Mixing
Continuity and differentiability of $f(x,y)$
$begingroup$
Explore continuity and differentiability of the function
$f(x,y)=begin{cases}
dfrac {xysin x}{x^2+y^2} & text{$(x,y)not=(0,0)$} \
0 & text{$(x,y)=(0,0)$}
end{cases}.$
In $(x,y)not=(0,0)$, function is continuous, as
quotient of two continuous functions.
I am checking in $(x,y)=(0,0)$.
Well, I think this function is not continuous, because
we can see that
$leftlvert dfrac {xysin x}{x^2+y^2}rightrvert le leftlvert dfrac {xy}{x^2+y^2}rightrvert.$
Let $left(frac{1}{k},frac{1}{k}right) rightarrow (0,0)$, where $krightarrow infty.$
And we get
$lim_{krightarrow infty} {dfrac{frac{1}{k^2}}{frac{1}{k^2}+frac{1}{k^2}}} = frac{1}{2} not= 0=f(0,0).$
If a function is discontinuous, then it is not differentiable.
Is that correct?
analysis
edited Jan 28 at 22:50
John Wayland Bales
15.1k21238
asked Jan 28 at 19:24
VictorVictor
464
$endgroup$
add a comment |
$begingroup$
Explore continuity and differentiability of the function
$f(x,y)=begin{cases}
dfrac {xysin x}{x^2+y^2} & text{$(x,y)not=(0,0)$} \
0 & text{$(x,y)=(0,0)$}
end{cases}.$
In $(x,y)not=(0,0)$, function is continuous, as
quotient of two continuous functions.
I am checking in $(x,y)=(0,0)$.
Well, I think this function is not continuous, because
we can see that
$leftlvert dfrac {xysin x}{x^2+y^2}rightrvert le leftlvert dfrac {xy}{x^2+y^2}rightrvert.$
Let $left(frac{1}{k},frac{1}{k}right) rightarrow (0,0)$, where $krightarrow infty.$
And we get
$lim_{krightarrow infty} {dfrac{frac{1}{k^2}}{frac{1}{k^2}+frac{1}{k^2}}} = frac{1}{2} not= 0=f(0,0).$
If a function is discontinuous, then it is not differentiable.
Is that correct?
analysis
edited Jan 28 at 22:50
John Wayland Bales
15.1k21238
asked Jan 28 at 19:24
VictorVictor
464
$endgroup$
$begingroup$
The function is continuous at and around $(0,0)$. Making the shift to polar coordinates,we find that $lim_{(x,y)to(0,0)} = f(x,y) = 0$.
$endgroup$
– Hyperion
Jan 28 at 19:32
$begingroup$
$frac{r^2(cosx)(sin^2x)}{r^2} = (cosx)(sin^2x) = 0 $ where $x rightarrow 0$ ??
$endgroup$
– Victor
Jan 28 at 19:36
1
$begingroup$
Not quite. $sin(x) = sin(rcos(theta))$.
$endgroup$
– Hyperion
Jan 28 at 21:36
add a comment |
$begingroup$
Explore continuity and differentiability of the function
$f(x,y)=begin{cases}
dfrac {xysin x}{x^2+y^2} & text{$(x,y)not=(0,0)$} \
0 & text{$(x,y)=(0,0)$}
end{cases}.$
In $(x,y)not=(0,0)$, function is continuous, as
quotient of two continuous functions.
I am checking in $(x,y)=(0,0)$.
Well, I think this function is not continuous, because
we can see that
$leftlvert dfrac {xysin x}{x^2+y^2}rightrvert le leftlvert dfrac {xy}{x^2+y^2}rightrvert.$
Let $left(frac{1}{k},frac{1}{k}right) rightarrow (0,0)$, where $krightarrow infty.$
And we get
$lim_{krightarrow infty} {dfrac{frac{1}{k^2}}{frac{1}{k^2}+frac{1}{k^2}}} = frac{1}{2} not= 0=f(0,0).$
If a function is discontinuous, then it is not differentiable.
