Continuity of $f(x,y) = frac{x}{y}sin(x^2+y^2)$ if $yneq 0$,$ f(x,0) = 0$
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This function is continuous in $c=(x,y)$ when $yneq0$ and it is not continuous when $y=0$ and $sin x^2 neq 0$. I think it is also not continuous when $y=0$ and $sin(x^2) = 0$ , particularly at $(0,0)$. However, in this case, I can´t find a subset for which the limit is different from $0$. I know there are other ways to prove this, but it would be much easier if I could find a proper set. Can you help me?
calculus continuity
asked Nov 24 at 16:32
Seven
829
add a comment |
up vote
1
down vote
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This function is continuous in $c=(x,y)$ when $yneq0$ and it is not continuous when $y=0$ and $sin x^2 neq 0$. I think it is also not continuous when $y=0$ and $sin(x^2) = 0$ , particularly at $(0,0)$. However, in this case, I can´t find a subset for which the limit is different from $0$. I know there are other ways to prove this, but it would be much easier if I could find a proper set. Can you help me?
calculus continuity
asked Nov 24 at 16:32
Seven
829
1
I might be wrong, but I think the function is not continuous when $x=sqrt{pi/2}$ (meaning $sin^2xne 0$) and $yto 0$
– Andrei
Nov 24 at 16:43
Look again, the square is over the $x$ not the $sin$. I'll add some parentheses to make it clearer though
– Seven
Nov 24 at 16:55
1
So at that point $x^2=pi/2$, the sine term is equal to $1$, and $x/yto (pi/2)/0=infty$
– Andrei
Nov 24 at 18:58
Yeah sorry, then it's also discontinuous. I will edit
– Seven
Nov 24 at 20:04
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This function is continuous in $c=(x,y)$ when $yneq0$ and it is not continuous when $y=0$ and $sin x^2 neq 0$. I think it is also not continuous when $y=0$ and $sin(x^2) = 0$ , particularly at $(0,0)$. However, in this case, I can´t find a subset for which the limit is different from $0$. I know there are other ways to prove this, but it would be much easier if I could find a proper set. Can you help me?
calculus continuity
asked Nov 24 at 16:32
Seven
829
This function is continuous in $c=(x,y)$ when $yneq0$ and it is not continuous when $y=0$ and $sin x^2 neq 0$. I think it is also not continuous when $y=0$ and $sin(x^2) = 0$ , particularly at $(0,0)$. However, in this case, I can´t find a subset for which the limit is different from $0$. I know there are other ways to prove this, but it would be much easier if I could find a proper set. Can you help me?
calculus continuity
calculus continuity
asked Nov 24 at 16:32
Seven
829
asked Nov 24 at 16:32
Seven
829
edited Nov 24 at 20:05
asked Nov 24 at 16:32
Seven
829
asked Nov 24 at 16:32
Seven
829
asked Nov 24 at 16:32
Seven
829
829
1
I might be wrong, but I think the function is not continuous when $x=sqrt{pi/2}$ (meaning $sin^2xne 0$) and $yto 0$
– Andrei
Nov 24 at 16:43
Look again, the square is over the $x$ not the $sin$. I'll add some parentheses to make it clearer though
– Seven
Nov 24 at 16:55
1
So at that point $x^2=pi/2$, the sine term is equal to $1$, and $x/yto (pi/2)/0=infty$
– Andrei
Nov 24 at 18:58
Yeah sorry, then it's also discontinuous. I will edit
– Seven
Nov 24 at 20:04
add a comment |
1
I might be wrong, but I think the function is not continuous when $x=sqrt{pi/2}$ (meaning $sin^2xne 0$) and $yto 0$
– Andrei
Nov 24 at 16:43
Look again, the square is over the $x$ not the $sin$. I'll add some parentheses to make it clearer though
– Seven
Nov 24 at 16:55
1
So at that point $x^2=pi/2$, the sine term is equal to $1$, and $x/yto (pi/2)/0=infty$
– Andrei
Nov 24 at 18:58
Yeah sorry, then it's also discontinuous. I will edit
– Seven
Nov 24 at 20:04
1
1
I might be wrong, but I think the function is not continuous when $x=sqrt{pi/2}$ (meaning $sin^2xne 0$) and $yto 0$
– Andrei
Nov 24 at 16:43
I might be wrong, but I think the function is not continuous when $x=sqrt{pi/2}$ (meaning $sin^2xne 0$) and $yto 0$
– Andrei
Nov 24 at 16:43
Look again, the square is over the $x$ not the $sin$. I'll add some parentheses to make it clearer though
– Seven
Nov 24 at 16:55
Look again, the square is over the $x$ not the $sin$. I'll add some parentheses to make it clearer though
– Seven
Nov 24 at 16:55
1
1
So at that point $x^2=pi/2$, the sine term is equal to $1$, and $x/yto (pi/2)/0=infty$
– Andrei
Nov 24 at 18:58
So at that point $x^2=pi/2$, the sine term is equal to $1$, and $x/yto (pi/2)/0=infty$
– Andrei
Nov 24 at 18:58
Yeah sorry, then it's also discontinuous. I will edit
– Seven
Nov 24 at 20:04
Yeah sorry, then it's also discontinuous. I will edit
– Seven
Nov 24 at 20:04
add a comment |
1 Answer
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hint
$$f(x,y)=frac xysin(x^2+y^2)=$$
$$xsin(x^2)frac{cos(y^2)}{y}+xfrac{sin(y^2)}{y}cos(x^2)$$
$$f(x,x^3)to 1$$ and $$f(x,-x^3)to -1$$
answered Nov 24 at 17:17
hamam_Abdallah
37.8k21634
This is OK for (0,0), what should I do in other cases?
