Evaluate $lim_(x,y) to (0,0), x+y neq 0fracln(1-x-y)x+y $ The 2019 Stack Overflow Developer Survey Results Are InLimits for multivariate functions $displaystyle lim_(x,y) to (0,0)fracxy^2x^2+y^4$Multivariate limit $lim_(x,y) to (0,0) fracxy^2x^2 + y^4 = 0$Finding $lim_(x,y)to (0,0) frac3x^2sin^2y2x^4+2sin y^4$$lim_(x,y) rightarrow (0,0) fracxyx$ when $(x,y)neq (0,0)$ and $f(0,0)=0$.Limit $lim_(x,y)to (0,0)fracsin(xy)xy$Evaluate $lim_(x,y)to (0,0)fracx+sin yx+y$Tricky limits question - If $lim_(x,y) rightarrow(0,0) f(0,y)=0$ then $lim_(x,y) rightarrow(0,0) f(x,y)=0$. True or False?Computing the limit or showing that the limit does not exist: $lim_(x,y)to(0,0)frac x ^ 2 + 3 x - 4 y x - y$Show that the limit does not exist $lim_(x, y) to (0,0)frac5x^2x^2 + y^2$
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Evaluate $lim_(x,y) to (0,0), x+y neq 0fracln(1-x-y)x+y $
The 2019 Stack Overflow Developer Survey Results Are InLimits for multivariate functions $displaystyle lim_(x,y) to (0,0)fracxy^2x^2+y^4}$Multivariate limit $lim_(x,y) to (0,0) frac{xy^2x^2 + y^4 = 0$Finding $lim_(x,y)to (0,0) frac3x^2sin^2y2x^4+2sin y^4$$lim_(x,y) rightarrow (0,0) fracxy$$f(x,y)= e^-frac1x^2+y^2$ when $(x,y)neq (0,0)$ and $f(0,0)=0$.Limit $lim_(x,y)to (0,0)fracsin(xy)xy$Evaluate $lim_(x,y)to (0,0)fracx+sin yx+y$Tricky limits question - If $lim_(x,y) rightarrow(0,0) f(0,y)=0$ then $lim_(x,y) rightarrow(0,0) f(x,y)=0$. True or False?Computing the limit or showing that the limit does not exist: $lim_(x,y)to(0,0)frac x ^ 2 + 3 x - 4 y x - y$Show that the limit does not exist $lim_(x, y) to (0,0)frac5x^2x^2 + y^2$
$begingroup$
This was the question of a test. My question is if my attempt to solve it is correct, and if it is, why is it correct.
$$lim_(x,y) to (0,0), x+y neq 0fracln(1-x-y)x+y $$
My attempt:
Let $xi = -x-y $. Then $xi to 0$ whenever $(x,y) to (0,0)$ and $x+y neq 0 iff xi neq 0$. (Is it then correct to say that the previous limit exists and is equal to the following iff the following exists? And why?):
$$lim_xi to 0, xi neq 0fracln(1+xi )-xi$$
If it is correct, then the limit exists and is $-1$. If it is correct, why is it correct?
limits multivariable-calculus proof-verification
edited Apr 6 at 17:09
StubbornAtom
6,37831440
asked Apr 6 at 16:57
RUBEN GONÇALO MOROUÇORUBEN GONÇALO MOROUÇO
895
$endgroup$
|
show 2 more comments
$begingroup$
This was the question of a test. My question is if my attempt to solve it is correct, and if it is, why is it correct.
$$lim_(x,y) to (0,0), x+y neq 0fracln(1-x-y)x+y $$
My attempt:
Let $xi = -x-y $. Then $xi to 0$ whenever $(x,y) to (0,0)$ and $x+y neq 0 iff xi neq 0$. (Is it then correct to say that the previous limit exists and is equal to the following iff the following exists? And why?):
$$lim_xi to 0, xi neq 0fracln(1+xi )-xi$$
If it is correct, then the limit exists and is $-1$. If it is correct, why is it correct?
limits multivariable-calculus proof-verification
edited Apr 6 at 17:09
StubbornAtom
6,37831440
asked Apr 6 at 16:57
RUBEN GONÇALO MOROUÇORUBEN GONÇALO MOROUÇO
895
$endgroup$
$begingroup$
If you can show from the $epsilon-delta$-definition (take $x^2+y^2<delta$) that the function is continuous, then it means that the limit exists and is unique from any path.
