Urgent with the continuity: Existence of directional derivatives at all directions of a $f:mathbb{R}^{2} to...
$begingroup$
Let $f(x,y)=
begin{cases}
frac{2x^2y}{x^2+y^2} & if & (x,y) neq (0,0) \
0 & if & (x,y)=(0,0) .
end{cases}$
(i) Prove the directional derivatives of $f$ exist in any direction at the point $(0,0)$.
(ii)¿Is $f$ continuous on (0,0)?
For (i) I took $u=(u_{1},u_{2}) in mathbb{R^{2}}$ such $||u||=1$ and $0=(0,0)$. So $$lim_{t to 0} frac{f(0+ t u)-f(0)}{t}=lim_{t to 0}frac{2(tu_{1})^2 (tu_{2})}{(tu_{1})^4+(tu_{2})^2}=lim_{t to 0}frac{(t^2 u_{1}^2)(t^2 u_{1}^{2}) tu_{2}}{t^{4}u_{1}^{4}+t^{2} u_{2}^{2}}$$. But I cannot find the limit when $t$ aproaches to $0$. Which by finding them I would prove the directional derivative exist at every direction right?
For (ii) I got the intuition $f$ is not continuous at $(0,0)$ so I took $lbrace (frac{1}{k},frac{1}{k}) rbrace_{k in mathbb{N}}$ and $lbrace (frac{1}{k},0) rbrace_{k in mathbb{N}}$ which are two different sequences in $mathbb{R}^{2}$ converging to $(0,0)$. However, $lbrace f(frac{1}{k},frac{1}{k}) rbrace_{k in mathbb{N}}=lbrace frac{1}{k^{2}} rbrace_{n in mathbb{N}} to 0$ and $lbrace f(frac{1}{k},0) rbrace_{k in mathbb{N}} to 0$. So maybe my intuition was not right? Can anyone help me end the proof of continuity or not continuity , please?
multivariable-calculus continuity
edited Nov 26 '18 at 9:01
Yadati Kiran
1,751619
asked Nov 25 '18 at 5:48
CosCos
1326
$endgroup$
|
show 3 more comments
$begingroup$
Let $f(x,y)=
begin{cases}
frac{2x^2y}{x^2+y^2} & if & (x,y) neq (0,0) \
0 & if & (x,y)=(0,0) .
end{cases}$
(i) Prove the directional derivatives of $f$ exist in any direction at the point $(0,0)$.
(ii)¿Is $f$ continuous on (0,0)?
For (i) I took $u=(u_{1},u_{2}) in mathbb{R^{2}}$ such $||u||=1$ and $0=(0,0)$. So $$lim_{t to 0} frac{f(0+ t u)-f(0)}{t}=lim_{t to 0}frac{2(tu_{1})^2 (tu_{2})}{(tu_{1})^4+(tu_{2})^2}=lim_{t to 0}frac{(t^2 u_{1}^2)(t^2 u_{1}^{2}) tu_{2}}{t^{4}u_{1}^{4}+t^{2} u_{2}^{2}}$$. But I cannot find the limit when $t$ aproaches to $0$. Which by finding them I would prove the directional derivative exist at every direction right?
For (ii) I got the intuition $f$ is not continuous at $(0,0)$ so I took $lbrace (frac{1}{k},frac{1}{k}) rbrace_{k in mathbb{N}}$ and $lbrace (frac{1}{k},0) rbrace_{k in mathbb{N}}$ which are two different sequences in $mathbb{R}^{2}$ converging to $(0,0)$. However, $lbrace f(frac{1}{k},frac{1}{k}) rbrace_{k in mathbb{N}}=lbrace frac{1}{k^{2}} rbrace_{n in mathbb{N}} to 0$ and $lbrace f(frac{1}{k},0) rbrace_{k in mathbb{N}} to 0$. So maybe my intuition was not right? Can anyone help me end the proof of continuity or not continuity , please?
multivariable-calculus continuity
edited Nov 26 '18 at 9:01
Yadati Kiran
1,751619
asked Nov 25 '18 at 5:48
CosCos
1326
$endgroup$
$begingroup$
The powers in your very first line do not seem to match those in which you have the limits. If they were the same, you could use your user name, i.e. take $u_1=costheta$ and $u_2=sintheta$.
$endgroup$
– marmot
Nov 25 '18 at 5:51
1
$begingroup$
Has not this question being asked a bazillion times? (i) Is is $x^2 + y^2$ or $x^4+y^2$ in the denominator? In the latter case, I believe there are not directional derivatives. (ii) Consider something of the form $x = sqrt{t}, y = t.$
$endgroup$
– Will M.
