Mixing
Discontinuity of multivariate functions
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Let $f(x,y)=frac2(x^3+y^3)x^2+2y$ , when $(x,y) neq (0,0)$ and $f(x,y)=0 $ when $(x,y)=(0,0)$
We are required to prove the discontinuity of this function at $(0,0)$.
So, I put $y=mx$ where $x rightarrow 0$, to get the limit to be $0$.
But I can't find another instance where the limiti is different.
Can you guys help?
Thanks
calculus real-analysis multivariable-calculus continuity
asked 2 hours ago
Legend Killer
1,510523
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up vote
2
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Let $f(x,y)=frac2(x^3+y^3)x^2+2y$ , when $(x,y) neq (0,0)$ and $f(x,y)=0 $ when $(x,y)=(0,0)$
We are required to prove the discontinuity of this function at $(0,0)$.
So, I put $y=mx$ where $x rightarrow 0$, to get the limit to be $0$.
But I can't find another instance where the limiti is different.
Can you guys help?
Thanks
calculus real-analysis multivariable-calculus continuity
asked 2 hours ago
Legend Killer
1,510523
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $f(x,y)=frac2(x^3+y^3)x^2+2y$ , when $(x,y) neq (0,0)$ and $f(x,y)=0 $ when $(x,y)=(0,0)$
We are required to prove the discontinuity of this function at $(0,0)$.
So, I put $y=mx$ where $x rightarrow 0$, to get the limit to be $0$.
But I can't find another instance where the limiti is different.
Can you guys help?
Thanks
calculus real-analysis multivariable-calculus continuity
asked 2 hours ago
Legend Killer
1,510523
Let $f(x,y)=frac2(x^3+y^3)x^2+2y$ , when $(x,y) neq (0,0)$ and $f(x,y)=0 $ when $(x,y)=(0,0)$
We are required to prove the discontinuity of this function at $(0,0)$.
So, I put $y=mx$ where $x rightarrow 0$, to get the limit to be $0$.
But I can't find another instance where the limiti is different.
Can you guys help?
Thanks
calculus real-analysis multivariable-calculus continuity
calculus real-analysis multivariable-calculus continuity
asked 2 hours ago
Legend Killer
1,510523
asked 2 hours ago
Legend Killer
1,510523
asked 2 hours ago
Legend Killer
1,510523
asked 2 hours ago
Legend Killer
1,510523
asked 2 hours ago
Legend Killer
1,510523
1,510523
add a comment |Â
add a comment |Â
2 Answers
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up vote
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Credit: Carmeister's solution.
I am just curious about the behavior of other similar path.
Let $y = -fracx^22+x^n$ where $n >0$,then we have
beginalignlim_x to 0frac2(x^3+(-fracx^22+x^n)^3)2x^n &= lim_x to 0 frac(x^3+(-fracx^22+x^n)^3)x^n \
&= lim_x to 0 left[ x^3-n+left(-fracx^2+fracn32+x^frac4n3 right)^3right]endalign
If $n in (0,3)$, the limit is $0$.
If $n>3$, it is undefined.
If $n=3$, beginalign lim_x to 0 left[ x^3-n+left(-fracx^2+fracn32+x^frac4n3 right)^3right]=1endalign
answered 1 hour ago
Siong Thye Goh
83.5k1456104
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up vote
2
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Note that $f$ is undefined when $y=-x^2/2$, and if $y$ is very close to $-x^2/2$ then $f$ gets very large. So we should try something like $y=-x^2/2+x^3$.
answered 2 hours ago
Carmeister
2,3291720
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Credit: Carmeister's solution.
I am just curious about the behavior of other similar path.
Let $y = -fracx^22+x^n$ where $n >0$,then we have
beginalignlim_x to 0frac2(x^3+(-fracx^22+x^n)^3)2x^n &= lim_x to 0 frac(x^3+(-fracx^22+x^n)^3)x^n \
&= lim_x to 0 left[ x^3-n+left(-fracx^2+fracn32+x^frac4n3 right)^3right]endalign
If $n in (0,3)$, the limit is $0$.
If $n>3$, it is undefined.
