Mixing
Continuity and differentiability at (0,0)
Clash Royale CLAN TAG#URR8PPP
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My problem is the following:
Let $lambda>0$. Consider the function on $R^2$:
$f(x,y)=fracx^4+2y^2(x^2+y^2)^lambda$ for $(x,y)ne (0,0)$ and $f(x,y)=0$ at $(x,y)=(0,0)$.
I now want to find the values of $lambda>0$ such that $f(x,y)$ is continuous at $(0,0)$.
Correct me if I am wrong, but if I have understood it correctly, in order for $f(x,y)$ to be conti. at $(0,0)$, the following three must be true:
1, $f(0,0)$ is defined, (here given by $0$).
2, $lim_(x,y)rightarrow (0,0)f(x,y)$ exists.
3, $lim_(x,y)rightarrow (0,0)f(x,y)=f(0,0)(=0)$.
I first consider $lambdain ]0,1]$
I now choose a path $(t,mcdot t)$, and see if the limit exists.
Calculations give that
$lim_(x,y)rightarrow (0,0)f(x,y)$=$lim_trightarrow 0f(t,mcdot t)$=$fract^4+2m^2t^2(t^2+t^2)^lambdaleq lim_trightarrow 0fract^4+2m^2t^2(1+m^2)t^2=frac2m^21+m^2$.
Therefore I concluded that there dosent exist a unique limit for the function $f$ in $lambda in ]0,1]$. Therefore I concluded that cannot be continuous at $(0,0)$ Is this correct concluded?
The second question I have, is to determine the $lambda>0$ such that $f(x,y)$ becomes differentiable at $(0,0)$. I have no idea how to start on this, can anybody help me with that? If I find that there does not exist any $lambda$ for which $f$ is continuous at $(0,0)$, is it then also fair to assume that there does not exist any $lambda>0$ for which $f$ is differentiable at $(0,0)$?
Thanks in advance.
continuity
asked Aug 6 at 8:48
Jonathan Kiersch
708
add a comment |Â
up vote
1
down vote
favorite
My problem is the following:
Let $lambda>0$. Consider the function on $R^2$:
$f(x,y)=fracx^4+2y^2(x^2+y^2)^lambda$ for $(x,y)ne (0,0)$ and $f(x,y)=0$ at $(x,y)=(0,0)$.
I now want to find the values of $lambda>0$ such that $f(x,y)$ is continuous at $(0,0)$.
Correct me if I am wrong, but if I have understood it correctly, in order for $f(x,y)$ to be conti. at $(0,0)$, the following three must be true:
1, $f(0,0)$ is defined, (here given by $0$).
2, $lim_(x,y)rightarrow (0,0)f(x,y)$ exists.
3, $lim_(x,y)rightarrow (0,0)f(x,y)=f(0,0)(=0)$.
I first consider $lambdain ]0,1]$
I now choose a path $(t,mcdot t)$, and see if the limit exists.
Calculations give that
$lim_(x,y)rightarrow (0,0)f(x,y)$=$lim_trightarrow 0f(t,mcdot t)$=$fract^4+2m^2t^2(t^2+t^2)^lambdaleq lim_trightarrow 0fract^4+2m^2t^2(1+m^2)t^2=frac2m^21+m^2$.
Therefore I concluded that there dosent exist a unique limit for the function $f$ in $lambda in ]0,1]$. Therefore I concluded that cannot be continuous at $(0,0)$ Is this correct concluded?
The second question I have, is to determine the $lambda>0$ such that $f(x,y)$ becomes differentiable at $(0,0)$. I have no idea how to start on this, can anybody help me with that? If I find that there does not exist any $lambda$ for which $f$ is continuous at $(0,0)$, is it then also fair to assume that there does not exist any $lambda>0$ for which $f$ is differentiable at $(0,0)$?
Thanks in advance.
continuity
asked Aug 6 at 8:48
Jonathan Kiersch
708
How can that be?
– Jonathan Kiersch
Aug 6 at 9:03
sorry, my earlier comment was wrong.
– Kavi Rama Murthy
Aug 6 at 9:12
I'd recommend using polar coordinates for such problems. It usually makes things a bit easier. For differentiability, you should use its deffinition (check: mathcounterexamples.net/…). Answering your second question; differentiability implies continuity so if there are no $lambda$ for which $f$ is continuous you can conclude there is no differentiability.
