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Show continuity at $(0,0)$ of $f(x,y)=fracxysqrtx^2+y^2$ for $(x,y)neq (0,0)$ and $f(0,0)=0$
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Show continuity at $(0,0)$ of $f(x,y)=fracxysqrtx^2+y^2$ for $(x,y)neq (0,0)$ and $f(0,0)=0$.
A solution I saw was to write
$$lim_(x,y)to 0xdfracysqrtx^2+y^2$$
$$lim_(x,y)to 0x cdot lim_(x,y)to 0dfracysqrtx^2+y^2$$
The term on the left tends to $0$ and the term on the right has absolute value less than $1$.
My Question
Originally I tried to use polar coordinates to solve.
$$f(x,y)=f(rcostheta,rsintheta)$$
$$f(x,y)=dfracxysqrtx^2+y^2= rcostheta sintheta$$
But I wasn't sure how to proceed. I'm not comfortable working in polar form but I'm thinking it's obvious that $rcosthetasintheta to 0$ as $(rcostheta,rsintheta)$.
Is it the case that for $(rcostheta,rsintheta) to (0,0)$ we must have that $r to 0$ since when $sintheta=0 Rightarrow costheta neq 0$.
real-analysis analysis multivariable-calculus continuity proof-writing
edited Jul 28 at 16:26
Did
242k23208441
asked Jul 28 at 14:21
john fowles
1,088817
add a comment |Â
up vote
1
down vote
favorite
Show continuity at $(0,0)$ of $f(x,y)=fracxysqrtx^2+y^2$ for $(x,y)neq (0,0)$ and $f(0,0)=0$.
A solution I saw was to write
$$lim_(x,y)to 0xdfracysqrtx^2+y^2$$
$$lim_(x,y)to 0x cdot lim_(x,y)to 0dfracysqrtx^2+y^2$$
The term on the left tends to $0$ and the term on the right has absolute value less than $1$.
My Question
Originally I tried to use polar coordinates to solve.
$$f(x,y)=f(rcostheta,rsintheta)$$
$$f(x,y)=dfracxysqrtx^2+y^2= rcostheta sintheta$$
But I wasn't sure how to proceed. I'm not comfortable working in polar form but I'm thinking it's obvious that $rcosthetasintheta to 0$ as $(rcostheta,rsintheta)$.
Is it the case that for $(rcostheta,rsintheta) to (0,0)$ we must have that $r to 0$ since when $sintheta=0 Rightarrow costheta neq 0$.
real-analysis analysis multivariable-calculus continuity proof-writing
edited Jul 28 at 16:26
Did
242k23208441
asked Jul 28 at 14:21
john fowles
1,088817
Nodfracin titles please (and actually, anywhere).
– Did
Jul 28 at 14:44
9 minutes. $ $ $ $
– Did
Jul 28 at 14:45
What's wrong with dfrac?
– john fowles
Jul 28 at 15:17
Should be $(x,y)to (0,0).$
– zhw.
Jul 28 at 15:48
Too much vertical space taken in the list of questions, with neither need nor use.
– Did
Jul 28 at 16:26
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Show continuity at $(0,0)$ of $f(x,y)=fracxysqrtx^2+y^2$ for $(x,y)neq (0,0)$ and $f(0,0)=0$.
A solution I saw was to write
$$lim_(x,y)to 0xdfracysqrtx^2+y^2$$
$$lim_(x,y)to 0x cdot lim_(x,y)to 0dfracysqrtx^2+y^2$$
The term on the left tends to $0$ and the term on the right has absolute value less than $1$.
My Question
Originally I tried to use polar coordinates to solve.
$$f(x,y)=f(rcostheta,rsintheta)$$
$$f(x,y)=dfracxysqrtx^2+y^2= rcostheta sintheta$$
But I wasn't sure how to proceed. I'm not comfortable working in polar form but I'm thinking it's obvious that $rcosthetasintheta to 0$ as $(rcostheta,rsintheta)$.
Is it the case that for $(rcostheta,rsintheta) to (0,0)$ we must have that $r to 0$ since when $sintheta=0 Rightarrow costheta neq 0$.
real-analysis analysis multivariable-calculus continuity proof-writing
edited Jul 28 at 16:26
Did
242k23208441
asked Jul 28 at 14:21
john fowles
1,088817
Show continuity at $(0,0)$ of $f(x,y)=fracxysqrtx^2+y^2$ for $(x,y)neq (0,0)$ and $f(0,0)=0$.