Is that correct?
analysis
edited Jan 28 at 22:50
John Wayland Bales
15.1k21238
asked Jan 28 at 19:24
VictorVictor
464
$endgroup$
Explore continuity and differentiability of the function
$f(x,y)=begin{cases}
dfrac {xysin x}{x^2+y^2} & text{$(x,y)not=(0,0)$} \
0 & text{$(x,y)=(0,0)$}
end{cases}.$
In $(x,y)not=(0,0)$, function is continuous, as
quotient of two continuous functions.
I am checking in $(x,y)=(0,0)$.
Well, I think this function is not continuous, because
we can see that
$leftlvert dfrac {xysin x}{x^2+y^2}rightrvert le leftlvert dfrac {xy}{x^2+y^2}rightrvert.$
Let $left(frac{1}{k},frac{1}{k}right) rightarrow (0,0)$, where $krightarrow infty.$
And we get
$lim_{krightarrow infty} {dfrac{frac{1}{k^2}}{frac{1}{k^2}+frac{1}{k^2}}} = frac{1}{2} not= 0=f(0,0).$
If a function is discontinuous, then it is not differentiable.
Is that correct?
analysis
analysis
edited Jan 28 at 22:50
John Wayland Bales
15.1k21238
asked Jan 28 at 19:24
VictorVictor
464
edited Jan 28 at 22:50
John Wayland Bales
15.1k21238
asked Jan 28 at 19:24
VictorVictor
464
edited Jan 28 at 22:50
John Wayland Bales
15.1k21238
edited Jan 28 at 22:50
John Wayland Bales
15.1k21238
edited Jan 28 at 22:50
John Wayland Bales
15.1k21238
15.1k21238
asked Jan 28 at 19:24
VictorVictor
464
asked Jan 28 at 19:24
VictorVictor
464
asked Jan 28 at 19:24
VictorVictor
464
464
$begingroup$
The function is continuous at and around $(0,0)$. Making the shift to polar coordinates,we find that $lim_{(x,y)to(0,0)} = f(x,y) = 0$.
$endgroup$
– Hyperion
Jan 28 at 19:32
$begingroup$
$frac{r^2(cosx)(sin^2x)}{r^2} = (cosx)(sin^2x) = 0 $ where $x rightarrow 0$ ??
$endgroup$
– Victor
Jan 28 at 19:36
1
$begingroup$
Not quite. $sin(x) = sin(rcos(theta))$.
$endgroup$
– Hyperion
Jan 28 at 21:36
add a comment |
$begingroup$
The function is continuous at and around $(0,0)$. Making the shift to polar coordinates,we find that $lim_{(x,y)to(0,0)} = f(x,y) = 0$.
$endgroup$
– Hyperion
Jan 28 at 19:32
$begingroup$
$frac{r^2(cosx)(sin^2x)}{r^2} = (cosx)(sin^2x) = 0 $ where $x rightarrow 0$ ??
$endgroup$
– Victor
Jan 28 at 19:36
1
$begingroup$
Not quite. $sin(x) = sin(rcos(theta))$.
$endgroup$
– Hyperion
Jan 28 at 21:36
$begingroup$
The function is continuous at and around $(0,0)$. Making the shift to polar coordinates,we find that $lim_{(x,y)to(0,0)} = f(x,y) = 0$.
$endgroup$
– Hyperion
Jan 28 at 19:32
$begingroup$
The function is continuous at and around $(0,0)$. Making the shift to polar coordinates,we find that $lim_{(x,y)to(0,0)} = f(x,y) = 0$.
$endgroup$
– Hyperion
Jan 28 at 19:32
$begingroup$
$frac{r^2(cosx)(sin^2x)}{r^2} = (cosx)(sin^2x) = 0 $ where $x rightarrow 0$ ??
$endgroup$
– Victor
Jan 28 at 19:36
$begingroup$
$frac{r^2(cosx)(sin^2x)}{r^2} = (cosx)(sin^2x) = 0 $ where $x rightarrow 0$ ??
$endgroup$
– Victor
Jan 28 at 19:36
1
1
$begingroup$
Not quite. $sin(x) = sin(rcos(theta))$.
