Mixing
Proof for continuity: $f (x) = left{ begin{array}{ll} frac{2xy}{x^2+y^2} & (x,y)neq(0,0) \ 0 & ,...
$begingroup$
I have to prove the continuity of the following function:
$f (x) = left{
begin{array}{ll}
frac{2xy}{x^2+y^2} & (x,y)neq(0,0) \
0 & , (x,y)=(0,0) \
end{array}
right. $
Case 1: $(x,y)neq(0,0)$ is obviously continuous
Case 2: $(x,y)=(0,0)$, use $epsilon$-$delta$-criteria
$|f(x,y)-f(0,0)|=|f(x,y)|=|frac{2xy}{x^2+y^2}|=frac{|2xy|}{x^2+y^2}leq2|xy|$
Look at a $delta$-environment of $(0,0)$
$|(x,y)-(0,0)|<delta Rightarrow x^2+y^2<delta^2$
$|x|<delta, |y|<delta Rightarrow |xy|<delta^2$, from where we can see
$|f(x,y)|leq 2|xy|<2delta^2leqepsilon^2 Rightarrow delta < frac{epsilon}{sqrt{2}}$
Therefore the function has to be continuous for all $(x,y)inmathbb{R}^2$
My question is:
Is the proof right and what can I improve in the clarity of my proof?
multivariable-calculus continuity
edited Jan 20 at 13:44
Martin Sleziak
44.8k10119272
asked Jan 20 at 13:40
DeDuTotDeDuTot
11
$endgroup$
add a comment |
$begingroup$
I have to prove the continuity of the following function:
$f (x) = left{
begin{array}{ll}
frac{2xy}{x^2+y^2} & (x,y)neq(0,0) \
0 & , (x,y)=(0,0) \
end{array}
right. $
Case 1: $(x,y)neq(0,0)$ is obviously continuous
Case 2: $(x,y)=(0,0)$, use $epsilon$-$delta$-criteria
$|f(x,y)-f(0,0)|=|f(x,y)|=|frac{2xy}{x^2+y^2}|=frac{|2xy|}{x^2+y^2}leq2|xy|$
Look at a $delta$-environment of $(0,0)$
$|(x,y)-(0,0)|<delta Rightarrow x^2+y^2<delta^2$
$|x|<delta, |y|<delta Rightarrow |xy|<delta^2$, from where we can see
$|f(x,y)|leq 2|xy|<2delta^2leqepsilon^2 Rightarrow delta < frac{epsilon}{sqrt{2}}$
Therefore the function has to be continuous for all $(x,y)inmathbb{R}^2$
My question is:
Is the proof right and what can I improve in the clarity of my proof?
multivariable-calculus continuity
edited Jan 20 at 13:44
Martin Sleziak
44.8k10119272
asked Jan 20 at 13:40
DeDuTotDeDuTot
11
$endgroup$
2
$begingroup$
How do you get $|f(x,y)|le 2|xy|$??
$endgroup$
– Lord Shark the Unknown
Jan 20 at 13:45
1
$begingroup$
You can probably find several post on this site about continuity of this (or a very similar) function, such as Show discontinuity of $frac{xy}{x^2+y^2}$ or Prove that $frac{xy}{x^2+y^2}$ is not continuous If you are mainly asking about checking your own attempt - as opposed to finding any solution of the problem - you should add the (proof-verification) tag (see the tag-info.)
$endgroup$
– Martin Sleziak
Jan 20 at 13:45
add a comment |
$begingroup$
I have to prove the continuity of the following function:
$f (x) = left{
begin{array}{ll}
frac{2xy}{x^2+y^2} & (x,y)neq(0,0) \
0 & , (x,y)=(0,0) \
end{array}
right. $
Case 1: $(x,y)neq(0,0)$ is obviously continuous
Case 2: $(x,y)=(0,0)$, use $epsilon$-$delta$-criteria
$|f(x,y)-f(0,0)|=|f(x,y)|=|frac{2xy}{x^2+y^2}|=frac{|2xy|}{x^2+y^2}leq2|xy|$
Look at a $delta$-environment of $(0,0)$
$|(x,y)-(0,0)|<delta Rightarrow x^2+y^2<delta^2$
$|x|<delta, |y|<delta Rightarrow |xy|<delta^2$, from where we can see
$|f(x,y)|leq 2|xy|<2delta^2leqepsilon^2 Rightarrow delta < frac{epsilon}{sqrt{2}}$
Therefore the function has to be continuous for all $(x,y)inmathbb{R}^2$
My question is:
Is the proof right and what can I improve in the clarity of my proof?