A solution I saw was to write
$$lim_(x,y)to 0xdfracysqrtx^2+y^2$$
$$lim_(x,y)to 0x cdot lim_(x,y)to 0dfracysqrtx^2+y^2$$
The term on the left tends to $0$ and the term on the right has absolute value less than $1$.
My Question
Originally I tried to use polar coordinates to solve.
$$f(x,y)=f(rcostheta,rsintheta)$$
$$f(x,y)=dfracxysqrtx^2+y^2= rcostheta sintheta$$
But I wasn't sure how to proceed. I'm not comfortable working in polar form but I'm thinking it's obvious that $rcosthetasintheta to 0$ as $(rcostheta,rsintheta)$.
Is it the case that for $(rcostheta,rsintheta) to (0,0)$ we must have that $r to 0$ since when $sintheta=0 Rightarrow costheta neq 0$.
real-analysis analysis multivariable-calculus continuity proof-writing
edited Jul 28 at 16:26
Did
242k23208441
asked Jul 28 at 14:21
john fowles
1,088817
edited Jul 28 at 16:26
Did
242k23208441
edited Jul 28 at 16:26
Did
242k23208441
edited Jul 28 at 16:26
Did
242k23208441
242k23208441
asked Jul 28 at 14:21
john fowles
1,088817
asked Jul 28 at 14:21
john fowles
1,088817
asked Jul 28 at 14:21
john fowles
1,088817
1,088817
Nodfracin titles please (and actually, anywhere).
– Did
Jul 28 at 14:44
9 minutes. $ $ $ $
– Did
Jul 28 at 14:45
What's wrong with dfrac?
– john fowles
Jul 28 at 15:17
Should be $(x,y)to (0,0).$
– zhw.
Jul 28 at 15:48
Too much vertical space taken in the list of questions, with neither need nor use.
– Did
Jul 28 at 16:26
add a comment |Â
Nodfracin titles please (and actually, anywhere).
– Did
Jul 28 at 14:44
9 minutes. $ $ $ $
– Did
Jul 28 at 14:45
What's wrong with dfrac?
– john fowles
Jul 28 at 15:17
Should be $(x,y)to (0,0).$
– zhw.
Jul 28 at 15:48
Too much vertical space taken in the list of questions, with neither need nor use.
– Did
Jul 28 at 16:26
No
dfrac in titles please (and actually, anywhere).– Did
Jul 28 at 14:44
No
dfrac in titles please (and actually, anywhere).– Did
Jul 28 at 14:44
9 minutes. $ $ $ $
– Did
Jul 28 at 14:45
9 minutes. $ $ $ $
– Did
Jul 28 at 14:45
What's wrong with dfrac?
– john fowles
Jul 28 at 15:17
What's wrong with dfrac?
– john fowles
Jul 28 at 15:17
Should be $(x,y)to (0,0).$
– zhw.
Jul 28 at 15:48
Should be $(x,y)to (0,0).$
– zhw.
Jul 28 at 15:48
Too much vertical space taken in the list of questions, with neither need nor use.
– Did
Jul 28 at 16:26
Too much vertical space taken in the list of questions, with neither need nor use.
– Did
Jul 28 at 16:26
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
1
down vote
accepted
You can use $|sintheta|leq1$ and $|costheta|leq1$ to make
$$rsinthetacosthetato0$$
as $rto0$. Like Descartes coordinates here we have both $rto0$ and $thetato0$ for evaluating the limit. Then
$$lim_(x,y)to(0,0)f(x,y)=lim_(r,theta)to(0,0)rcosthetasintheta to0$$
means the function is continuous in origin.
answered Jul 28 at 14:30
user 108128
19k41544
1
Why do we need to let $theta to 0$ ?
– ComplexYetTrivial
Jul 28 at 15:12
I said !$$
– user 108128
Jul 28 at 15:14
But then we only approach the origin from one direction (the $x$-axis). If we consider $f(x,y) = fracx^2+y^2y = fracrsin theta$ (for $y neq 0$), we get the correct result $lim_r to 0 fracrsin theta = 0 = lim_(x,y) to (0,0) f(x,y)$, but $lim_(r,theta) to (0,0) fracrsin theta$ does not exist. So even though it makes no difference in the particular problem from the question, I think we should only let $r to 0$ in general.