Nov 25 '18 at 5:57
$begingroup$
Is $x^{2}+y^{2}$ in the denominator? I have seen other similar functions but still cannot solve for this specific one :( .For (ii) you mean those values as sequences? @WillM.
$endgroup$
– Cos
Nov 25 '18 at 6:07
$begingroup$
By substituting $x=sqrt{t}$ and $y=t$ in $f(x,y)=f(sqrt{t}, t)$ i got $f(sqrt t, t)=1$, so How can i conclude f is not continuous at (0,0)? @WillM.
$endgroup$
– Cos
Nov 25 '18 at 6:15
$begingroup$
$f(0,0)=0$ ${}{}{}{}$
$endgroup$
– Will M.
Nov 25 '18 at 6:47
|
show 3 more comments
$begingroup$
Let $f(x,y)=
begin{cases}
frac{2x^2y}{x^2+y^2} & if & (x,y) neq (0,0) \
0 & if & (x,y)=(0,0) .
end{cases}$
(i) Prove the directional derivatives of $f$ exist in any direction at the point $(0,0)$.
(ii)¿Is $f$ continuous on (0,0)?
For (i) I took $u=(u_{1},u_{2}) in mathbb{R^{2}}$ such $||u||=1$ and $0=(0,0)$. So $$lim_{t to 0} frac{f(0+ t u)-f(0)}{t}=lim_{t to 0}frac{2(tu_{1})^2 (tu_{2})}{(tu_{1})^4+(tu_{2})^2}=lim_{t to 0}frac{(t^2 u_{1}^2)(t^2 u_{1}^{2}) tu_{2}}{t^{4}u_{1}^{4}+t^{2} u_{2}^{2}}$$. But I cannot find the limit when $t$ aproaches to $0$. Which by finding them I would prove the directional derivative exist at every direction right?
For (ii) I got the intuition $f$ is not continuous at $(0,0)$ so I took $lbrace (frac{1}{k},frac{1}{k}) rbrace_{k in mathbb{N}}$ and $lbrace (frac{1}{k},0) rbrace_{k in mathbb{N}}$ which are two different sequences in $mathbb{R}^{2}$ converging to $(0,0)$. However, $lbrace f(frac{1}{k},frac{1}{k}) rbrace_{k in mathbb{N}}=lbrace frac{1}{k^{2}} rbrace_{n in mathbb{N}} to 0$ and $lbrace f(frac{1}{k},0) rbrace_{k in mathbb{N}} to 0$. So maybe my intuition was not right? Can anyone help me end the proof of continuity or not continuity , please?
multivariable-calculus continuity
edited Nov 26 '18 at 9:01
Yadati Kiran
1,751619
asked Nov 25 '18 at 5:48
CosCos
1326
$endgroup$
Let $f(x,y)=
begin{cases}
frac{2x^2y}{x^2+y^2} & if & (x,y) neq (0,0) \
0 & if & (x,y)=(0,0) .
end{cases}$
(i) Prove the directional derivatives of $f$ exist in any direction at the point $(0,0)$.
(ii)¿Is $f$ continuous on (0,0)?
For (i) I took $u=(u_{1},u_{2}) in mathbb{R^{2}}$ such $||u||=1$ and $0=(0,0)$. So $$lim_{t to 0} frac{f(0+ t u)-f(0)}{t}=lim_{t to 0}frac{2(tu_{1})^2 (tu_{2})}{(tu_{1})^4+(tu_{2})^2}=lim_{t to 0}frac{(t^2 u_{1}^2)(t^2 u_{1}^{2}) tu_{2}}{t^{4}u_{1}^{4}+t^{2} u_{2}^{2}}$$. But I cannot find the limit when $t$ aproaches to $0$. Which by finding them I would prove the directional derivative exist at every direction right?
For (ii) I got the intuition $f$ is not continuous at $(0,0)$ so I took $lbrace (frac{1}{k},frac{1}{k}) rbrace_{k in mathbb{N}}$ and $lbrace (frac{1}{k},0) rbrace_{k in mathbb{N}}$ which are two different sequences in $mathbb{R}^{2}$ converging to $(0,0)$. However, $lbrace f(frac{1}{k},frac{1}{k}) rbrace_{k in mathbb{N}}=lbrace frac{1}{k^{2}} rbrace_{n in mathbb{N}} to 0$ and $lbrace f(frac{1}{k},0) rbrace_{k in mathbb{N}} to 0$. So maybe my intuition was not right? Can anyone help me end the proof of continuity or not continuity , please?