$endgroup$
– D.B.
Apr 6 at 17:20
$begingroup$
@D.B. the function is elementary, so I can assume it is continuous in it's domain.. however it is undefined in the point $(0,0)$.
$endgroup$
– RUBEN GONÇALO MOROUÇO
Apr 6 at 17:30
$begingroup$
Right, it may not be continuous at $0$.
$endgroup$
– D.B.
Apr 6 at 17:33
$begingroup$
I haven't had Calculus III since 2008, but from what I recall from 2 variable functions like this, you can try to graph them to visually find the answer. Just go to wolframalpha.com and input "3d plot Log[1-x-y]/(x+y)" (without quotes) into the search bar. You will find that as x approaches 0, z = f(x,y) approaches 1. As y approaches 0, z approaches negative infinity. So the limit probably does not exist.
$endgroup$
– Christopher Mowla
Apr 6 at 21:04
$begingroup$
@ChristopherMowla curiously if I ask wolfram to evaluate that limit it says it's -1.. But wolfram alpha is not reliable for multivariable limits. It often lies. Although so, it would be strange if it didn't exist, since the question explicity tells me to calculate it's value. And I don't think they'd put limits that are too hard to solve in the test.
$endgroup$
– RUBEN GONÇALO MOROUÇO
Apr 6 at 21:13
|
show 2 more comments
$begingroup$
This was the question of a test. My question is if my attempt to solve it is correct, and if it is, why is it correct.
$$lim_(x,y) to (0,0), x+y neq 0fracln(1-x-y)x+y $$
My attempt:
Let $xi = -x-y $. Then $xi to 0$ whenever $(x,y) to (0,0)$ and $x+y neq 0 iff xi neq 0$. (Is it then correct to say that the previous limit exists and is equal to the following iff the following exists? And why?):
$$lim_xi to 0, xi neq 0fracln(1+xi )-xi$$
If it is correct, then the limit exists and is $-1$. If it is correct, why is it correct?
limits multivariable-calculus proof-verification
edited Apr 6 at 17:09
StubbornAtom
6,37831440
asked Apr 6 at 16:57
RUBEN GONÇALO MOROUÇORUBEN GONÇALO MOROUÇO
895
$endgroup$
This was the question of a test. My question is if my attempt to solve it is correct, and if it is, why is it correct.
$$lim_(x,y) to (0,0), x+y neq 0fracln(1-x-y)x+y $$
My attempt:
Let $xi = -x-y $. Then $xi to 0$ whenever $(x,y) to (0,0)$ and $x+y neq 0 iff xi neq 0$. (Is it then correct to say that the previous limit exists and is equal to the following iff the following exists? And why?):
$$lim_xi to 0, xi neq 0fracln(1+xi )-xi$$
If it is correct, then the limit exists and is $-1$. If it is correct, why is it correct?
limits multivariable-calculus proof-verification
limits multivariable-calculus proof-verification
edited Apr 6 at 17:09
StubbornAtom
6,37831440
asked Apr 6 at 16:57
RUBEN GONÇALO MOROUÇORUBEN GONÇALO MOROUÇO
895
edited Apr 6 at 17:09
StubbornAtom
6,37831440
asked Apr 6 at 16:57
RUBEN GONÇALO MOROUÇORUBEN GONÇALO MOROUÇO
895
edited Apr 6 at 17:09
StubbornAtom
6,37831440
edited Apr 6 at 17:09
StubbornAtom
6,37831440
edited Apr 6 at 17:09
StubbornAtom
6,37831440
6,37831440
asked Apr 6 at 16:57
RUBEN GONÇALO MOROUÇORUBEN GONÇALO MOROUÇO
895
asked Apr 6 at 16:57
RUBEN GONÇALO MOROUÇORUBEN GONÇALO MOROUÇO
895
asked Apr 6 at 16:57
RUBEN GONÇALO MOROUÇORUBEN GONÇALO MOROUÇO
895
895
$begingroup$
If you can show from the $epsilon-delta$-definition (take $x^2+y^2<delta$) that the function is continuous, then it means that the limit exists and is unique from any path.
