Mixing
Continuity of partial derivatives at (0,0)
$begingroup$
For the function $$f = begin {cases} frac{xy}{x^2+y^2}, text{if } (x,y) neq (0,0) \ 0, text{if } (x,y) = (0,0) end {cases}$$
show that $frac{partial f}{partial x}$ and $frac{partial f}{partial y}$ are not continuous at (0,0).
My attempt:
I found $$frac{partial f}{partial x}(x,y)=frac{y^3-x^2y}{(x^2+y^2)^2}$$
At $(x,y)=(0,0)$ $$frac{partial f}{partial x}(0,0)=lim_{trightarrow 0}frac{f(0+t,0)-f(0,0)}{t}=0.$$
Similarly $frac{partial f}{partial y}(0,0)=0$
Then $$lim_{(x,y)rightarrow(0,0)}frac{partial f}{partial x}(x,y)=lim_{(x,y)rightarrow(0,0)}frac{y^3-x^2y}{(x^2+y^2)^2}= text{ (consider restriction to the line } x = my) = lim_{yrightarrow 0}frac{y^3-m^2y^3}{(m^2y^2+y^2)^2}=lim_{yrightarrow 0}frac{y^3(1-m^2)}{(m^2+1)^2y^4}=infty neq frac{partial f}{partial x}(0,0).$$
Thus $frac{partial f}{partial x}$ is not continuous at (0,0). Similarly, for $frac{partial f}{partial y}.$
Is it correct? Also is it sufficient to show the inequality of the limit and the value of the partial, considering only one restriction to the line (in this case $x=my$)? Or should we consider other restrictions as well?
real-analysis multivariable-calculus partial-derivative
asked Feb 1 at 19:13
dxdydzdxdydz
49110
$endgroup$
add a comment |
$begingroup$
For the function $$f = begin {cases} frac{xy}{x^2+y^2}, text{if } (x,y) neq (0,0) \ 0, text{if } (x,y) = (0,0) end {cases}$$
show that $frac{partial f}{partial x}$ and $frac{partial f}{partial y}$ are not continuous at (0,0).
My attempt:
I found $$frac{partial f}{partial x}(x,y)=frac{y^3-x^2y}{(x^2+y^2)^2}$$
At $(x,y)=(0,0)$ $$frac{partial f}{partial x}(0,0)=lim_{trightarrow 0}frac{f(0+t,0)-f(0,0)}{t}=0.$$
Similarly $frac{partial f}{partial y}(0,0)=0$
Then $$lim_{(x,y)rightarrow(0,0)}frac{partial f}{partial x}(x,y)=lim_{(x,y)rightarrow(0,0)}frac{y^3-x^2y}{(x^2+y^2)^2}= text{ (consider restriction to the line } x = my) = lim_{yrightarrow 0}frac{y^3-m^2y^3}{(m^2y^2+y^2)^2}=lim_{yrightarrow 0}frac{y^3(1-m^2)}{(m^2+1)^2y^4}=infty neq frac{partial f}{partial x}(0,0).$$
Thus $frac{partial f}{partial x}$ is not continuous at (0,0). Similarly, for $frac{partial f}{partial y}.$
Is it correct? Also is it sufficient to show the inequality of the limit and the value of the partial, considering only one restriction to the line (in this case $x=my$)? Or should we consider other restrictions as well?
real-analysis multivariable-calculus partial-derivative
asked Feb 1 at 19:13
dxdydzdxdydz
49110
$endgroup$
$begingroup$
Since the limit should be the same irrespective of the approach chosen, it is enough to consider limit along $x=my$ here.
$endgroup$
– Exp ikx
Feb 1 at 19:20
add a comment |
$begingroup$
For the function $$f = begin {cases} frac{xy}{x^2+y^2}, text{if } (x,y) neq (0,0) \ 0, text{if } (x,y) = (0,0) end {cases}$$
show that $frac{partial f}{partial x}$ and $frac{partial f}{partial y}$ are not continuous at (0,0).