multivariable-calculus continuity
edited Jan 20 at 13:44
Martin Sleziak
44.8k10119272
asked Jan 20 at 13:40
DeDuTotDeDuTot
11
$endgroup$
I have to prove the continuity of the following function:
$f (x) = left{
begin{array}{ll}
frac{2xy}{x^2+y^2} & (x,y)neq(0,0) \
0 & , (x,y)=(0,0) \
end{array}
right. $
Case 1: $(x,y)neq(0,0)$ is obviously continuous
Case 2: $(x,y)=(0,0)$, use $epsilon$-$delta$-criteria
$|f(x,y)-f(0,0)|=|f(x,y)|=|frac{2xy}{x^2+y^2}|=frac{|2xy|}{x^2+y^2}leq2|xy|$
Look at a $delta$-environment of $(0,0)$
$|(x,y)-(0,0)|<delta Rightarrow x^2+y^2<delta^2$
$|x|<delta, |y|<delta Rightarrow |xy|<delta^2$, from where we can see
$|f(x,y)|leq 2|xy|<2delta^2leqepsilon^2 Rightarrow delta < frac{epsilon}{sqrt{2}}$
Therefore the function has to be continuous for all $(x,y)inmathbb{R}^2$
My question is:
Is the proof right and what can I improve in the clarity of my proof?
multivariable-calculus continuity
multivariable-calculus continuity
edited Jan 20 at 13:44
Martin Sleziak
44.8k10119272
asked Jan 20 at 13:40
DeDuTotDeDuTot
11
edited Jan 20 at 13:44
Martin Sleziak
44.8k10119272
asked Jan 20 at 13:40
DeDuTotDeDuTot
11
edited Jan 20 at 13:44
Martin Sleziak
44.8k10119272
edited Jan 20 at 13:44
Martin Sleziak
44.8k10119272
edited Jan 20 at 13:44
Martin Sleziak
44.8k10119272
44.8k10119272
asked Jan 20 at 13:40
DeDuTotDeDuTot
11
asked Jan 20 at 13:40
DeDuTotDeDuTot
11
asked Jan 20 at 13:40
DeDuTotDeDuTot
11
11
2
$begingroup$
How do you get $|f(x,y)|le 2|xy|$??
$endgroup$
– Lord Shark the Unknown
Jan 20 at 13:45
1
$begingroup$
You can probably find several post on this site about continuity of this (or a very similar) function, such as Show discontinuity of $frac{xy}{x^2+y^2}$ or Prove that $frac{xy}{x^2+y^2}$ is not continuous If you are mainly asking about checking your own attempt - as opposed to finding any solution of the problem - you should add the (proof-verification) tag (see the tag-info.)
$endgroup$
– Martin Sleziak
Jan 20 at 13:45
add a comment |
2
$begingroup$
How do you get $|f(x,y)|le 2|xy|$??
$endgroup$
– Lord Shark the Unknown
Jan 20 at 13:45
1
$begingroup$
You can probably find several post on this site about continuity of this (or a very similar) function, such as Show discontinuity of $frac{xy}{x^2+y^2}$ or Prove that $frac{xy}{x^2+y^2}$ is not continuous If you are mainly asking about checking your own attempt - as opposed to finding any solution of the problem - you should add the (proof-verification) tag (see the tag-info.)
$endgroup$
– Martin Sleziak
Jan 20 at 13:45
2
2
$begingroup$
How do you get $|f(x,y)|le 2|xy|$??
$endgroup$
– Lord Shark the Unknown
Jan 20 at 13:45
$begingroup$
How do you get $|f(x,y)|le 2|xy|$??