– ComplexYetTrivial
Jul 28 at 15:28
You omit $foralltheta$, $sinleq1$, in your's the function isn't bounded in origin. Our case is exactly squeez theorem.
– user 108128
Jul 28 at 15:32
1
Yes, since the function in the original problem is a bounded function of $theta$, letting $theta to 0$ does not change the result in this case. However, it is not necessary and even wrong in other cases.
– ComplexYetTrivial
Jul 28 at 15:39
add a comment |Â
up vote
2
down vote
If you are not comfortable with polar coordinates, your remark
"...and the term on the right has absolute value less than 1"
suffices to show that $left|x fracysqrtx^2+y^2 right|leq left| x right| leq sqrtx^2+y^2$ and thus $f(x,y) rightarrow0$ as $(x,y)rightarrow 0$.
answered Jul 28 at 14:45
SEBASTIAN VARGAS LOAIZA
1464
add a comment |Â
up vote
0
down vote
We can also use that
$$fracxysqrtx^2+y^2le fracxysqrt2sqrtxy=frac1sqrt2|xy|^1/2$$
answered Jul 28 at 16:06
Dr. Sonnhard Graubner
66.7k32659
add a comment |Â
up vote
0
down vote
Your argument is correct: $(r cos theta, r sin theta) = (0,0)$ holds if and only if $r = 0$ , so approaching the origin in polar coordinates indeed corresponds to letting $r to 0$ . You can, however, approach the origin from different directions, which is something you need to be careful about when working in polar coordinates! The limit $ lim_(x,y) to (0,0) f(x,y)$ exists if and only if the limits
$$ lim_r to 0 f(r cos vartheta(r), r sin vartheta(r))$$
exist and agree for every parametrisation $vartheta: (0,1) to [0,2pi)$ of the polar angle.
In your problem this works out:
$$ lim_rto 0 f(r cos vartheta(r), r sin vartheta(r)) = lim_rto 0 , r cos vartheta(r) sin vartheta(r) = 0$$
holds for every $vartheta$ since $cos$ and $sin$ are bounded by $1$ .
If we consider $g(x,y) = fracxsqrtx^2 + y^2$ for $(x,y) neq (0,0)$ instead, things are quite different. We have
$$ g(r cos(theta), r sin(theta)) = cos theta ,$$
so
$$ lim_r to 0 g(r cos(0), r sin(0)) = 1 neq -1 = lim_r to 0 g(r cos(pi), r sin(pi)) , .$$
Therefore the limit $lim_(x,y) to (0,0) g(x,y)$ does not exist and we cannot choose a value for $g(0,0)$ in order to make $g$ continuous.
A better example of the issue is provided by this question. Consider
$$ h(x,y) = fracx^2 yx^4 + y^2 = fracr cos^2 theta sin thetar^2 cos^4 theta + sin^2 theta , . $$
At first glance one would probably expect the limit $0$ at the origin, given the above representation in polar coordinates. However, along the parabolic path $y = x^2$ the limit is equal to $frac12$, so the limit $lim_(x,y) to (0,0)h(x,y)$ does not exist.
In polar coordinates this path is parametrised implicitly by
$$fracsin vartheta(r)cos^2 vartheta(r) = r , $$
or explicitly by
$$ sin vartheta(r) = fracsqrt1+4 r^2-12r , , $$
which is not easy to spot. This illustrates the fact that using polar coordinates is possible, but not always advisable when working on these problems.
answered Jul 28 at 15:11
ComplexYetTrivial
2,787624
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You can use $|sintheta|leq1$ and $|costheta|leq1$ to make
$$rsinthetacosthetato0$$
as $rto0$. Like Descartes coordinates here we have both $rto0$ and $thetato0$ for evaluating the limit. Then
$$lim_(x,y)to(0,0)f(x,y)=lim_(r,theta)to(0,0)rcosthetasintheta to0$$
means the function is continuous in origin.
answered Jul 28 at 14:30
user 108128
19k41544
1
Why do we need to let $theta to 0$ ?