– Seven
Nov 24 at 18:52
@Seven As you said, its continuous if $yne $ and at $(x,0) $ if $xne 0$.
– hamam_Abdallah
Nov 24 at 19:00
And when $sin(x^2)= 0$?
– Seven
Nov 24 at 20:01
You can't take $y=x^3$ in the general case because the point has to be a cluster point of the set chosen
– Seven
Nov 24 at 20:09
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
hint
$$f(x,y)=frac xysin(x^2+y^2)=$$
$$xsin(x^2)frac{cos(y^2)}{y}+xfrac{sin(y^2)}{y}cos(x^2)$$
$$f(x,x^3)to 1$$ and $$f(x,-x^3)to -1$$
answered Nov 24 at 17:17
hamam_Abdallah
37.8k21634
This is OK for (0,0), what should I do in other cases?
– Seven
Nov 24 at 18:52
@Seven As you said, its continuous if $yne $ and at $(x,0) $ if $xne 0$.
– hamam_Abdallah
Nov 24 at 19:00
And when $sin(x^2)= 0$?
– Seven
Nov 24 at 20:01
You can't take $y=x^3$ in the general case because the point has to be a cluster point of the set chosen
– Seven
Nov 24 at 20:09
add a comment |
up vote
1
down vote
hint
$$f(x,y)=frac xysin(x^2+y^2)=$$
$$xsin(x^2)frac{cos(y^2)}{y}+xfrac{sin(y^2)}{y}cos(x^2)$$
$$f(x,x^3)to 1$$ and $$f(x,-x^3)to -1$$
answered Nov 24 at 17:17
hamam_Abdallah
37.8k21634
This is OK for (0,0), what should I do in other cases?
– Seven
Nov 24 at 18:52
@Seven As you said, its continuous if $yne $ and at $(x,0) $ if $xne 0$.
– hamam_Abdallah
Nov 24 at 19:00
And when $sin(x^2)= 0$?
– Seven
Nov 24 at 20:01
You can't take $y=x^3$ in the general case because the point has to be a cluster point of the set chosen
– Seven
Nov 24 at 20:09
add a comment |
up vote
1
down vote
up vote
1
down vote
hint
$$f(x,y)=frac xysin(x^2+y^2)=$$
$$xsin(x^2)frac{cos(y^2)}{y}+xfrac{sin(y^2)}{y}cos(x^2)$$
$$f(x,x^3)to 1$$ and $$f(x,-x^3)to -1$$
answered Nov 24 at 17:17
hamam_Abdallah
37.8k21634
hint
$$f(x,y)=frac xysin(x^2+y^2)=$$
$$xsin(x^2)frac{cos(y^2)}{y}+xfrac{sin(y^2)}{y}cos(x^2)$$
$$f(x,x^3)to 1$$ and $$f(x,-x^3)to -1$$
answered Nov 24 at 17:17
hamam_Abdallah
37.8k21634
edited Nov 24 at 17:25
answered Nov 24 at 17:17
hamam_Abdallah
37.8k21634
answered Nov 24 at 17:17
hamam_Abdallah
37.8k21634
answered Nov 24 at 17:17
hamam_Abdallah
37.8k21634
37.8k21634
This is OK for (0,0), what should I do in other cases?
– Seven
Nov 24 at 18:52
@Seven As you said, its continuous if $yne $ and at $(x,0) $ if $xne 0$.
– hamam_Abdallah
Nov 24 at 19:00
And when $sin(x^2)= 0$?
– Seven
Nov 24 at 20:01
You can't take $y=x^3$ in the general case because the point has to be a cluster point of the set chosen
– Seven
Nov 24 at 20:09
add a comment |
This is OK for (0,0), what should I do in other cases?
– Seven
Nov 24 at 18:52
@Seven As you said, its continuous if $yne $ and at $(x,0) $ if $xne 0$.
– hamam_Abdallah
Nov 24 at 19:00
And when $sin(x^2)= 0$?
– Seven
Nov 24 at 20:01
You can't take $y=x^3$ in the general case because the point has to be a cluster point of the set chosen
– Seven
Nov 24 at 20:09
This is OK for (0,0), what should I do in other cases?
– Seven
Nov 24 at 18:52
This is OK for (0,0), what should I do in other cases?
– Seven
Nov 24 at 18:52
@Seven As you said, its continuous if $yne $ and at $(x,0) $ if $xne 0$.
– hamam_Abdallah
Nov 24 at 19:00
@Seven As you said, its continuous if $yne $ and at $(x,0) $ if $xne 0$.
– hamam_Abdallah
Nov 24 at 19:00
And when $sin(x^2)= 0$?
– Seven
Nov 24 at 20:01
And when $sin(x^2)= 0$?
– Seven
Nov 24 at 20:01
You can't take $y=x^3$ in the general case because the point has to be a cluster point of the set chosen
– Seven
Nov 24 at 20:09
You can't take $y=x^3$ in the general case because the point has to be a cluster point of the set chosen
– Seven
Nov 24 at 20:09
add a comment |
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1
I might be wrong, but I think the function is not continuous when $x=sqrt{pi/2}$ (meaning $sin^2xne 0$) and $yto 0$
– Andrei
Nov 24 at 16:43
Look again, the square is over the $x$ not the $sin$. I'll add some parentheses to make it clearer though
– Seven
Nov 24 at 16:55
1
So at that point $x^2=pi/2$, the sine term is equal to $1$, and $x/yto (pi/2)/0=infty$
– Andrei
Nov 24 at 18:58
Yeah sorry, then it's also discontinuous. I will edit
– Seven
Nov 24 at 20:04