$endgroup$
– Lord Shark the Unknown
Jan 20 at 13:45
1
1
$begingroup$
You can probably find several post on this site about continuity of this (or a very similar) function, such as Show discontinuity of $frac{xy}{x^2+y^2}$ or Prove that $frac{xy}{x^2+y^2}$ is not continuous If you are mainly asking about checking your own attempt - as opposed to finding any solution of the problem - you should add the (proof-verification) tag (see the tag-info.)
$endgroup$
– Martin Sleziak
Jan 20 at 13:45
$begingroup$
You can probably find several post on this site about continuity of this (or a very similar) function, such as Show discontinuity of $frac{xy}{x^2+y^2}$ or Prove that $frac{xy}{x^2+y^2}$ is not continuous If you are mainly asking about checking your own attempt - as opposed to finding any solution of the problem - you should add the (proof-verification) tag (see the tag-info.)
$endgroup$
– Martin Sleziak
Jan 20 at 13:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The function is not continuous at $(0,0).$ Note that
$$lim_{xto 0} f(x,lambda x)=lim_{xto 0}dfrac{2lambda x^2}{(1+lambda^2)x^2}=dfrac{2lambda}{1+lambda^2}.$$
You use the inequality $|f(x,y)|le 2|xy|$ but note that:
$$frac{2|xy|}{x^2+y^2}=|f(x,y)|leq 2|xy|iff x^2+y^2ge 1.$$ Since you are working on a neighbourhood of $(0,0)$ such inequality is not valid.
answered Jan 20 at 13:45
mflmfl
26.7k12142
$endgroup$
$begingroup$
Mh, what should I get from that "note"?
$endgroup$
– DeDuTot
Jan 20 at 14:02
$begingroup$
If you refer to the first one you should get that the limit depends on the way the function approaches to $(0,0).$ If you refer to the second one you should get that the inequality doesn't hold. Indeed, it is $|f(x,y)|>2|xy|$ if $x^2+y^2<1$ and $(x,y)ne (0,0).$
$endgroup$
– mfl
Jan 20 at 14:08
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
The function is not continuous at $(0,0).$ Note that
$$lim_{xto 0} f(x,lambda x)=lim_{xto 0}dfrac{2lambda x^2}{(1+lambda^2)x^2}=dfrac{2lambda}{1+lambda^2}.$$
You use the inequality $|f(x,y)|le 2|xy|$ but note that:
$$frac{2|xy|}{x^2+y^2}=|f(x,y)|leq 2|xy|iff x^2+y^2ge 1.$$ Since you are working on a neighbourhood of $(0,0)$ such inequality is not valid.
answered Jan 20 at 13:45
mflmfl
26.7k12142
$endgroup$
$begingroup$
Mh, what should I get from that "note"?
$endgroup$
– DeDuTot
Jan 20 at 14:02
$begingroup$
If you refer to the first one you should get that the limit depends on the way the function approaches to $(0,0).$ If you refer to the second one you should get that the inequality doesn't hold. Indeed, it is $|f(x,y)|>2|xy|$ if $x^2+y^2<1$ and $(x,y)ne (0,0).$
$endgroup$
– mfl
Jan 20 at 14:08
add a comment |
$begingroup$
The function is not continuous at $(0,0).$ Note that
$$lim_{xto 0} f(x,lambda x)=lim_{xto 0}dfrac{2lambda x^2}{(1+lambda^2)x^2}=dfrac{2lambda}{1+lambda^2}.$$
You use the inequality $|f(x,y)|le 2|xy|$ but note that:
$$frac{2|xy|}{x^2+y^2}=|f(x,y)|leq 2|xy|iff x^2+y^2ge 1.$$ Since you are working on a neighbourhood of $(0,0)$ such inequality is not valid.
answered Jan 20 at 13:45
mflmfl
26.7k12142
$endgroup$
$begingroup$
Mh, what should I get from that "note"?