– ComplexYetTrivial
Jul 28 at 15:12
I said !$$
– user 108128
Jul 28 at 15:14
But then we only approach the origin from one direction (the $x$-axis). If we consider $f(x,y) = fracx^2+y^2y = fracrsin theta$ (for $y neq 0$), we get the correct result $lim_r to 0 fracrsin theta = 0 = lim_(x,y) to (0,0) f(x,y)$, but $lim_(r,theta) to (0,0) fracrsin theta$ does not exist. So even though it makes no difference in the particular problem from the question, I think we should only let $r to 0$ in general.
– ComplexYetTrivial
Jul 28 at 15:28
You omit $foralltheta$, $sinleq1$, in your's the function isn't bounded in origin. Our case is exactly squeez theorem.
– user 108128
Jul 28 at 15:32
1
Yes, since the function in the original problem is a bounded function of $theta$, letting $theta to 0$ does not change the result in this case. However, it is not necessary and even wrong in other cases.
– ComplexYetTrivial
Jul 28 at 15:39
add a comment |Â
up vote
1
down vote
accepted
You can use $|sintheta|leq1$ and $|costheta|leq1$ to make
$$rsinthetacosthetato0$$
as $rto0$. Like Descartes coordinates here we have both $rto0$ and $thetato0$ for evaluating the limit. Then
$$lim_(x,y)to(0,0)f(x,y)=lim_(r,theta)to(0,0)rcosthetasintheta to0$$
means the function is continuous in origin.
answered Jul 28 at 14:30
user 108128
19k41544
1
Why do we need to let $theta to 0$ ?
– ComplexYetTrivial
Jul 28 at 15:12
I said !$$
– user 108128
Jul 28 at 15:14
But then we only approach the origin from one direction (the $x$-axis). If we consider $f(x,y) = fracx^2+y^2y = fracrsin theta$ (for $y neq 0$), we get the correct result $lim_r to 0 fracrsin theta = 0 = lim_(x,y) to (0,0) f(x,y)$, but $lim_(r,theta) to (0,0) fracrsin theta$ does not exist. So even though it makes no difference in the particular problem from the question, I think we should only let $r to 0$ in general.
– ComplexYetTrivial
Jul 28 at 15:28
You omit $foralltheta$, $sinleq1$, in your's the function isn't bounded in origin. Our case is exactly squeez theorem.
– user 108128
Jul 28 at 15:32
1
Yes, since the function in the original problem is a bounded function of $theta$, letting $theta to 0$ does not change the result in this case. However, it is not necessary and even wrong in other cases.
– ComplexYetTrivial
Jul 28 at 15:39
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You can use $|sintheta|leq1$ and $|costheta|leq1$ to make
$$rsinthetacosthetato0$$
as $rto0$. Like Descartes coordinates here we have both $rto0$ and $thetato0$ for evaluating the limit. Then
$$lim_(x,y)to(0,0)f(x,y)=lim_(r,theta)to(0,0)rcosthetasintheta to0$$
means the function is continuous in origin.
answered Jul 28 at 14:30
user 108128
19k41544
You can use $|sintheta|leq1$ and $|costheta|leq1$ to make
$$rsinthetacosthetato0$$
as $rto0$. Like Descartes coordinates here we have both $rto0$ and $thetato0$ for evaluating the limit. Then
$$lim_(x,y)to(0,0)f(x,y)=lim_(r,theta)to(0,0)rcosthetasintheta to0$$
means the function is continuous in origin.
answered Jul 28 at 14:30
user 108128
19k41544
edited Jul 28 at 14:41
answered Jul 28 at 14:30
user 108128
19k41544
answered Jul 28 at 14:30
user 108128
19k41544
answered Jul 28 at 14:30
user 108128
19k41544
19k41544
1
Why do we need to let $theta to 0$ ?
– ComplexYetTrivial
Jul 28 at 15:12
I said !$$
– user 108128
Jul 28 at 15:14
But then we only approach the origin from one direction (the $x$-axis). If we consider $f(x,y) = fracx^2+y^2y = fracrsin theta$ (for $y neq 0$), we get the correct result $lim_r to 0 fracrsin theta = 0 = lim_(x,y) to (0,0) f(x,y)$, but $lim_(r,theta) to (0,0) fracrsin theta$ does not exist. So even though it makes no difference in the particular problem from the question, I think we should only let $r to 0$ in general.