My attempt:
I found $$frac{partial f}{partial x}(x,y)=frac{y^3-x^2y}{(x^2+y^2)^2}$$
At $(x,y)=(0,0)$ $$frac{partial f}{partial x}(0,0)=lim_{trightarrow 0}frac{f(0+t,0)-f(0,0)}{t}=0.$$
Similarly $frac{partial f}{partial y}(0,0)=0$
Then $$lim_{(x,y)rightarrow(0,0)}frac{partial f}{partial x}(x,y)=lim_{(x,y)rightarrow(0,0)}frac{y^3-x^2y}{(x^2+y^2)^2}= text{ (consider restriction to the line } x = my) = lim_{yrightarrow 0}frac{y^3-m^2y^3}{(m^2y^2+y^2)^2}=lim_{yrightarrow 0}frac{y^3(1-m^2)}{(m^2+1)^2y^4}=infty neq frac{partial f}{partial x}(0,0).$$
Thus $frac{partial f}{partial x}$ is not continuous at (0,0). Similarly, for $frac{partial f}{partial y}.$
Is it correct? Also is it sufficient to show the inequality of the limit and the value of the partial, considering only one restriction to the line (in this case $x=my$)? Or should we consider other restrictions as well?
real-analysis multivariable-calculus partial-derivative
asked Feb 1 at 19:13
dxdydzdxdydz
49110
$endgroup$
For the function $$f = begin {cases} frac{xy}{x^2+y^2}, text{if } (x,y) neq (0,0) \ 0, text{if } (x,y) = (0,0) end {cases}$$
show that $frac{partial f}{partial x}$ and $frac{partial f}{partial y}$ are not continuous at (0,0).
My attempt:
I found $$frac{partial f}{partial x}(x,y)=frac{y^3-x^2y}{(x^2+y^2)^2}$$
At $(x,y)=(0,0)$ $$frac{partial f}{partial x}(0,0)=lim_{trightarrow 0}frac{f(0+t,0)-f(0,0)}{t}=0.$$
Similarly $frac{partial f}{partial y}(0,0)=0$
Then $$lim_{(x,y)rightarrow(0,0)}frac{partial f}{partial x}(x,y)=lim_{(x,y)rightarrow(0,0)}frac{y^3-x^2y}{(x^2+y^2)^2}= text{ (consider restriction to the line } x = my) = lim_{yrightarrow 0}frac{y^3-m^2y^3}{(m^2y^2+y^2)^2}=lim_{yrightarrow 0}frac{y^3(1-m^2)}{(m^2+1)^2y^4}=infty neq frac{partial f}{partial x}(0,0).$$
Thus $frac{partial f}{partial x}$ is not continuous at (0,0). Similarly, for $frac{partial f}{partial y}.$
Is it correct? Also is it sufficient to show the inequality of the limit and the value of the partial, considering only one restriction to the line (in this case $x=my$)? Or should we consider other restrictions as well?
real-analysis multivariable-calculus partial-derivative
real-analysis multivariable-calculus partial-derivative
asked Feb 1 at 19:13
dxdydzdxdydz
49110
asked Feb 1 at 19:13
dxdydzdxdydz
49110
asked Feb 1 at 19:13
dxdydzdxdydz
49110
asked Feb 1 at 19:13
dxdydzdxdydz
49110
asked Feb 1 at 19:13
dxdydzdxdydz
49110
49110
$begingroup$
Since the limit should be the same irrespective of the approach chosen, it is enough to consider limit along $x=my$ here.
$endgroup$
– Exp ikx
Feb 1 at 19:20
add a comment |
$begingroup$
Since the limit should be the same irrespective of the approach chosen, it is enough to consider limit along $x=my$ here.
$endgroup$
– Exp ikx
Feb 1 at 19:20
$begingroup$
Since the limit should be the same irrespective of the approach chosen, it is enough to consider limit along $x=my$ here.