– BBC3
Aug 6 at 10:22
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
My problem is the following:
Let $lambda>0$. Consider the function on $R^2$:
$f(x,y)=fracx^4+2y^2(x^2+y^2)^lambda$ for $(x,y)ne (0,0)$ and $f(x,y)=0$ at $(x,y)=(0,0)$.
I now want to find the values of $lambda>0$ such that $f(x,y)$ is continuous at $(0,0)$.
Correct me if I am wrong, but if I have understood it correctly, in order for $f(x,y)$ to be conti. at $(0,0)$, the following three must be true:
1, $f(0,0)$ is defined, (here given by $0$).
2, $lim_(x,y)rightarrow (0,0)f(x,y)$ exists.
3, $lim_(x,y)rightarrow (0,0)f(x,y)=f(0,0)(=0)$.
I first consider $lambdain ]0,1]$
I now choose a path $(t,mcdot t)$, and see if the limit exists.
Calculations give that
$lim_(x,y)rightarrow (0,0)f(x,y)$=$lim_trightarrow 0f(t,mcdot t)$=$fract^4+2m^2t^2(t^2+t^2)^lambdaleq lim_trightarrow 0fract^4+2m^2t^2(1+m^2)t^2=frac2m^21+m^2$.
Therefore I concluded that there dosent exist a unique limit for the function $f$ in $lambda in ]0,1]$. Therefore I concluded that cannot be continuous at $(0,0)$ Is this correct concluded?
The second question I have, is to determine the $lambda>0$ such that $f(x,y)$ becomes differentiable at $(0,0)$. I have no idea how to start on this, can anybody help me with that? If I find that there does not exist any $lambda$ for which $f$ is continuous at $(0,0)$, is it then also fair to assume that there does not exist any $lambda>0$ for which $f$ is differentiable at $(0,0)$?
Thanks in advance.
continuity
asked Aug 6 at 8:48
Jonathan Kiersch
708
My problem is the following:
Let $lambda>0$. Consider the function on $R^2$:
$f(x,y)=fracx^4+2y^2(x^2+y^2)^lambda$ for $(x,y)ne (0,0)$ and $f(x,y)=0$ at $(x,y)=(0,0)$.
I now want to find the values of $lambda>0$ such that $f(x,y)$ is continuous at $(0,0)$.
Correct me if I am wrong, but if I have understood it correctly, in order for $f(x,y)$ to be conti. at $(0,0)$, the following three must be true:
1, $f(0,0)$ is defined, (here given by $0$).
2, $lim_(x,y)rightarrow (0,0)f(x,y)$ exists.
3, $lim_(x,y)rightarrow (0,0)f(x,y)=f(0,0)(=0)$.
I first consider $lambdain ]0,1]$
I now choose a path $(t,mcdot t)$, and see if the limit exists.
Calculations give that
$lim_(x,y)rightarrow (0,0)f(x,y)$=$lim_trightarrow 0f(t,mcdot t)$=$fract^4+2m^2t^2(t^2+t^2)^lambdaleq lim_trightarrow 0fract^4+2m^2t^2(1+m^2)t^2=frac2m^21+m^2$.
Therefore I concluded that there dosent exist a unique limit for the function $f$ in $lambda in ]0,1]$. Therefore I concluded that cannot be continuous at $(0,0)$ Is this correct concluded?
The second question I have, is to determine the $lambda>0$ such that $f(x,y)$ becomes differentiable at $(0,0)$. I have no idea how to start on this, can anybody help me with that? If I find that there does not exist any $lambda$ for which $f$ is continuous at $(0,0)$, is it then also fair to assume that there does not exist any $lambda>0$ for which $f$ is differentiable at $(0,0)$?
Thanks in advance.
continuity
asked Aug 6 at 8:48
Jonathan Kiersch
708
asked Aug 6 at 8:48
Jonathan Kiersch
708
asked Aug 6 at 8:48
Jonathan Kiersch
708
asked Aug 6 at 8:48
Jonathan Kiersch
708
708
How can that be?
– Jonathan Kiersch
Aug 6 at 9:03
sorry, my earlier comment was wrong.