– ComplexYetTrivial
Jul 28 at 15:28
You omit $foralltheta$, $sinleq1$, in your's the function isn't bounded in origin. Our case is exactly squeez theorem.
– user 108128
Jul 28 at 15:32
1
Yes, since the function in the original problem is a bounded function of $theta$, letting $theta to 0$ does not change the result in this case. However, it is not necessary and even wrong in other cases.
– ComplexYetTrivial
Jul 28 at 15:39
add a comment |Â
1
Why do we need to let $theta to 0$ ?
– ComplexYetTrivial
Jul 28 at 15:12
I said !$$
– user 108128
Jul 28 at 15:14
But then we only approach the origin from one direction (the $x$-axis). If we consider $f(x,y) = fracx^2+y^2y = fracrsin theta$ (for $y neq 0$), we get the correct result $lim_r to 0 fracrsin theta = 0 = lim_(x,y) to (0,0) f(x,y)$, but $lim_(r,theta) to (0,0) fracrsin theta$ does not exist. So even though it makes no difference in the particular problem from the question, I think we should only let $r to 0$ in general.
– ComplexYetTrivial
Jul 28 at 15:28
You omit $foralltheta$, $sinleq1$, in your's the function isn't bounded in origin. Our case is exactly squeez theorem.
– user 108128
Jul 28 at 15:32
1
Yes, since the function in the original problem is a bounded function of $theta$, letting $theta to 0$ does not change the result in this case. However, it is not necessary and even wrong in other cases.
– ComplexYetTrivial
Jul 28 at 15:39
1
1
Why do we need to let $theta to 0$ ?
– ComplexYetTrivial
Jul 28 at 15:12
Why do we need to let $theta to 0$ ?
– ComplexYetTrivial
Jul 28 at 15:12
I said !$$
– user 108128
Jul 28 at 15:14
I said !$$
– user 108128
Jul 28 at 15:14
But then we only approach the origin from one direction (the $x$-axis). If we consider $f(x,y) = fracx^2+y^2y = fracrsin theta$ (for $y neq 0$), we get the correct result $lim_r to 0 fracrsin theta = 0 = lim_(x,y) to (0,0) f(x,y)$, but $lim_(r,theta) to (0,0) fracrsin theta$ does not exist. So even though it makes no difference in the particular problem from the question, I think we should only let $r to 0$ in general.
– ComplexYetTrivial
Jul 28 at 15:28
But then we only approach the origin from one direction (the $x$-axis). If we consider $f(x,y) = fracx^2+y^2y = fracrsin theta$ (for $y neq 0$), we get the correct result $lim_r to 0 fracrsin theta = 0 = lim_(x,y) to (0,0) f(x,y)$, but $lim_(r,theta) to (0,0) fracrsin theta$ does not exist. So even though it makes no difference in the particular problem from the question, I think we should only let $r to 0$ in general.
– ComplexYetTrivial
Jul 28 at 15:28
You omit $foralltheta$, $sinleq1$, in your's the function isn't bounded in origin. Our case is exactly squeez theorem.
– user 108128
Jul 28 at 15:32
You omit $foralltheta$, $sinleq1$, in your's the function isn't bounded in origin. Our case is exactly squeez theorem.
– user 108128
Jul 28 at 15:32
1
1
Yes, since the function in the original problem is a bounded function of $theta$, letting $theta to 0$ does not change the result in this case. However, it is not necessary and even wrong in other cases.
– ComplexYetTrivial
Jul 28 at 15:39
Yes, since the function in the original problem is a bounded function of $theta$, letting $theta to 0$ does not change the result in this case. However, it is not necessary and even wrong in other cases.