$endgroup$
– D.B.
Apr 6 at 17:20
$begingroup$
@D.B. the function is elementary, so I can assume it is continuous in it's domain.. however it is undefined in the point $(0,0)$.
$endgroup$
– RUBEN GONÇALO MOROUÇO
Apr 6 at 17:30
$begingroup$
Right, it may not be continuous at $0$.
$endgroup$
– D.B.
Apr 6 at 17:33
$begingroup$
I haven't had Calculus III since 2008, but from what I recall from 2 variable functions like this, you can try to graph them to visually find the answer. Just go to wolframalpha.com and input "3d plot Log[1-x-y]/(x+y)" (without quotes) into the search bar. You will find that as x approaches 0, z = f(x,y) approaches 1. As y approaches 0, z approaches negative infinity. So the limit probably does not exist.
$endgroup$
– Christopher Mowla
Apr 6 at 21:04
$begingroup$
@ChristopherMowla curiously if I ask wolfram to evaluate that limit it says it's -1.. But wolfram alpha is not reliable for multivariable limits. It often lies. Although so, it would be strange if it didn't exist, since the question explicity tells me to calculate it's value. And I don't think they'd put limits that are too hard to solve in the test.
$endgroup$
– RUBEN GONÇALO MOROUÇO
Apr 6 at 21:13
|
show 2 more comments
$begingroup$
If you can show from the $epsilon-delta$-definition (take $x^2+y^2<delta$) that the function is continuous, then it means that the limit exists and is unique from any path.
$endgroup$
– D.B.
Apr 6 at 17:20
$begingroup$
@D.B. the function is elementary, so I can assume it is continuous in it's domain.. however it is undefined in the point $(0,0)$.
$endgroup$
– RUBEN GONÇALO MOROUÇO
Apr 6 at 17:30
$begingroup$
Right, it may not be continuous at $0$.
$endgroup$
– D.B.
Apr 6 at 17:33
$begingroup$
I haven't had Calculus III since 2008, but from what I recall from 2 variable functions like this, you can try to graph them to visually find the answer. Just go to wolframalpha.com and input "3d plot Log[1-x-y]/(x+y)" (without quotes) into the search bar. You will find that as x approaches 0, z = f(x,y) approaches 1. As y approaches 0, z approaches negative infinity. So the limit probably does not exist.
$endgroup$
– Christopher Mowla
Apr 6 at 21:04
$begingroup$
@ChristopherMowla curiously if I ask wolfram to evaluate that limit it says it's -1.. But wolfram alpha is not reliable for multivariable limits. It often lies. Although so, it would be strange if it didn't exist, since the question explicity tells me to calculate it's value. And I don't think they'd put limits that are too hard to solve in the test.
$endgroup$
– RUBEN GONÇALO MOROUÇO
Apr 6 at 21:13
$begingroup$
If you can show from the $epsilon-delta$-definition (take $x^2+y^2<delta$) that the function is continuous, then it means that the limit exists and is unique from any path.
$endgroup$
– D.B.
Apr 6 at 17:20
$begingroup$
If you can show from the $epsilon-delta$-definition (take $x^2+y^2<delta$) that the function is continuous, then it means that the limit exists and is unique from any path.
$endgroup$
– D.B.