$endgroup$
– Hyperion
Jan 28 at 21:36
$begingroup$
Not quite. $sin(x) = sin(rcos(theta))$.
$endgroup$
– Hyperion
Jan 28 at 21:36
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You wrote
$$leftlvert dfrac {xysin x}{x^2+y^2}rightrvert le leftlvert dfrac {xy}{x^2+y^2}rightrvert.$$
But clearly we have a better estimate: $$leftlvert dfrac {xysin x}{x^2+y^2}rightrvert le |sin x|leftlvert dfrac {xy}{x^2+y^2}rightrvert.$$
That implies continuity at $(0,0)$.
Added later: Suppose $f$ is differentiable at $(0,0).$ Then
$$f(x,y) = f(0,0) + nabla f (0,0)cdot (x,y) + o((x^2 +y^2)^{1/2}).$$
But note $f(0,0)=0$ and $nabla f (0,0)=(0,0).$ On the ray $y=x, x>0$ we then have
$$frac{x^2sin x}{2x^2} = (sin x)/2 = o((2x^2)^{1/2}) = o(x).$$
Since $(sin x)/xto 1,$ we have a contradiction.
answered Jan 28 at 23:04
zhw.zhw.
74.8k43175
$endgroup$
$begingroup$
I added a bit about differentiability.
$endgroup$
– zhw.
Jan 29 at 22:26
add a comment |
$begingroup$
No. $frac{xy}{x^2+y^2}$ is bounded and $sin xto 0$ as $(x,y)to (0,0)$. Hence the function is continuous at $(0,0)$. You definitely need to check whether it is differentiable at $(0,0)$.
answered Jan 28 at 19:38
Eclipse SunEclipse Sun
7,9851438
$endgroup$
$begingroup$
$frac{df}{dx}(0,0) = lim t rightarrow 0 frac{f(t,0)-f(0)}{t}=0 , frac{df}{dy}(0,0) = lim t rightarrow 0 frac{f(0,t)-f(0)}{t}=0$ Both partial derivatives are continuous, so I have to check $frac{df}{dx}(x,y)$ and $frac{df}{dy}(x,y)$ ? Or from the differential mapping (sorry I don't know how it is in english, I hope u will understand) definition?
$endgroup$
– Victor
Jan 28 at 19:45
$begingroup$
You only showed both the partial derivatives exist. I would be surprised if they are continuous at $(0,0)$. You need to refer to the definition of differentiability directly.
$endgroup$
– Eclipse Sun
Jan 28 at 20:03
$begingroup$
Yes indeed, so $lim h=(h1,h2) rightarrow 0 frac{lVert f(h1,h2)-f(0,0)-A(h1,h2)lVert}{lVert h lVert}= ?? $ I don't know what next, how to calculate.... $A=gradf(0,0)=(0,0)$
$endgroup$
– Victor
Jan 28 at 20:11
$begingroup$
Let us suppose $f$ is differentiable. Then its differential $A$ has to be zero as you calculated the partial derivatives. But then we get a contradiction since $lim_{hto 0}frac{|f(h)|}{|h|}$ does not exist.
$endgroup$
– Eclipse Sun
Jan 28 at 20:28
$begingroup$
Basically the denominator and the numerator are of the same order. So along different directions the limit will be different.
$endgroup$
– Eclipse Sun
Jan 28 at 20:53
|
show 4 more comments
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You wrote
$$leftlvert dfrac {xysin x}{x^2+y^2}rightrvert le leftlvert dfrac {xy}{x^2+y^2}rightrvert.$$
But clearly we have a better estimate: $$leftlvert dfrac {xysin x}{x^2+y^2}rightrvert le |sin x|leftlvert dfrac {xy}{x^2+y^2}rightrvert.$$
That implies continuity at $(0,0)$.