– ComplexYetTrivial
Jul 28 at 15:39
add a comment |Â
up vote
2
down vote
If you are not comfortable with polar coordinates, your remark
"...and the term on the right has absolute value less than 1"
suffices to show that $left|x fracysqrtx^2+y^2 right|leq left| x right| leq sqrtx^2+y^2$ and thus $f(x,y) rightarrow0$ as $(x,y)rightarrow 0$.
answered Jul 28 at 14:45
SEBASTIAN VARGAS LOAIZA
1464
add a comment |Â
up vote
2
down vote
If you are not comfortable with polar coordinates, your remark
"...and the term on the right has absolute value less than 1"
suffices to show that $left|x fracysqrtx^2+y^2 right|leq left| x right| leq sqrtx^2+y^2$ and thus $f(x,y) rightarrow0$ as $(x,y)rightarrow 0$.
answered Jul 28 at 14:45
SEBASTIAN VARGAS LOAIZA
1464
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If you are not comfortable with polar coordinates, your remark
"...and the term on the right has absolute value less than 1"
suffices to show that $left|x fracysqrtx^2+y^2 right|leq left| x right| leq sqrtx^2+y^2$ and thus $f(x,y) rightarrow0$ as $(x,y)rightarrow 0$.
answered Jul 28 at 14:45
SEBASTIAN VARGAS LOAIZA
1464
If you are not comfortable with polar coordinates, your remark
"...and the term on the right has absolute value less than 1"
suffices to show that $left|x fracysqrtx^2+y^2 right|leq left| x right| leq sqrtx^2+y^2$ and thus $f(x,y) rightarrow0$ as $(x,y)rightarrow 0$.
answered Jul 28 at 14:45
SEBASTIAN VARGAS LOAIZA
1464
answered Jul 28 at 14:45
SEBASTIAN VARGAS LOAIZA
1464
answered Jul 28 at 14:45
SEBASTIAN VARGAS LOAIZA
1464
answered Jul 28 at 14:45
SEBASTIAN VARGAS LOAIZA
1464
1464
add a comment |Â
add a comment |Â
up vote
0
down vote
We can also use that
$$fracxysqrtx^2+y^2le fracxysqrt2sqrtxy=frac1sqrt2|xy|^1/2$$
answered Jul 28 at 16:06
Dr. Sonnhard Graubner
66.7k32659
add a comment |Â
up vote
0
down vote
We can also use that
$$fracxysqrtx^2+y^2le fracxysqrt2sqrtxy=frac1sqrt2|xy|^1/2$$
answered Jul 28 at 16:06
Dr. Sonnhard Graubner
66.7k32659
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We can also use that
$$fracxysqrtx^2+y^2le fracxysqrt2sqrtxy=frac1sqrt2|xy|^1/2$$
answered Jul 28 at 16:06
Dr. Sonnhard Graubner
66.7k32659
We can also use that
$$fracxysqrtx^2+y^2le fracxysqrt2sqrtxy=frac1sqrt2|xy|^1/2$$
answered Jul 28 at 16:06
Dr. Sonnhard Graubner
66.7k32659
answered Jul 28 at 16:06
Dr. Sonnhard Graubner
66.7k32659
answered Jul 28 at 16:06
Dr. Sonnhard Graubner
66.7k32659
answered Jul 28 at 16:06
Dr. Sonnhard Graubner
66.7k32659
66.7k32659
add a comment |Â
add a comment |Â
up vote
0
down vote
Your argument is correct: $(r cos theta, r sin theta) = (0,0)$ holds if and only if $r = 0$ , so approaching the origin in polar coordinates indeed corresponds to letting $r to 0$ . You can, however, approach the origin from different directions, which is something you need to be careful about when working in polar coordinates! The limit $ lim_(x,y) to (0,0) f(x,y)$ exists if and only if the limits
$$ lim_r to 0 f(r cos vartheta(r), r sin vartheta(r))$$
exist and agree for every parametrisation $vartheta: (0,1) to [0,2pi)$ of the polar angle.
In your problem this works out:
$$ lim_rto 0 f(r cos vartheta(r), r sin vartheta(r)) = lim_rto 0 , r cos vartheta(r) sin vartheta(r) = 0$$
holds for every $vartheta$ since $cos$ and $sin$ are bounded by $1$ .
If we consider $g(x,y) = fracxsqrtx^2 + y^2$ for $(x,y) neq (0,0)$ instead, things are quite different. We have
$$ g(r cos(theta), r sin(theta)) = cos theta ,$$
so
$$ lim_r to 0 g(r cos(0), r sin(0)) = 1 neq -1 = lim_r to 0 g(r cos(pi), r sin(pi)) , .$$
Therefore the limit $lim_(x,y) to (0,0) g(x,y)$ does not exist and we cannot choose a value for $g(0,0)$ in order to make $g$ continuous.