multivariable-calculus continuity
multivariable-calculus continuity
edited Nov 26 '18 at 9:01
Yadati Kiran
1,751619
asked Nov 25 '18 at 5:48
CosCos
1326
edited Nov 26 '18 at 9:01
Yadati Kiran
1,751619
asked Nov 25 '18 at 5:48
CosCos
1326
edited Nov 26 '18 at 9:01
Yadati Kiran
1,751619
edited Nov 26 '18 at 9:01
Yadati Kiran
1,751619
edited Nov 26 '18 at 9:01
Yadati Kiran
1,751619
1,751619
asked Nov 25 '18 at 5:48
CosCos
1326
asked Nov 25 '18 at 5:48
CosCos
1326
asked Nov 25 '18 at 5:48
CosCos
1326
1326
$begingroup$
The powers in your very first line do not seem to match those in which you have the limits. If they were the same, you could use your user name, i.e. take $u_1=costheta$ and $u_2=sintheta$.
$endgroup$
– marmot
Nov 25 '18 at 5:51
1
$begingroup$
Has not this question being asked a bazillion times? (i) Is is $x^2 + y^2$ or $x^4+y^2$ in the denominator? In the latter case, I believe there are not directional derivatives. (ii) Consider something of the form $x = sqrt{t}, y = t.$
$endgroup$
– Will M.
Nov 25 '18 at 5:57
$begingroup$
Is $x^{2}+y^{2}$ in the denominator? I have seen other similar functions but still cannot solve for this specific one :( .For (ii) you mean those values as sequences? @WillM.
$endgroup$
– Cos
Nov 25 '18 at 6:07
$begingroup$
By substituting $x=sqrt{t}$ and $y=t$ in $f(x,y)=f(sqrt{t}, t)$ i got $f(sqrt t, t)=1$, so How can i conclude f is not continuous at (0,0)? @WillM.
$endgroup$
– Cos
Nov 25 '18 at 6:15
$begingroup$
$f(0,0)=0$ ${}{}{}{}$
$endgroup$
– Will M.
Nov 25 '18 at 6:47
|
show 3 more comments
$begingroup$
The powers in your very first line do not seem to match those in which you have the limits. If they were the same, you could use your user name, i.e. take $u_1=costheta$ and $u_2=sintheta$.
$endgroup$
– marmot
Nov 25 '18 at 5:51
1
$begingroup$
Has not this question being asked a bazillion times? (i) Is is $x^2 + y^2$ or $x^4+y^2$ in the denominator? In the latter case, I believe there are not directional derivatives. (ii) Consider something of the form $x = sqrt{t}, y = t.$
$endgroup$
– Will M.
Nov 25 '18 at 5:57
$begingroup$
Is $x^{2}+y^{2}$ in the denominator? I have seen other similar functions but still cannot solve for this specific one :( .For (ii) you mean those values as sequences? @WillM.
$endgroup$
– Cos
Nov 25 '18 at 6:07
$begingroup$
By substituting $x=sqrt{t}$ and $y=t$ in $f(x,y)=f(sqrt{t}, t)$ i got $f(sqrt t, t)=1$, so How can i conclude f is not continuous at (0,0)? @WillM.
$endgroup$
– Cos
Nov 25 '18 at 6:15
$begingroup$
$f(0,0)=0$ ${}{}{}{}$
$endgroup$
– Will M.
Nov 25 '18 at 6:47
$begingroup$
The powers in your very first line do not seem to match those in which you have the limits. If they were the same, you could use your user name, i.e. take $u_1=costheta$ and $u_2=sintheta$.
$endgroup$
– marmot
Nov 25 '18 at 5:51
$begingroup$
The powers in your very first line do not seem to match those in which you have the limits. If they were the same, you could use your user name, i.e. take $u_1=costheta$ and $u_2=sintheta$.
$endgroup$
– marmot
Nov 25 '18 at 5:51
1
1
$begingroup$
Has not this question being asked a bazillion times? (i) Is is $x^2 + y^2$ or $x^4+y^2$ in the denominator? In the latter case, I believe there are not directional derivatives. (ii) Consider something of the form $x = sqrt{t}, y = t.$
$endgroup$
– Will M.
Nov 25 '18 at 5:57
$begingroup$
Has not this question being asked a bazillion times? (i) Is is $x^2 + y^2$ or $x^4+y^2$ in the denominator? In the latter case, I believe there are not directional derivatives. (ii) Consider something of the form $x = sqrt{t}, y = t.$
$endgroup$
– Will M.
Nov 25 '18 at 5:57
$begingroup$
Is $x^{2}+y^{2}$ in the denominator? I have seen other similar functions but still cannot solve for this specific one :( .For (ii) you mean those values as sequences? @WillM.