Added later: Suppose $f$ is differentiable at $(0,0).$ Then
$$f(x,y) = f(0,0) + nabla f (0,0)cdot (x,y) + o((x^2 +y^2)^{1/2}).$$
But note $f(0,0)=0$ and $nabla f (0,0)=(0,0).$ On the ray $y=x, x>0$ we then have
$$frac{x^2sin x}{2x^2} = (sin x)/2 = o((2x^2)^{1/2}) = o(x).$$
Since $(sin x)/xto 1,$ we have a contradiction.
answered Jan 28 at 23:04
zhw.zhw.
74.8k43175
$endgroup$
$begingroup$
I added a bit about differentiability.
$endgroup$
– zhw.
Jan 29 at 22:26
add a comment |
$begingroup$
You wrote
$$leftlvert dfrac {xysin x}{x^2+y^2}rightrvert le leftlvert dfrac {xy}{x^2+y^2}rightrvert.$$
But clearly we have a better estimate: $$leftlvert dfrac {xysin x}{x^2+y^2}rightrvert le |sin x|leftlvert dfrac {xy}{x^2+y^2}rightrvert.$$
That implies continuity at $(0,0)$.
Added later: Suppose $f$ is differentiable at $(0,0).$ Then
$$f(x,y) = f(0,0) + nabla f (0,0)cdot (x,y) + o((x^2 +y^2)^{1/2}).$$
But note $f(0,0)=0$ and $nabla f (0,0)=(0,0).$ On the ray $y=x, x>0$ we then have
$$frac{x^2sin x}{2x^2} = (sin x)/2 = o((2x^2)^{1/2}) = o(x).$$
Since $(sin x)/xto 1,$ we have a contradiction.
answered Jan 28 at 23:04
zhw.zhw.
74.8k43175
$endgroup$
$begingroup$
I added a bit about differentiability.
$endgroup$
– zhw.
Jan 29 at 22:26
add a comment |
$begingroup$
You wrote
$$leftlvert dfrac {xysin x}{x^2+y^2}rightrvert le leftlvert dfrac {xy}{x^2+y^2}rightrvert.$$
But clearly we have a better estimate: $$leftlvert dfrac {xysin x}{x^2+y^2}rightrvert le |sin x|leftlvert dfrac {xy}{x^2+y^2}rightrvert.$$
That implies continuity at $(0,0)$.
Added later: Suppose $f$ is differentiable at $(0,0).$ Then
$$f(x,y) = f(0,0) + nabla f (0,0)cdot (x,y) + o((x^2 +y^2)^{1/2}).$$
But note $f(0,0)=0$ and $nabla f (0,0)=(0,0).$ On the ray $y=x, x>0$ we then have
$$frac{x^2sin x}{2x^2} = (sin x)/2 = o((2x^2)^{1/2}) = o(x).$$
Since $(sin x)/xto 1,$ we have a contradiction.
answered Jan 28 at 23:04
zhw.zhw.
74.8k43175
$endgroup$
You wrote
$$leftlvert dfrac {xysin x}{x^2+y^2}rightrvert le leftlvert dfrac {xy}{x^2+y^2}rightrvert.$$
But clearly we have a better estimate: $$leftlvert dfrac {xysin x}{x^2+y^2}rightrvert le |sin x|leftlvert dfrac {xy}{x^2+y^2}rightrvert.$$
That implies continuity at $(0,0)$.
Added later: Suppose $f$ is differentiable at $(0,0).$ Then
$$f(x,y) = f(0,0) + nabla f (0,0)cdot (x,y) + o((x^2 +y^2)^{1/2}).$$
But note $f(0,0)=0$ and $nabla f (0,0)=(0,0).$ On the ray $y=x, x>0$ we then have
$$frac{x^2sin x}{2x^2} = (sin x)/2 = o((2x^2)^{1/2}) = o(x).$$
Since $(sin x)/xto 1,$ we have a contradiction.
answered Jan 28 at 23:04
zhw.zhw.
74.8k43175
edited Jan 29 at 22:26
answered Jan 28 at 23:04
zhw.zhw.
74.8k43175
answered Jan 28 at 23:04
zhw.zhw.
74.8k43175
answered Jan 28 at 23:04
zhw.zhw.