A better example of the issue is provided by this question. Consider
$$ h(x,y) = fracx^2 yx^4 + y^2 = fracr cos^2 theta sin thetar^2 cos^4 theta + sin^2 theta , . $$
At first glance one would probably expect the limit $0$ at the origin, given the above representation in polar coordinates. However, along the parabolic path $y = x^2$ the limit is equal to $frac12$, so the limit $lim_(x,y) to (0,0)h(x,y)$ does not exist.
In polar coordinates this path is parametrised implicitly by
$$fracsin vartheta(r)cos^2 vartheta(r) = r , $$
or explicitly by
$$ sin vartheta(r) = fracsqrt1+4 r^2-12r , , $$
which is not easy to spot. This illustrates the fact that using polar coordinates is possible, but not always advisable when working on these problems.
answered Jul 28 at 15:11
ComplexYetTrivial
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Your argument is correct: $(r cos theta, r sin theta) = (0,0)$ holds if and only if $r = 0$ , so approaching the origin in polar coordinates indeed corresponds to letting $r to 0$ . You can, however, approach the origin from different directions, which is something you need to be careful about when working in polar coordinates! The limit $ lim_(x,y) to (0,0) f(x,y)$ exists if and only if the limits
$$ lim_r to 0 f(r cos vartheta(r), r sin vartheta(r))$$
exist and agree for every parametrisation $vartheta: (0,1) to [0,2pi)$ of the polar angle.
In your problem this works out:
$$ lim_rto 0 f(r cos vartheta(r), r sin vartheta(r)) = lim_rto 0 , r cos vartheta(r) sin vartheta(r) = 0$$
holds for every $vartheta$ since $cos$ and $sin$ are bounded by $1$ .
If we consider $g(x,y) = fracxsqrtx^2 + y^2$ for $(x,y) neq (0,0)$ instead, things are quite different. We have
$$ g(r cos(theta), r sin(theta)) = cos theta ,$$
so
$$ lim_r to 0 g(r cos(0), r sin(0)) = 1 neq -1 = lim_r to 0 g(r cos(pi), r sin(pi)) , .$$
Therefore the limit $lim_(x,y) to (0,0) g(x,y)$ does not exist and we cannot choose a value for $g(0,0)$ in order to make $g$ continuous.
A better example of the issue is provided by this question. Consider
$$ h(x,y) = fracx^2 yx^4 + y^2 = fracr cos^2 theta sin thetar^2 cos^4 theta + sin^2 theta , . $$
At first glance one would probably expect the limit $0$ at the origin, given the above representation in polar coordinates. However, along the parabolic path $y = x^2$ the limit is equal to $frac12$, so the limit $lim_(x,y) to (0,0)h(x,y)$ does not exist.
In polar coordinates this path is parametrised implicitly by
$$fracsin vartheta(r)cos^2 vartheta(r) = r , $$
or explicitly by
$$ sin vartheta(r) = fracsqrt1+4 r^2-12r , , $$
which is not easy to spot. This illustrates the fact that using polar coordinates is possible, but not always advisable when working on these problems.
answered Jul 28 at 15:11
ComplexYetTrivial
2,787624
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Your argument is correct: $(r cos theta, r sin theta) = (0,0)$ holds if and only if $r = 0$ , so approaching the origin in polar coordinates indeed corresponds to letting $r to 0$ . You can, however, approach the origin from different directions, which is something you need to be careful about when working in polar coordinates! The limit $ lim_(x,y) to (0,0) f(x,y)$ exists if and only if the limits
$$ lim_r to 0 f(r cos vartheta(r), r sin vartheta(r))$$
exist and agree for every parametrisation $vartheta: (0,1) to [0,2pi)$ of the polar angle.
In your problem this works out:
$$ lim_rto 0 f(r cos vartheta(r), r sin vartheta(r)) = lim_rto 0 , r cos vartheta(r) sin vartheta(r) = 0$$
holds for every $vartheta$ since $cos$ and $sin$ are bounded by $1$ .