– Kavi Rama Murthy
Aug 6 at 9:12
I'd recommend using polar coordinates for such problems. It usually makes things a bit easier. For differentiability, you should use its deffinition (check: mathcounterexamples.net/…). Answering your second question; differentiability implies continuity so if there are no $lambda$ for which $f$ is continuous you can conclude there is no differentiability.
– BBC3
Aug 6 at 10:22
add a comment |Â
How can that be?
– Jonathan Kiersch
Aug 6 at 9:03
sorry, my earlier comment was wrong.
– Kavi Rama Murthy
Aug 6 at 9:12
I'd recommend using polar coordinates for such problems. It usually makes things a bit easier. For differentiability, you should use its deffinition (check: mathcounterexamples.net/…). Answering your second question; differentiability implies continuity so if there are no $lambda$ for which $f$ is continuous you can conclude there is no differentiability.
– BBC3
Aug 6 at 10:22
How can that be?
– Jonathan Kiersch
Aug 6 at 9:03
How can that be?
– Jonathan Kiersch
Aug 6 at 9:03
sorry, my earlier comment was wrong.
– Kavi Rama Murthy
Aug 6 at 9:12
sorry, my earlier comment was wrong.
– Kavi Rama Murthy
Aug 6 at 9:12
I'd recommend using polar coordinates for such problems. It usually makes things a bit easier. For differentiability, you should use its deffinition (check: mathcounterexamples.net/…). Answering your second question; differentiability implies continuity so if there are no $lambda$ for which $f$ is continuous you can conclude there is no differentiability.
– BBC3
Aug 6 at 10:22
I'd recommend using polar coordinates for such problems. It usually makes things a bit easier. For differentiability, you should use its deffinition (check: mathcounterexamples.net/…). Answering your second question; differentiability implies continuity so if there are no $lambda$ for which $f$ is continuous you can conclude there is no differentiability.
– BBC3
Aug 6 at 10:22
add a comment |Â
1 Answer
1
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votes
up vote
1
down vote
accepted
If $lambda >1$ then $f(0,y)=frac 2 y^2(lambda -1)$ so $|f(0,y)| to infty $ as $y to 0$. Hence $f$ is not continuous for $lambda >1$.
answered Aug 6 at 9:11
Kavi Rama Murthy
21k2830
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If $lambda >1$ then $f(0,y)=frac 2 y^2(lambda -1)$ so $|f(0,y)| to infty $ as $y to 0$. Hence $f$ is not continuous for $lambda >1$.
answered Aug 6 at 9:11
Kavi Rama Murthy
21k2830
add a comment |Â
up vote
1
down vote
accepted
If $lambda >1$ then $f(0,y)=frac 2 y^2(lambda -1)$ so $|f(0,y)| to infty $ as $y to 0$. Hence $f$ is not continuous for $lambda >1$.
answered Aug 6 at 9:11
Kavi Rama Murthy
21k2830
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If $lambda >1$ then $f(0,y)=frac 2 y^2(lambda -1)$ so $|f(0,y)| to infty $ as $y to 0$. Hence $f$ is not continuous for $lambda >1$.
answered Aug 6 at 9:11
Kavi Rama Murthy
21k2830
If $lambda >1$ then $f(0,y)=frac 2 y^2(lambda -1)$ so $|f(0,y)| to infty $ as $y to 0$. Hence $f$ is not continuous for $lambda >1$.
answered Aug 6 at 9:11
Kavi Rama Murthy
21k2830
answered Aug 6 at 9:11
Kavi Rama Murthy
21k2830
answered Aug 6 at 9:11
Kavi Rama Murthy
21k2830
answered Aug 6 at 9:11
Kavi Rama Murthy
21k2830
21k2830
add a comment |Â
add a comment |Â
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How can that be?
– Jonathan Kiersch
Aug 6 at 9:03
sorry, my earlier comment was wrong.
– Kavi Rama Murthy
Aug 6 at 9:12
I'd recommend using polar coordinates for such problems. It usually makes things a bit easier. For differentiability, you should use its deffinition (check: mathcounterexamples.net/…). Answering your second question; differentiability implies continuity so if there are no $lambda$ for which $f$ is continuous you can conclude there is no differentiability.
– BBC3
Aug 6 at 10:22