74.8k43175
74.8k43175
$begingroup$
I added a bit about differentiability.
$endgroup$
– zhw.
Jan 29 at 22:26
add a comment |
$begingroup$
I added a bit about differentiability.
$endgroup$
– zhw.
Jan 29 at 22:26
$begingroup$
I added a bit about differentiability.
$endgroup$
– zhw.
Jan 29 at 22:26
$begingroup$
I added a bit about differentiability.
$endgroup$
– zhw.
Jan 29 at 22:26
add a comment |
$begingroup$
No. $frac{xy}{x^2+y^2}$ is bounded and $sin xto 0$ as $(x,y)to (0,0)$. Hence the function is continuous at $(0,0)$. You definitely need to check whether it is differentiable at $(0,0)$.
answered Jan 28 at 19:38
Eclipse SunEclipse Sun
7,9851438
$endgroup$
$begingroup$
$frac{df}{dx}(0,0) = lim t rightarrow 0 frac{f(t,0)-f(0)}{t}=0 , frac{df}{dy}(0,0) = lim t rightarrow 0 frac{f(0,t)-f(0)}{t}=0$ Both partial derivatives are continuous, so I have to check $frac{df}{dx}(x,y)$ and $frac{df}{dy}(x,y)$ ? Or from the differential mapping (sorry I don't know how it is in english, I hope u will understand) definition?
$endgroup$
– Victor
Jan 28 at 19:45
$begingroup$
You only showed both the partial derivatives exist. I would be surprised if they are continuous at $(0,0)$. You need to refer to the definition of differentiability directly.
$endgroup$
– Eclipse Sun
Jan 28 at 20:03
$begingroup$
Yes indeed, so $lim h=(h1,h2) rightarrow 0 frac{lVert f(h1,h2)-f(0,0)-A(h1,h2)lVert}{lVert h lVert}= ?? $ I don't know what next, how to calculate.... $A=gradf(0,0)=(0,0)$
$endgroup$
– Victor
Jan 28 at 20:11
$begingroup$
Let us suppose $f$ is differentiable. Then its differential $A$ has to be zero as you calculated the partial derivatives. But then we get a contradiction since $lim_{hto 0}frac{|f(h)|}{|h|}$ does not exist.
$endgroup$
– Eclipse Sun
Jan 28 at 20:28
$begingroup$
Basically the denominator and the numerator are of the same order. So along different directions the limit will be different.
$endgroup$
– Eclipse Sun
Jan 28 at 20:53
|
show 4 more comments
$begingroup$
No. $frac{xy}{x^2+y^2}$ is bounded and $sin xto 0$ as $(x,y)to (0,0)$. Hence the function is continuous at $(0,0)$. You definitely need to check whether it is differentiable at $(0,0)$.
answered Jan 28 at 19:38
Eclipse SunEclipse Sun
7,9851438
$endgroup$
$begingroup$
$frac{df}{dx}(0,0) = lim t rightarrow 0 frac{f(t,0)-f(0)}{t}=0 , frac{df}{dy}(0,0) = lim t rightarrow 0 frac{f(0,t)-f(0)}{t}=0$ Both partial derivatives are continuous, so I have to check $frac{df}{dx}(x,y)$ and $frac{df}{dy}(x,y)$ ? Or from the differential mapping (sorry I don't know how it is in english, I hope u will understand) definition?
$endgroup$
– Victor
Jan 28 at 19:45
$begingroup$
You only showed both the partial derivatives exist. I would be surprised if they are continuous at $(0,0)$. You need to refer to the definition of differentiability directly.
$endgroup$
– Eclipse Sun
Jan 28 at 20:03
$begingroup$
Yes indeed, so $lim h=(h1,h2) rightarrow 0 frac{lVert f(h1,h2)-f(0,0)-A(h1,h2)lVert}{lVert h lVert}= ?? $ I don't know what next, how to calculate.... $A=gradf(0,0)=(0,0)$
$endgroup$
– Victor
Jan 28 at 20:11
$begingroup$
Let us suppose $f$ is differentiable. Then its differential $A$ has to be zero as you calculated the partial derivatives. But then we get a contradiction since $lim_{hto 0}frac{|f(h)|}{|h|}$ does not exist.