If we consider $g(x,y) = fracxsqrtx^2 + y^2$ for $(x,y) neq (0,0)$ instead, things are quite different. We have
$$ g(r cos(theta), r sin(theta)) = cos theta ,$$
so
$$ lim_r to 0 g(r cos(0), r sin(0)) = 1 neq -1 = lim_r to 0 g(r cos(pi), r sin(pi)) , .$$
Therefore the limit $lim_(x,y) to (0,0) g(x,y)$ does not exist and we cannot choose a value for $g(0,0)$ in order to make $g$ continuous.
A better example of the issue is provided by this question. Consider
$$ h(x,y) = fracx^2 yx^4 + y^2 = fracr cos^2 theta sin thetar^2 cos^4 theta + sin^2 theta , . $$
At first glance one would probably expect the limit $0$ at the origin, given the above representation in polar coordinates. However, along the parabolic path $y = x^2$ the limit is equal to $frac12$, so the limit $lim_(x,y) to (0,0)h(x,y)$ does not exist.
In polar coordinates this path is parametrised implicitly by
$$fracsin vartheta(r)cos^2 vartheta(r) = r , $$
or explicitly by
$$ sin vartheta(r) = fracsqrt1+4 r^2-12r , , $$
which is not easy to spot. This illustrates the fact that using polar coordinates is possible, but not always advisable when working on these problems.
answered Jul 28 at 15:11
ComplexYetTrivial
2,787624
Your argument is correct: $(r cos theta, r sin theta) = (0,0)$ holds if and only if $r = 0$ , so approaching the origin in polar coordinates indeed corresponds to letting $r to 0$ . You can, however, approach the origin from different directions, which is something you need to be careful about when working in polar coordinates! The limit $ lim_(x,y) to (0,0) f(x,y)$ exists if and only if the limits
$$ lim_r to 0 f(r cos vartheta(r), r sin vartheta(r))$$
exist and agree for every parametrisation $vartheta: (0,1) to [0,2pi)$ of the polar angle.
In your problem this works out:
$$ lim_rto 0 f(r cos vartheta(r), r sin vartheta(r)) = lim_rto 0 , r cos vartheta(r) sin vartheta(r) = 0$$
holds for every $vartheta$ since $cos$ and $sin$ are bounded by $1$ .
If we consider $g(x,y) = fracxsqrtx^2 + y^2$ for $(x,y) neq (0,0)$ instead, things are quite different. We have
$$ g(r cos(theta), r sin(theta)) = cos theta ,$$
so
$$ lim_r to 0 g(r cos(0), r sin(0)) = 1 neq -1 = lim_r to 0 g(r cos(pi), r sin(pi)) , .$$
Therefore the limit $lim_(x,y) to (0,0) g(x,y)$ does not exist and we cannot choose a value for $g(0,0)$ in order to make $g$ continuous.
A better example of the issue is provided by this question. Consider
$$ h(x,y) = fracx^2 yx^4 + y^2 = fracr cos^2 theta sin thetar^2 cos^4 theta + sin^2 theta , . $$
At first glance one would probably expect the limit $0$ at the origin, given the above representation in polar coordinates. However, along the parabolic path $y = x^2$ the limit is equal to $frac12$, so the limit $lim_(x,y) to (0,0)h(x,y)$ does not exist.
In polar coordinates this path is parametrised implicitly by
$$fracsin vartheta(r)cos^2 vartheta(r) = r , $$
or explicitly by
$$ sin vartheta(r) = fracsqrt1+4 r^2-12r , , $$
which is not easy to spot. This illustrates the fact that using polar coordinates is possible, but not always advisable when working on these problems.
answered Jul 28 at 15:11
ComplexYetTrivial
2,787624
edited Jul 28 at 16:30
answered Jul 28 at 15:11
ComplexYetTrivial
2,787624
answered Jul 28 at 15:11
ComplexYetTrivial
2,787624
answered Jul 28 at 15:11
ComplexYetTrivial
2,787624
2,787624
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No
dfracin titles please (and actually, anywhere).– Did
Jul 28 at 14:44
9 minutes. $ $ $ $
– Did
Jul 28 at 14:45
What's wrong with dfrac?
– john fowles
Jul 28 at 15:17
Should be $(x,y)to (0,0).$
– zhw.
Jul 28 at 15:48
Too much vertical space taken in the list of questions, with neither need nor use.
– Did
Jul 28 at 16:26