How to prove $lim_{(x,y) to (0,0)} frac{xy}{x^4+y^4}$ does not exist
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$$lim_{(x,y)to(0,0)} frac{xy}{x^4+y^4}$$
Anyone can teach me how to solve this question? I have tried so many times but still unable to solve it.
calculus limits
edited Dec 2 at 4:20
Tianlalu
2,9451936
asked Dec 2 at 3:35
Alvin
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up vote
1
down vote
favorite
$$lim_{(x,y)to(0,0)} frac{xy}{x^4+y^4}$$
Anyone can teach me how to solve this question? I have tried so many times but still unable to solve it.
calculus limits
edited Dec 2 at 4:20
Tianlalu
2,9451936
asked Dec 2 at 3:35
Alvin
61
New contributor
Alvin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
Work out what happens as $xto0$ with $y$ held fixed at zero. Then work out what happens as you approach the origin along the curve $y=x^4$. Then write up what you have found, and post it as an answer.
– Gerry Myerson
Dec 2 at 3:39
Try to choose a suitable path depending on some variable constant.
– Anik Bhowmick
Dec 2 at 4:00
Well the fact that the expression has transposable variables (i.e. $f(x,y) equiv f(y,x)$) means that you just need to find $lim_{xto 0} f(x,x)$, or more precisely: $$lim_{xto 0}{2x^{-2}}$$ I think
– Rhys Hughes
Dec 2 at 4:21
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$$lim_{(x,y)to(0,0)} frac{xy}{x^4+y^4}$$
Anyone can teach me how to solve this question? I have tried so many times but still unable to solve it.
calculus limits
edited Dec 2 at 4:20
Tianlalu
2,9451936
asked Dec 2 at 3:35
Alvin
61
New contributor
Alvin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$$lim_{(x,y)to(0,0)} frac{xy}{x^4+y^4}$$
Anyone can teach me how to solve this question? I have tried so many times but still unable to solve it.
calculus limits
calculus limits
edited Dec 2 at 4:20
Tianlalu
2,9451936
asked Dec 2 at 3:35
Alvin
61
New contributor
Alvin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 2 at 4:20
Tianlalu
2,9451936
asked Dec 2 at 3:35
Alvin
61
New contributor
Alvin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 2 at 4:20
Tianlalu
2,9451936
edited Dec 2 at 4:20
Tianlalu
2,9451936
edited Dec 2 at 4:20
Tianlalu
2,9451936
2,9451936
asked Dec 2 at 3:35
Alvin
61
New contributor
Alvin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 2 at 3:35
Alvin
61
asked Dec 2 at 3:35
Alvin
61
61
New contributor
Alvin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Alvin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Alvin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
Work out what happens as $xto0$ with $y$ held fixed at zero. Then work out what happens as you approach the origin along the curve $y=x^4$. Then write up what you have found, and post it as an answer.
– Gerry Myerson
Dec 2 at 3:39
Try to choose a suitable path depending on some variable constant.
– Anik Bhowmick
Dec 2 at 4:00
Well the fact that the expression has transposable variables (i.e. $f(x,y) equiv f(y,x)$) means that you just need to find $lim_{xto 0} f(x,x)$, or more precisely: $$lim_{xto 0}{2x^{-2}}$$ I think
– Rhys Hughes
Dec 2 at 4:21
add a comment |
3
Work out what happens as $xto0$ with $y$ held fixed at zero. Then work out what happens as you approach the origin along the curve $y=x^4$. Then write up what you have found, and post it as an answer.
– Gerry Myerson
Dec 2 at 3:39
Try to choose a suitable path depending on some variable constant.
– Anik Bhowmick
Dec 2 at 4:00
Well the fact that the expression has transposable variables (i.e. $f(x,y) equiv f(y,x)$) means that you just need to find $lim_{xto 0} f(x,x)$, or more precisely: $$lim_{xto 0}{2x^{-2}}$$ I think
– Rhys Hughes
Dec 2 at 4:21
3
3
Work out what happens as $xto0$ with $y$ held fixed at zero. Then work out what happens as you approach the origin along the curve $y=x^4$. Then write up what you have found, and post it as an answer.
– Gerry Myerson
Dec 2 at 3:39
Work out what happens as $xto0$ with $y$ held fixed at zero. Then work out what happens as you approach the origin along the curve $y=x^4$. Then write up what you have found, and post it as an answer.
– Gerry Myerson
Dec 2 at 3:39
Try to choose a suitable path depending on some variable constant.
– Anik Bhowmick
Dec 2 at 4:00
Try to choose a suitable path depending on some variable constant.
– Anik Bhowmick
Dec 2 at 4:00
Well the fact that the expression has transposable variables (i.e. $f(x,y) equiv f(y,x)$) means that you just need to find $lim_{xto 0} f(x,x)$, or more precisely: $$lim_{xto 0}{2x^{-2}}$$ I think
– Rhys Hughes
Dec 2 at 4:21
Well the fact that the expression has transposable variables (i.e. $f(x,y) equiv f(y,x)$) means that you just need to find $lim_{xto 0} f(x,x)$, or more precisely: $$lim_{xto 0}{2x^{-2}}$$ I think
– Rhys Hughes
Dec 2 at 4:21
add a comment |
1 Answer
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Suppose the limit $lim_{(x,y)to(0,0)} frac{xy}{x^4+y^4}$ exits then in some deleted nbd of $(0,0)$ the function $f:Bbb R^2-{(0,0)}rightarrow Bbb R,f(x,y)=frac{xy}{x^4+y^4}$ will be bounded . But notice that on the set $S:={(x,y)in Bbb R^2-{(0,0)}: x=y}$ the function $f$ is of the form $frac{x^2}{2x^4}=frac{1}{2x^2}$ i.e. in every deleted nbd of $(0,0)$ the function is unbounded.