$endgroup$
– Eclipse Sun
Jan 28 at 20:28
$begingroup$
Basically the denominator and the numerator are of the same order. So along different directions the limit will be different.
$endgroup$
– Eclipse Sun
Jan 28 at 20:53
|
show 4 more comments
$begingroup$
No. $frac{xy}{x^2+y^2}$ is bounded and $sin xto 0$ as $(x,y)to (0,0)$. Hence the function is continuous at $(0,0)$. You definitely need to check whether it is differentiable at $(0,0)$.
answered Jan 28 at 19:38
Eclipse SunEclipse Sun
7,9851438
$endgroup$
No. $frac{xy}{x^2+y^2}$ is bounded and $sin xto 0$ as $(x,y)to (0,0)$. Hence the function is continuous at $(0,0)$. You definitely need to check whether it is differentiable at $(0,0)$.
answered Jan 28 at 19:38
Eclipse SunEclipse Sun
7,9851438
answered Jan 28 at 19:38
Eclipse SunEclipse Sun
7,9851438
answered Jan 28 at 19:38
Eclipse SunEclipse Sun
7,9851438
answered Jan 28 at 19:38
Eclipse SunEclipse Sun
7,9851438
7,9851438
$begingroup$
$frac{df}{dx}(0,0) = lim t rightarrow 0 frac{f(t,0)-f(0)}{t}=0 , frac{df}{dy}(0,0) = lim t rightarrow 0 frac{f(0,t)-f(0)}{t}=0$ Both partial derivatives are continuous, so I have to check $frac{df}{dx}(x,y)$ and $frac{df}{dy}(x,y)$ ? Or from the differential mapping (sorry I don't know how it is in english, I hope u will understand) definition?
$endgroup$
– Victor
Jan 28 at 19:45
$begingroup$
You only showed both the partial derivatives exist. I would be surprised if they are continuous at $(0,0)$. You need to refer to the definition of differentiability directly.
$endgroup$
– Eclipse Sun
Jan 28 at 20:03
$begingroup$
Yes indeed, so $lim h=(h1,h2) rightarrow 0 frac{lVert f(h1,h2)-f(0,0)-A(h1,h2)lVert}{lVert h lVert}= ?? $ I don't know what next, how to calculate.... $A=gradf(0,0)=(0,0)$
$endgroup$
– Victor
Jan 28 at 20:11
$begingroup$
Let us suppose $f$ is differentiable. Then its differential $A$ has to be zero as you calculated the partial derivatives. But then we get a contradiction since $lim_{hto 0}frac{|f(h)|}{|h|}$ does not exist.
$endgroup$
– Eclipse Sun
Jan 28 at 20:28
$begingroup$
Basically the denominator and the numerator are of the same order. So along different directions the limit will be different.
$endgroup$
– Eclipse Sun
Jan 28 at 20:53
|
show 4 more comments
$begingroup$
$frac{df}{dx}(0,0) = lim t rightarrow 0 frac{f(t,0)-f(0)}{t}=0 , frac{df}{dy}(0,0) = lim t rightarrow 0 frac{f(0,t)-f(0)}{t}=0$ Both partial derivatives are continuous, so I have to check $frac{df}{dx}(x,y)$ and $frac{df}{dy}(x,y)$ ? Or from the differential mapping (sorry I don't know how it is in english, I hope u will understand) definition?
$endgroup$
– Victor
Jan 28 at 19:45
$begingroup$
You only showed both the partial derivatives exist. I would be surprised if they are continuous at $(0,0)$. You need to refer to the definition of differentiability directly.