$endgroup$
– Cos
Nov 25 '18 at 6:07
$begingroup$
Is $x^{2}+y^{2}$ in the denominator? I have seen other similar functions but still cannot solve for this specific one :( .For (ii) you mean those values as sequences? @WillM.
$endgroup$
– Cos
Nov 25 '18 at 6:07
$begingroup$
By substituting $x=sqrt{t}$ and $y=t$ in $f(x,y)=f(sqrt{t}, t)$ i got $f(sqrt t, t)=1$, so How can i conclude f is not continuous at (0,0)? @WillM.
$endgroup$
– Cos
Nov 25 '18 at 6:15
$begingroup$
By substituting $x=sqrt{t}$ and $y=t$ in $f(x,y)=f(sqrt{t}, t)$ i got $f(sqrt t, t)=1$, so How can i conclude f is not continuous at (0,0)? @WillM.
$endgroup$
– Cos
Nov 25 '18 at 6:15
$begingroup$
$f(0,0)=0$ ${}{}{}{}$
$endgroup$
– Will M.
Nov 25 '18 at 6:47
$begingroup$
$f(0,0)=0$ ${}{}{}{}$
$endgroup$
– Will M.
Nov 25 '18 at 6:47
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Directional derivative is defined as $$D_vec{u}f(x,y)=lim_{hto0}dfrac{f(x+ah,y+bh)-f(x,y)}{h} $$ where $vec{u}=langle a,brangle$ is the unit vector along which rate of change of $f(x,y)$ is calculated.
$$displaystyle D_vec{u}f(0,0)=lim_{hto0}dfrac{dfrac{2(ah)^2(bh)}{(ah)^2+(bh)^2}-0}{h}= lim_{hto0}frac{2a^2b}{a^2+b^2}$$ which exists since $||vec{u}||=1implies a^2+b^2neq0$. So directional derivatives of $f$ exist in any direction at the point $(0,0)$ since limit is independent of how $h$ approaches $0$.
To calculate continuity let $(x,y)to(0,0)$ along $y=mx$ where $m$ is a constant.
$$displaystylelim_{(x,mxto(0,0))}dfrac{2x^2mx}{x^2(1+m^2)}=lim_{substack{xto 0\text{along y=mx}}}dfrac{2mx}{(1+m^2)}=0=f(0,0)$$
So $f$ is continuous at $(0,0)$.
answered Nov 26 '18 at 8:48
Yadati KiranYadati Kiran
1,751619
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Directional derivative is defined as $$D_vec{u}f(x,y)=lim_{hto0}dfrac{f(x+ah,y+bh)-f(x,y)}{h} $$ where $vec{u}=langle a,brangle$ is the unit vector along which rate of change of $f(x,y)$ is calculated.
$$displaystyle D_vec{u}f(0,0)=lim_{hto0}dfrac{dfrac{2(ah)^2(bh)}{(ah)^2+(bh)^2}-0}{h}= lim_{hto0}frac{2a^2b}{a^2+b^2}$$ which exists since $||vec{u}||=1implies a^2+b^2neq0$. So directional derivatives of $f$ exist in any direction at the point $(0,0)$ since limit is independent of how $h$ approaches $0$.
To calculate continuity let $(x,y)to(0,0)$ along $y=mx$ where $m$ is a constant.
$$displaystylelim_{(x,mxto(0,0))}dfrac{2x^2mx}{x^2(1+m^2)}=lim_{substack{xto 0\text{along y=mx}}}dfrac{2mx}{(1+m^2)}=0=f(0,0)$$
So $f$ is continuous at $(0,0)$.
answered Nov 26 '18 at 8:48
Yadati KiranYadati Kiran
1,751619
$endgroup$
add a comment |
$begingroup$
Directional derivative is defined as $$D_vec{u}f(x,y)=lim_{hto0}dfrac{f(x+ah,y+bh)-f(x,y)}{h} $$ where $vec{u}=langle a,brangle$ is the unit vector along which rate of change of $f(x,y)$ is calculated.
$$displaystyle D_vec{u}f(0,0)=lim_{hto0}dfrac{dfrac{2(ah)^2(bh)}{(ah)^2+(bh)^2}-0}{h}= lim_{hto0}frac{2a^2b}{a^2+b^2}$$ which exists since $||vec{u}||=1implies a^2+b^2neq0$. So directional derivatives of $f$ exist in any direction at the point $(0,0)$ since limit is independent of how $h$ approaches $0$.
