Continuity of $f(x,y) = frac{x}{y}sin(x^2+y^2)$ if $yneq 0$,$ f(x,0) = 0$












1














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







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    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














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







share|cite|improve this question




















  • 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








1







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







share|cite|improve this question















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






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share|cite|improve this question













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edited Nov 24 at 20:05

























asked Nov 24 at 16:32









Seven

949




949








  • 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




    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










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 $$f(x,-x^3)to -1$$






share|cite|improve this answer























  • 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











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














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 $$f(x,-x^3)to -1$$






share|cite|improve this answer























  • 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
















1














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 $$f(x,-x^3)to -1$$






share|cite|improve this answer























  • 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














1












1








1






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 $$f(x,-x^3)to -1$$






share|cite|improve this answer














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 $$f(x,-x^3)to -1$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 24 at 17:25

























answered Nov 24 at 17:17









hamam_Abdallah

38k21634




38k21634












  • 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










  • @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


















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