$lim_{(x,y)rightarrow (0,0)} f(x,y)=limplies$ for $epsilon=1$ there is a $delta>0$ such that $0<x^2+y^2<deltaimplies |f(x,y)-l|<1implies l-1<f(x,y)<l+1 , forall (x,y)$ with $0<x^2+y^2<delta$ i.e. $f$ is bounded in a deleted nbd of $(0,0)$.
answered Dec 2 at 4:35
UserS
1,441112
add a comment |
1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
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up vote
1
down vote
Suppose the limit $lim_{(x,y)to(0,0)} frac{xy}{x^4+y^4}$ exits then in some deleted nbd of $(0,0)$ the function $f:Bbb R^2-{(0,0)}rightarrow Bbb R,f(x,y)=frac{xy}{x^4+y^4}$ will be bounded . But notice that on the set $S:={(x,y)in Bbb R^2-{(0,0)}: x=y}$ the function $f$ is of the form $frac{x^2}{2x^4}=frac{1}{2x^2}$ i.e. in every deleted nbd of $(0,0)$ the function is unbounded.
$lim_{(x,y)rightarrow (0,0)} f(x,y)=limplies$ for $epsilon=1$ there is a $delta>0$ such that $0<x^2+y^2<deltaimplies |f(x,y)-l|<1implies l-1<f(x,y)<l+1 , forall (x,y)$ with $0<x^2+y^2<delta$ i.e. $f$ is bounded in a deleted nbd of $(0,0)$.
answered Dec 2 at 4:35
UserS
1,441112
add a comment |
up vote
1
down vote
Suppose the limit $lim_{(x,y)to(0,0)} frac{xy}{x^4+y^4}$ exits then in some deleted nbd of $(0,0)$ the function $f:Bbb R^2-{(0,0)}rightarrow Bbb R,f(x,y)=frac{xy}{x^4+y^4}$ will be bounded . But notice that on the set $S:={(x,y)in Bbb R^2-{(0,0)}: x=y}$ the function $f$ is of the form $frac{x^2}{2x^4}=frac{1}{2x^2}$ i.e. in every deleted nbd of $(0,0)$ the function is unbounded.
$lim_{(x,y)rightarrow (0,0)} f(x,y)=limplies$ for $epsilon=1$ there is a $delta>0$ such that $0<x^2+y^2<deltaimplies |f(x,y)-l|<1implies l-1<f(x,y)<l+1 , forall (x,y)$ with $0<x^2+y^2<delta$ i.e. $f$ is bounded in a deleted nbd of $(0,0)$.
answered Dec 2 at 4:35
UserS
1,441112
add a comment |
up vote
1
down vote
up vote
1
down vote
Suppose the limit $lim_{(x,y)to(0,0)} frac{xy}{x^4+y^4}$ exits then in some deleted nbd of $(0,0)$ the function $f:Bbb R^2-{(0,0)}rightarrow Bbb R,f(x,y)=frac{xy}{x^4+y^4}$ will be bounded . But notice that on the set $S:={(x,y)in Bbb R^2-{(0,0)}: x=y}$ the function $f$ is of the form $frac{x^2}{2x^4}=frac{1}{2x^2}$ i.e. in every deleted nbd of $(0,0)$ the function is unbounded.
$lim_{(x,y)rightarrow (0,0)} f(x,y)=limplies$ for $epsilon=1$ there is a $delta>0$ such that $0<x^2+y^2<deltaimplies |f(x,y)-l|<1implies l-1<f(x,y)<l+1 , forall (x,y)$ with $0<x^2+y^2<delta$ i.e. $f$ is bounded in a deleted nbd of $(0,0)$.
answered Dec 2 at 4:35
UserS
1,441112
Suppose the limit $lim_{(x,y)to(0,0)} frac{xy}{x^4+y^4}$ exits then in some deleted nbd of $(0,0)$ the function $f:Bbb R^2-{(0,0)}rightarrow Bbb R,f(x,y)=frac{xy}{x^4+y^4}$ will be bounded . But notice that on the set $S:={(x,y)in Bbb R^2-{(0,0)}: x=y}$ the function $f$ is of the form $frac{x^2}{2x^4}=frac{1}{2x^2}$ i.e. in every deleted nbd of $(0,0)$ the function is unbounded.
$lim_{(x,y)rightarrow (0,0)} f(x,y)=limplies$ for $epsilon=1$ there is a $delta>0$ such that $0<x^2+y^2<deltaimplies |f(x,y)-l|<1implies l-1<f(x,y)<l+1 , forall (x,y)$ with $0<x^2+y^2<delta$ i.e. $f$ is bounded in a deleted nbd of $(0,0)$.
answered Dec 2 at 4:35
UserS
1,441112
edited Dec 2 at 4:45
answered Dec 2 at 4:35
UserS
1,441112
answered Dec 2 at 4:35
UserS
1,441112
answered Dec 2 at 4:35
UserS
1,441112
1,441112
add a comment |
add a comment |
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3
Work out what happens as $xto0$ with $y$ held fixed at zero. Then work out what happens as you approach the origin along the curve $y=x^4$. Then write up what you have found, and post it as an answer.
– Gerry Myerson
Dec 2 at 3:39
Try to choose a suitable path depending on some variable constant.
– Anik Bhowmick
Dec 2 at 4:00
Well the fact that the expression has transposable variables (i.e. $f(x,y) equiv f(y,x)$) means that you just need to find $lim_{xto 0} f(x,x)$, or more precisely: $$lim_{xto 0}{2x^{-2}}$$ I think
– Rhys Hughes
Dec 2 at 4:21