If $n=3$, beginalign lim_x to 0 left[ x^3-n+left(-fracx^2+fracn32+x^frac4n3 right)^3right]=1endalign
answered 1 hour ago
Siong Thye Goh
83.5k1456104
add a comment |Â
up vote
4
down vote
Credit: Carmeister's solution.
I am just curious about the behavior of other similar path.
Let $y = -fracx^22+x^n$ where $n >0$,then we have
beginalignlim_x to 0frac2(x^3+(-fracx^22+x^n)^3)2x^n &= lim_x to 0 frac(x^3+(-fracx^22+x^n)^3)x^n \
&= lim_x to 0 left[ x^3-n+left(-fracx^2+fracn32+x^frac4n3 right)^3right]endalign
If $n in (0,3)$, the limit is $0$.
If $n>3$, it is undefined.
If $n=3$, beginalign lim_x to 0 left[ x^3-n+left(-fracx^2+fracn32+x^frac4n3 right)^3right]=1endalign
answered 1 hour ago
Siong Thye Goh
83.5k1456104
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Credit: Carmeister's solution.
I am just curious about the behavior of other similar path.
Let $y = -fracx^22+x^n$ where $n >0$,then we have
beginalignlim_x to 0frac2(x^3+(-fracx^22+x^n)^3)2x^n &= lim_x to 0 frac(x^3+(-fracx^22+x^n)^3)x^n \
&= lim_x to 0 left[ x^3-n+left(-fracx^2+fracn32+x^frac4n3 right)^3right]endalign
If $n in (0,3)$, the limit is $0$.
If $n>3$, it is undefined.
If $n=3$, beginalign lim_x to 0 left[ x^3-n+left(-fracx^2+fracn32+x^frac4n3 right)^3right]=1endalign
answered 1 hour ago
Siong Thye Goh
83.5k1456104
Credit: Carmeister's solution.
I am just curious about the behavior of other similar path.
Let $y = -fracx^22+x^n$ where $n >0$,then we have
beginalignlim_x to 0frac2(x^3+(-fracx^22+x^n)^3)2x^n &= lim_x to 0 frac(x^3+(-fracx^22+x^n)^3)x^n \
&= lim_x to 0 left[ x^3-n+left(-fracx^2+fracn32+x^frac4n3 right)^3right]endalign
If $n in (0,3)$, the limit is $0$.
If $n>3$, it is undefined.
If $n=3$, beginalign lim_x to 0 left[ x^3-n+left(-fracx^2+fracn32+x^frac4n3 right)^3right]=1endalign
answered 1 hour ago
Siong Thye Goh
83.5k1456104
edited 1 hour ago
answered 1 hour ago
Siong Thye Goh
83.5k1456104
answered 1 hour ago
Siong Thye Goh
83.5k1456104
answered 1 hour ago
Siong Thye Goh
83.5k1456104
83.5k1456104
add a comment |Â
add a comment |Â
up vote
2
down vote
Note that $f$ is undefined when $y=-x^2/2$, and if $y$ is very close to $-x^2/2$ then $f$ gets very large. So we should try something like $y=-x^2/2+x^3$.
answered 2 hours ago
Carmeister
2,3291720
add a comment |Â
up vote
2
down vote
Note that $f$ is undefined when $y=-x^2/2$, and if $y$ is very close to $-x^2/2$ then $f$ gets very large. So we should try something like $y=-x^2/2+x^3$.
answered 2 hours ago
Carmeister
2,3291720
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Note that $f$ is undefined when $y=-x^2/2$, and if $y$ is very close to $-x^2/2$ then $f$ gets very large. So we should try something like $y=-x^2/2+x^3$.
answered 2 hours ago
Carmeister
2,3291720
Note that $f$ is undefined when $y=-x^2/2$, and if $y$ is very close to $-x^2/2$ then $f$ gets very large. So we should try something like $y=-x^2/2+x^3$.
answered 2 hours ago
Carmeister
2,3291720
answered 2 hours ago
Carmeister
2,3291720
answered 2 hours ago
Carmeister
2,3291720
answered 2 hours ago
Carmeister
2,3291720
2,3291720
add a comment |Â
add a comment |Â
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