To calculate continuity let $(x,y)to(0,0)$ along $y=mx$ where $m$ is a constant.
$$displaystylelim_{(x,mxto(0,0))}dfrac{2x^2mx}{x^2(1+m^2)}=lim_{substack{xto 0\text{along y=mx}}}dfrac{2mx}{(1+m^2)}=0=f(0,0)$$
So $f$ is continuous at $(0,0)$.
answered Nov 26 '18 at 8:48
Yadati KiranYadati Kiran
1,751619
$endgroup$
add a comment |
$begingroup$
Directional derivative is defined as $$D_vec{u}f(x,y)=lim_{hto0}dfrac{f(x+ah,y+bh)-f(x,y)}{h} $$ where $vec{u}=langle a,brangle$ is the unit vector along which rate of change of $f(x,y)$ is calculated.
$$displaystyle D_vec{u}f(0,0)=lim_{hto0}dfrac{dfrac{2(ah)^2(bh)}{(ah)^2+(bh)^2}-0}{h}= lim_{hto0}frac{2a^2b}{a^2+b^2}$$ which exists since $||vec{u}||=1implies a^2+b^2neq0$. So directional derivatives of $f$ exist in any direction at the point $(0,0)$ since limit is independent of how $h$ approaches $0$.
To calculate continuity let $(x,y)to(0,0)$ along $y=mx$ where $m$ is a constant.
$$displaystylelim_{(x,mxto(0,0))}dfrac{2x^2mx}{x^2(1+m^2)}=lim_{substack{xto 0\text{along y=mx}}}dfrac{2mx}{(1+m^2)}=0=f(0,0)$$
So $f$ is continuous at $(0,0)$.
answered Nov 26 '18 at 8:48
Yadati KiranYadati Kiran
1,751619
$endgroup$
Directional derivative is defined as $$D_vec{u}f(x,y)=lim_{hto0}dfrac{f(x+ah,y+bh)-f(x,y)}{h} $$ where $vec{u}=langle a,brangle$ is the unit vector along which rate of change of $f(x,y)$ is calculated.
$$displaystyle D_vec{u}f(0,0)=lim_{hto0}dfrac{dfrac{2(ah)^2(bh)}{(ah)^2+(bh)^2}-0}{h}= lim_{hto0}frac{2a^2b}{a^2+b^2}$$ which exists since $||vec{u}||=1implies a^2+b^2neq0$. So directional derivatives of $f$ exist in any direction at the point $(0,0)$ since limit is independent of how $h$ approaches $0$.
To calculate continuity let $(x,y)to(0,0)$ along $y=mx$ where $m$ is a constant.
$$displaystylelim_{(x,mxto(0,0))}dfrac{2x^2mx}{x^2(1+m^2)}=lim_{substack{xto 0\text{along y=mx}}}dfrac{2mx}{(1+m^2)}=0=f(0,0)$$
So $f$ is continuous at $(0,0)$.
answered Nov 26 '18 at 8:48
Yadati KiranYadati Kiran
1,751619
edited Nov 26 '18 at 8:58
answered Nov 26 '18 at 8:48
Yadati KiranYadati Kiran
1,751619
answered Nov 26 '18 at 8:48
Yadati KiranYadati Kiran
1,751619
answered Nov 26 '18 at 8:48
Yadati KiranYadati Kiran
1,751619
1,751619
add a comment |
add a comment |
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$begingroup$
The powers in your very first line do not seem to match those in which you have the limits. If they were the same, you could use your user name, i.e. take $u_1=costheta$ and $u_2=sintheta$.
$endgroup$
– marmot
Nov 25 '18 at 5:51
1
$begingroup$
Has not this question being asked a bazillion times? (i) Is is $x^2 + y^2$ or $x^4+y^2$ in the denominator? In the latter case, I believe there are not directional derivatives. (ii) Consider something of the form $x = sqrt{t}, y = t.$
$endgroup$
– Will M.
Nov 25 '18 at 5:57
$begingroup$
Is $x^{2}+y^{2}$ in the denominator? I have seen other similar functions but still cannot solve for this specific one :( .For (ii) you mean those values as sequences? @WillM.
$endgroup$
– Cos
Nov 25 '18 at 6:07
$begingroup$
By substituting $x=sqrt{t}$ and $y=t$ in $f(x,y)=f(sqrt{t}, t)$ i got $f(sqrt t, t)=1$, so How can i conclude f is not continuous at (0,0)? @WillM.
$endgroup$
– Cos
Nov 25 '18 at 6:15
$begingroup$
$f(0,0)=0$ ${}{}{}{}$
$endgroup$
– Will M.
Nov 25 '18 at 6:47