It is not a reason to conclude that the limit doesn't exist. What about, for example, for $lim_(x,y)to (0,0) fracsin xyxy$?
– gimusi
Jul 23 at 8:10
@Manuel Note also that, usually, when we refer to the existence of the limit we mean that the limit is the same (finite or infinite) for any path. Therefore to show that the limit doesn't exist it suffices to find (at least) two different paths with different limits.
– gimusi
Jul 23 at 8:11
I'd say your example is a different situation, because at the points where the denominator = 0, your function approaches 1.
– Manuel
Jul 23 at 8:21
@Manuel And what about $lim_(x,y)to (0,0) frac1x^2y^2$?
– gimusi
Jul 23 at 8:26
 |Â
show 2 more comments
up vote
1
down vote
up vote
1
down vote
Note that the points such that $3x+2y=0$ are simply excluded from the domain and thus we can't conclude that the limit doesn't exist from here.
To show that we need to find two different paths with different limits as $(x,y)to (0,0)$.
Notably in this case we have that
for $x=y$ we have $dfracx^3-y^33x+2y=0$
for $x=tto 0$ and $y=-frac32t+t^3$ we have $=dfracx^3-y^33x+2y=dfract^3-(-frac32t+t^3)^33t-3t+2t^3to Lin mathbbRsetminus0$
answered Jul 23 at 7:44
gimusi
65.2k73583
Note that the points such that $3x+2y=0$ are simply excluded from the domain and thus we can't conclude that the limit doesn't exist from here.
To show that we need to find two different paths with different limits as $(x,y)to (0,0)$.
Notably in this case we have that
for $x=y$ we have $dfracx^3-y^33x+2y=0$
for $x=tto 0$ and $y=-frac32t+t^3$ we have $=dfracx^3-y^33x+2y=dfract^3-(-frac32t+t^3)^33t-3t+2t^3to Lin mathbbRsetminus0$
answered Jul 23 at 7:44
gimusi
65.2k73583
edited Jul 23 at 7:50
answered Jul 23 at 7:44
gimusi
65.2k73583
answered Jul 23 at 7:44
gimusi
65.2k73583
answered Jul 23 at 7:44
gimusi
65.2k73583
65.2k73583
What I meant by 3x+2y=0 is that the function tends to + or -infinity near those points, so I assume the delta-epsilon definition would fail. Anyway, thank you very much. This helped a lot.
– Manuel
Jul 23 at 8:08
@Manuel It is not a reason to conclude that the limit doesn't exist. What about, for example, for $lim_(x,y)to (0,0) fracsin xyxy$?
– gimusi
Jul 23 at 8:10
@Manuel Note also that, usually, when we refer to the existence of the limit we mean that the limit is the same (finite or infinite) for any path. Therefore to show that the limit doesn't exist it suffices to find (at least) two different paths with different limits.
– gimusi
Jul 23 at 8:11
I'd say your example is a different situation, because at the points where the denominator = 0, your function approaches 1.
– Manuel
Jul 23 at 8:21
@Manuel And what about $lim_(x,y)to (0,0) frac1x^2y^2$?
– gimusi
Jul 23 at 8:26
 |Â
show 2 more comments
What I meant by 3x+2y=0 is that the function tends to + or -infinity near those points, so I assume the delta-epsilon definition would fail. Anyway, thank you very much. This helped a lot.
– Manuel
Jul 23 at 8:08
@Manuel It is not a reason to conclude that the limit doesn't exist. What about, for example, for $lim_(x,y)to (0,0) fracsin xyxy$?
– gimusi
Jul 23 at 8:10
@Manuel Note also that, usually, when we refer to the existence of the limit we mean that the limit is the same (finite or infinite) for any path. Therefore to show that the limit doesn't exist it suffices to find (at least) two different paths with different limits.
– gimusi
Jul 23 at 8:11
I'd say your example is a different situation, because at the points where the denominator = 0, your function approaches 1.
– Manuel
Jul 23 at 8:21
@Manuel And what about $lim_(x,y)to (0,0) frac1x^2y^2$?
– gimusi
Jul 23 at 8:26
What I meant by 3x+2y=0 is that the function tends to + or -infinity near those points, so I assume the delta-epsilon definition would fail. Anyway, thank you very much. This helped a lot.
– Manuel
Jul 23 at 8:08
What I meant by 3x+2y=0 is that the function tends to + or -infinity near those points, so I assume the delta-epsilon definition would fail. Anyway, thank you very much. This helped a lot.
– Manuel
Jul 23 at 8:08
@Manuel It is not a reason to conclude that the limit doesn't exist. What about, for example, for $lim_(x,y)to (0,0) fracsin xyxy$?
– gimusi
Jul 23 at 8:10
@Manuel It is not a reason to conclude that the limit doesn't exist. What about, for example, for $lim_(x,y)to (0,0) fracsin xyxy$?
– gimusi
Jul 23 at 8:10
@Manuel Note also that, usually, when we refer to the existence of the limit we mean that the limit is the same (finite or infinite) for any path. Therefore to show that the limit doesn't exist it suffices to find (at least) two different paths with different limits.
– gimusi
Jul 23 at 8:11
@Manuel Note also that, usually, when we refer to the existence of the limit we mean that the limit is the same (finite or infinite) for any path. Therefore to show that the limit doesn't exist it suffices to find (at least) two different paths with different limits.
– gimusi
Jul 23 at 8:11
I'd say your example is a different situation, because at the points where the denominator = 0, your function approaches 1.
– Manuel
Jul 23 at 8:21
I'd say your example is a different situation, because at the points where the denominator = 0, your function approaches 1.
– Manuel
Jul 23 at 8:21
@Manuel And what about $lim_(x,y)to (0,0) frac1x^2y^2$?
– gimusi
Jul 23 at 8:26
@Manuel And what about $lim_(x,y)to (0,0) frac1x^2y^2$?
– gimusi
Jul 23 at 8:26
 |Â
show 2 more comments
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The trick in these cases, when we want to show that the limit doesn't exist, is to approach along a path such that the main part of $3x+2y$ is equal to zero but we keep some remainder which determine the value for the denominator in such way that the limit takes a different values. In this case we can choose for $x=t to 0$and then $y=-frac32 t+t^n$. We need to choose $n$ in such way to obtain a different limit from the trivail one obtained for the path with $x=y$.
– gimusi
Jul 23 at 8:05
Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 10 at 23:26