Apr 6 at 17:20
$begingroup$
@D.B. the function is elementary, so I can assume it is continuous in it's domain.. however it is undefined in the point $(0,0)$.
$endgroup$
– RUBEN GONÇALO MOROUÇO
Apr 6 at 17:30
$begingroup$
@D.B. the function is elementary, so I can assume it is continuous in it's domain.. however it is undefined in the point $(0,0)$.
$endgroup$
– RUBEN GONÇALO MOROUÇO
Apr 6 at 17:30
$begingroup$
Right, it may not be continuous at $0$.
$endgroup$
– D.B.
Apr 6 at 17:33
$begingroup$
Right, it may not be continuous at $0$.
$endgroup$
– D.B.
Apr 6 at 17:33
$begingroup$
I haven't had Calculus III since 2008, but from what I recall from 2 variable functions like this, you can try to graph them to visually find the answer. Just go to wolframalpha.com and input "3d plot Log[1-x-y]/(x+y)" (without quotes) into the search bar. You will find that as x approaches 0, z = f(x,y) approaches 1. As y approaches 0, z approaches negative infinity. So the limit probably does not exist.
$endgroup$
– Christopher Mowla
Apr 6 at 21:04
$begingroup$
I haven't had Calculus III since 2008, but from what I recall from 2 variable functions like this, you can try to graph them to visually find the answer. Just go to wolframalpha.com and input "3d plot Log[1-x-y]/(x+y)" (without quotes) into the search bar. You will find that as x approaches 0, z = f(x,y) approaches 1. As y approaches 0, z approaches negative infinity. So the limit probably does not exist.
$endgroup$
– Christopher Mowla
Apr 6 at 21:04
$begingroup$
@ChristopherMowla curiously if I ask wolfram to evaluate that limit it says it's -1.. But wolfram alpha is not reliable for multivariable limits. It often lies. Although so, it would be strange if it didn't exist, since the question explicity tells me to calculate it's value. And I don't think they'd put limits that are too hard to solve in the test.
$endgroup$
– RUBEN GONÇALO MOROUÇO
Apr 6 at 21:13
$begingroup$
@ChristopherMowla curiously if I ask wolfram to evaluate that limit it says it's -1.. But wolfram alpha is not reliable for multivariable limits. It often lies. Although so, it would be strange if it didn't exist, since the question explicity tells me to calculate it's value. And I don't think they'd put limits that are too hard to solve in the test.
$endgroup$
– RUBEN GONÇALO MOROUÇO
Apr 6 at 21:13
|
show 2 more comments
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$begingroup$
If you can show from the $epsilon-delta$-definition (take $x^2+y^2<delta$) that the function is continuous, then it means that the limit exists and is unique from any path.
$endgroup$
– D.B.
Apr 6 at 17:20
$begingroup$
@D.B. the function is elementary, so I can assume it is continuous in it's domain.. however it is undefined in the point $(0,0)$.
$endgroup$
– RUBEN GONÇALO MOROUÇO
Apr 6 at 17:30
$begingroup$
Right, it may not be continuous at $0$.
$endgroup$
– D.B.
Apr 6 at 17:33
$begingroup$
I haven't had Calculus III since 2008, but from what I recall from 2 variable functions like this, you can try to graph them to visually find the answer. Just go to wolframalpha.com and input "3d plot Log[1-x-y]/(x+y)" (without quotes) into the search bar. You will find that as x approaches 0, z = f(x,y) approaches 1. As y approaches 0, z approaches negative infinity. So the limit probably does not exist.
$endgroup$
– Christopher Mowla
Apr 6 at 21:04
$begingroup$
@ChristopherMowla curiously if I ask wolfram to evaluate that limit it says it's -1.. But wolfram alpha is not reliable for multivariable limits. It often lies. Although so, it would be strange if it didn't exist, since the question explicity tells me to calculate it's value. And I don't think they'd put limits that are too hard to solve in the test.
$endgroup$
– RUBEN GONÇALO MOROUÇO
Apr 6 at 21:13