$endgroup$
– Exp ikx
Feb 1 at 19:20
$begingroup$
Since the limit should be the same irrespective of the approach chosen, it is enough to consider limit along $x=my$ here.
$endgroup$
– Exp ikx
Feb 1 at 19:20
add a comment |
1 Answer
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$begingroup$
You could also argue like this: Obviously, the partial derivatives are continuous in $mathbb R^2 setminus {(0,0)}$. If they were also continuous in $(0,0)$, the function would be continuously partially differentiable, hence totally differentiable and hence continuous. But $f(1/n,1/n)=1/2$ does not converge to $f(0,0)=0$ for $ntoinfty$.
To be exact, this only shows that $frac{partial f}{partial x}$ or $frac{partial f}{partial y}$ is discontinuous at $(0,0)$, but for symmetry reasons it cannot be only one of them.
answered Feb 2 at 1:40
Mars PlasticMars Plastic
1,455122
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add a comment |
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$begingroup$
You could also argue like this: Obviously, the partial derivatives are continuous in $mathbb R^2 setminus {(0,0)}$. If they were also continuous in $(0,0)$, the function would be continuously partially differentiable, hence totally differentiable and hence continuous. But $f(1/n,1/n)=1/2$ does not converge to $f(0,0)=0$ for $ntoinfty$.
To be exact, this only shows that $frac{partial f}{partial x}$ or $frac{partial f}{partial y}$ is discontinuous at $(0,0)$, but for symmetry reasons it cannot be only one of them.
answered Feb 2 at 1:40
Mars PlasticMars Plastic
1,455122
$endgroup$
add a comment |
$begingroup$
You could also argue like this: Obviously, the partial derivatives are continuous in $mathbb R^2 setminus {(0,0)}$. If they were also continuous in $(0,0)$, the function would be continuously partially differentiable, hence totally differentiable and hence continuous. But $f(1/n,1/n)=1/2$ does not converge to $f(0,0)=0$ for $ntoinfty$.
To be exact, this only shows that $frac{partial f}{partial x}$ or $frac{partial f}{partial y}$ is discontinuous at $(0,0)$, but for symmetry reasons it cannot be only one of them.
answered Feb 2 at 1:40
Mars PlasticMars Plastic
1,455122
$endgroup$
add a comment |
$begingroup$
You could also argue like this: Obviously, the partial derivatives are continuous in $mathbb R^2 setminus {(0,0)}$. If they were also continuous in $(0,0)$, the function would be continuously partially differentiable, hence totally differentiable and hence continuous. But $f(1/n,1/n)=1/2$ does not converge to $f(0,0)=0$ for $ntoinfty$.
To be exact, this only shows that $frac{partial f}{partial x}$ or $frac{partial f}{partial y}$ is discontinuous at $(0,0)$, but for symmetry reasons it cannot be only one of them.
answered Feb 2 at 1:40
Mars PlasticMars Plastic
1,455122
$endgroup$
You could also argue like this: Obviously, the partial derivatives are continuous in $mathbb R^2 setminus {(0,0)}$. If they were also continuous in $(0,0)$, the function would be continuously partially differentiable, hence totally differentiable and hence continuous. But $f(1/n,1/n)=1/2$ does not converge to $f(0,0)=0$ for $ntoinfty$.
To be exact, this only shows that $frac{partial f}{partial x}$ or $frac{partial f}{partial y}$ is discontinuous at $(0,0)$, but for symmetry reasons it cannot be only one of them.
answered Feb 2 at 1:40
Mars PlasticMars Plastic
1,455122
edited Feb 2 at 1:46
answered Feb 2 at 1:40
Mars PlasticMars Plastic
1,455122
answered Feb 2 at 1:40
Mars PlasticMars Plastic
1,455122
answered Feb 2 at 1:40
Mars PlasticMars Plastic
1,455122
1,455122
add a comment |
add a comment |
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$begingroup$
Since the limit should be the same irrespective of the approach chosen, it is enough to consider limit along $x=my$ here.
$endgroup$
– Exp ikx
Feb 1 at 19:20