$endgroup$
– Eclipse Sun
Jan 28 at 20:03
$begingroup$
Yes indeed, so $lim h=(h1,h2) rightarrow 0 frac{lVert f(h1,h2)-f(0,0)-A(h1,h2)lVert}{lVert h lVert}= ?? $ I don't know what next, how to calculate.... $A=gradf(0,0)=(0,0)$
$endgroup$
– Victor
Jan 28 at 20:11
$begingroup$
Let us suppose $f$ is differentiable. Then its differential $A$ has to be zero as you calculated the partial derivatives. But then we get a contradiction since $lim_{hto 0}frac{|f(h)|}{|h|}$ does not exist.
$endgroup$
– Eclipse Sun
Jan 28 at 20:28
$begingroup$
Basically the denominator and the numerator are of the same order. So along different directions the limit will be different.
$endgroup$
– Eclipse Sun
Jan 28 at 20:53
$begingroup$
$frac{df}{dx}(0,0) = lim t rightarrow 0 frac{f(t,0)-f(0)}{t}=0 , frac{df}{dy}(0,0) = lim t rightarrow 0 frac{f(0,t)-f(0)}{t}=0$ Both partial derivatives are continuous, so I have to check $frac{df}{dx}(x,y)$ and $frac{df}{dy}(x,y)$ ? Or from the differential mapping (sorry I don't know how it is in english, I hope u will understand) definition?
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– Victor
Jan 28 at 19:45
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$frac{df}{dx}(0,0) = lim t rightarrow 0 frac{f(t,0)-f(0)}{t}=0 , frac{df}{dy}(0,0) = lim t rightarrow 0 frac{f(0,t)-f(0)}{t}=0$ Both partial derivatives are continuous, so I have to check $frac{df}{dx}(x,y)$ and $frac{df}{dy}(x,y)$ ? Or from the differential mapping (sorry I don't know how it is in english, I hope u will understand) definition?
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– Victor
Jan 28 at 19:45
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You only showed both the partial derivatives exist. I would be surprised if they are continuous at $(0,0)$. You need to refer to the definition of differentiability directly.
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– Eclipse Sun
Jan 28 at 20:03
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You only showed both the partial derivatives exist. I would be surprised if they are continuous at $(0,0)$. You need to refer to the definition of differentiability directly.
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– Eclipse Sun
Jan 28 at 20:03
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Yes indeed, so $lim h=(h1,h2) rightarrow 0 frac{lVert f(h1,h2)-f(0,0)-A(h1,h2)lVert}{lVert h lVert}= ?? $ I don't know what next, how to calculate.... $A=gradf(0,0)=(0,0)$
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– Victor
Jan 28 at 20:11
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Yes indeed, so $lim h=(h1,h2) rightarrow 0 frac{lVert f(h1,h2)-f(0,0)-A(h1,h2)lVert}{lVert h lVert}= ?? $ I don't know what next, how to calculate.... $A=gradf(0,0)=(0,0)$
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– Victor
Jan 28 at 20:11
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Let us suppose $f$ is differentiable. Then its differential $A$ has to be zero as you calculated the partial derivatives. But then we get a contradiction since $lim_{hto 0}frac{|f(h)|}{|h|}$ does not exist.
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– Eclipse Sun
Jan 28 at 20:28
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Let us suppose $f$ is differentiable. Then its differential $A$ has to be zero as you calculated the partial derivatives. But then we get a contradiction since $lim_{hto 0}frac{|f(h)|}{|h|}$ does not exist.
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– Eclipse Sun
Jan 28 at 20:28
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Basically the denominator and the numerator are of the same order. So along different directions the limit will be different.
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– Eclipse Sun
Jan 28 at 20:53
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Basically the denominator and the numerator are of the same order. So along different directions the limit will be different.
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– Eclipse Sun
Jan 28 at 20:53
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The function is continuous at and around $(0,0)$. Making the shift to polar coordinates,we find that $lim_{(x,y)to(0,0)} = f(x,y) = 0$.
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– Hyperion
Jan 28 at 19:32
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$frac{r^2(cosx)(sin^2x)}{r^2} = (cosx)(sin^2x) = 0 $ where $x rightarrow 0$ ??
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– Victor
Jan 28 at 19:36
1
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Not quite. $sin(x) = sin(rcos(theta))$.
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– Hyperion
Jan 28 at 21:36