$endgroup$
– DeDuTot
Jan 20 at 14:02
$begingroup$
If you refer to the first one you should get that the limit depends on the way the function approaches to $(0,0).$ If you refer to the second one you should get that the inequality doesn't hold. Indeed, it is $|f(x,y)|>2|xy|$ if $x^2+y^2<1$ and $(x,y)ne (0,0).$
$endgroup$
– mfl
Jan 20 at 14:08
add a comment |
$begingroup$
The function is not continuous at $(0,0).$ Note that
$$lim_{xto 0} f(x,lambda x)=lim_{xto 0}dfrac{2lambda x^2}{(1+lambda^2)x^2}=dfrac{2lambda}{1+lambda^2}.$$
You use the inequality $|f(x,y)|le 2|xy|$ but note that:
$$frac{2|xy|}{x^2+y^2}=|f(x,y)|leq 2|xy|iff x^2+y^2ge 1.$$ Since you are working on a neighbourhood of $(0,0)$ such inequality is not valid.
answered Jan 20 at 13:45
mflmfl
26.7k12142
$endgroup$
The function is not continuous at $(0,0).$ Note that
$$lim_{xto 0} f(x,lambda x)=lim_{xto 0}dfrac{2lambda x^2}{(1+lambda^2)x^2}=dfrac{2lambda}{1+lambda^2}.$$
You use the inequality $|f(x,y)|le 2|xy|$ but note that:
$$frac{2|xy|}{x^2+y^2}=|f(x,y)|leq 2|xy|iff x^2+y^2ge 1.$$ Since you are working on a neighbourhood of $(0,0)$ such inequality is not valid.
answered Jan 20 at 13:45
mflmfl
26.7k12142
answered Jan 20 at 13:45
mflmfl
26.7k12142
answered Jan 20 at 13:45
mflmfl
26.7k12142
answered Jan 20 at 13:45
mflmfl
26.7k12142
26.7k12142
$begingroup$
Mh, what should I get from that "note"?
$endgroup$
– DeDuTot
Jan 20 at 14:02
$begingroup$
If you refer to the first one you should get that the limit depends on the way the function approaches to $(0,0).$ If you refer to the second one you should get that the inequality doesn't hold. Indeed, it is $|f(x,y)|>2|xy|$ if $x^2+y^2<1$ and $(x,y)ne (0,0).$
$endgroup$
– mfl
Jan 20 at 14:08
add a comment |
$begingroup$
Mh, what should I get from that "note"?
$endgroup$
– DeDuTot
Jan 20 at 14:02
$begingroup$
If you refer to the first one you should get that the limit depends on the way the function approaches to $(0,0).$ If you refer to the second one you should get that the inequality doesn't hold. Indeed, it is $|f(x,y)|>2|xy|$ if $x^2+y^2<1$ and $(x,y)ne (0,0).$
$endgroup$
– mfl
Jan 20 at 14:08
$begingroup$
Mh, what should I get from that "note"?
$endgroup$
– DeDuTot
Jan 20 at 14:02
$begingroup$
Mh, what should I get from that "note"?
$endgroup$
– DeDuTot
Jan 20 at 14:02
$begingroup$
If you refer to the first one you should get that the limit depends on the way the function approaches to $(0,0).$ If you refer to the second one you should get that the inequality doesn't hold. Indeed, it is $|f(x,y)|>2|xy|$ if $x^2+y^2<1$ and $(x,y)ne (0,0).$
$endgroup$
– mfl
Jan 20 at 14:08
$begingroup$
If you refer to the first one you should get that the limit depends on the way the function approaches to $(0,0).$ If you refer to the second one you should get that the inequality doesn't hold. Indeed, it is $|f(x,y)|>2|xy|$ if $x^2+y^2<1$ and $(x,y)ne (0,0).$
$endgroup$
– mfl
Jan 20 at 14:08
add a comment |
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2
$begingroup$
How do you get $|f(x,y)|le 2|xy|$??
$endgroup$
– Lord Shark the Unknown
Jan 20 at 13:45
1
$begingroup$
You can probably find several post on this site about continuity of this (or a very similar) function, such as Show discontinuity of $frac{xy}{x^2+y^2}$ or Prove that $frac{xy}{x^2+y^2}$ is not continuous If you are mainly asking about checking your own attempt - as opposed to finding any solution of the problem - you should add the (proof-verification) tag (see the tag-info.)
$endgroup$
– Martin Sleziak
Jan 20 at 13:45