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How to show that limit value exists?
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up vote
1
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Here I have function
$$f(x,y)=fracx^3-xy^2x^2+y^2$$
I know that $limlimits_(x,y) to (0,0)f(x,y)=0$ because if we put $y=mx$, then we have
$$f(x,mx)=fracx(1-m^2)1+m^2$$
$$Rightarrow~limlimits_x to 0f(x,mx)=0 $$
But how to show through $epsilon-delta$ definition that limiting value of the function exists.
Thanks in advance
calculus multivariable-calculus multivalued-functions
asked 1 hour ago
Noor Aslam
12110
add a comment |Â
up vote
1
down vote
favorite
Here I have function
$$f(x,y)=fracx^3-xy^2x^2+y^2$$
I know that $limlimits_(x,y) to (0,0)f(x,y)=0$ because if we put $y=mx$, then we have
$$f(x,mx)=fracx(1-m^2)1+m^2$$
$$Rightarrow~limlimits_x to 0f(x,mx)=0 $$
But how to show through $epsilon-delta$ definition that limiting value of the function exists.
Thanks in advance
calculus multivariable-calculus multivalued-functions
asked 1 hour ago
Noor Aslam
12110
1
This approach is not valid if you want to prove the existence of the limit. To obtain equal limit along some (even if they are numerous) paths does not suffice.
– user376343
23 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Here I have function
$$f(x,y)=fracx^3-xy^2x^2+y^2$$
I know that $limlimits_(x,y) to (0,0)f(x,y)=0$ because if we put $y=mx$, then we have
$$f(x,mx)=fracx(1-m^2)1+m^2$$
$$Rightarrow~limlimits_x to 0f(x,mx)=0 $$
But how to show through $epsilon-delta$ definition that limiting value of the function exists.
Thanks in advance
calculus multivariable-calculus multivalued-functions
asked 1 hour ago
Noor Aslam
12110
Here I have function
$$f(x,y)=fracx^3-xy^2x^2+y^2$$
I know that $limlimits_(x,y) to (0,0)f(x,y)=0$ because if we put $y=mx$, then we have
$$f(x,mx)=fracx(1-m^2)1+m^2$$
$$Rightarrow~limlimits_x to 0f(x,mx)=0 $$
But how to show through $epsilon-delta$ definition that limiting value of the function exists.
Thanks in advance
calculus multivariable-calculus multivalued-functions
calculus multivariable-calculus multivalued-functions
asked 1 hour ago
Noor Aslam
12110
asked 1 hour ago
Noor Aslam
12110
asked 1 hour ago
Noor Aslam
12110
asked 1 hour ago
Noor Aslam
12110
asked 1 hour ago
Noor Aslam
12110
12110
1
This approach is not valid if you want to prove the existence of the limit. To obtain equal limit along some (even if they are numerous) paths does not suffice.
– user376343
23 mins ago
add a comment |Â
1
This approach is not valid if you want to prove the existence of the limit. To obtain equal limit along some (even if they are numerous) paths does not suffice.
– user376343
23 mins ago
1
1
This approach is not valid if you want to prove the existence of the limit. To obtain equal limit along some (even if they are numerous) paths does not suffice.
– user376343
23 mins ago
This approach is not valid if you want to prove the existence of the limit. To obtain equal limit along some (even if they are numerous) paths does not suffice.
– user376343
23 mins ago
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
3
down vote
accepted
$|dfracx(x^2-y^2)x^2+y^2|le$
$|x| + |dfracx(-2y^2)x^2+y^2| le $
$|x| + |2x| = 3|x| =$
$3sqrtx^2 le 3sqrtx^2+y^2$.
Given $epsilon >0.$
Choose $delta =epsilon/3$.
answered 40 mins ago
Peter Szilas
8,9532719
add a comment |Â
up vote
5
down vote
Showing that on y=mx curve your limit is the same for every m isn't sufficient, because u can approach the point (0,0) through, let's say, y=sin(x) curve. (sin(x) tends to 0, as x tends to 0, everything's still okay).
But not to stray away from the topic, let's see:
$$fracx^3-xy^2x^2+y^2 = frac(x+y)(x-y)x^2+y^2x $$
Let's forget for a while about the last "x" that I've left from fraction.
Look at phrase:
$$ fracx^2-y^2x^2+y^2 = 1 - 2 fracy^2x^2+y^2 $$
Clearly, that cannot be either greater than 1 ( cause $fracy^2x^2+y^2 $ isn't negative, or be less than -1 ( cause $fracy^2x^2+y^2$ cannot be greater than 1)
So that we get VERY important inequality:
$$ 0 leq |fracx^3-xy^2x^2+y^2| leq |x| $$
Finally, let's conclude that x tends to 0 as (x,y) tends to (0,0).
By squeeze theorem ( I believe it is called so in English) you have what you wanted.
answered 42 mins ago
Dominik Kutek
513
New contributor
Dominik Kutek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Dominik.Very nice +.
– Peter Szilas
27 mins ago
add a comment |Â
up vote
1
down vote
I would write this equation in polar coordinates. $$x=rcostheta\y=rsintheta$$
Then $$lim_(x,y)to(0,0)f(x,y)=lim_rto 0fracr^3(cos^3theta-costhetasin^2theta)r^2$$
The absolute value of the expression in parenthesis is always less that $2$, so it's easy to write the $delta-varepsilon$ solution.
answered 54 mins ago
Andrei
8,6032923
You are right it's easy to show in polar coordinates but can we show in Cartesian coordinates?
– Noor Aslam
50 mins ago
add a comment |Â
up vote
1
down vote
For $x,yinmathbbR$ is $$0leq|x^2-y^2|leq x^2+y^2,$$ thus if $xy neq 0$ we get $$0leqfracx^2+y^2leq 1.$$ Then $$0<|x|fracx^2+y^2leq |x|$$ and the value $0$ of the limit follows immediately.
answered 26 mins ago
user376343
1,203614
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$|dfracx(x^2-y^2)x^2+y^2|le$
$|x| + |dfracx(-2y^2)x^2+y^2| le $
$|x| + |2x| = 3|x| =$
$3sqrtx^2 le 3sqrtx^2+y^2$.
Given $epsilon >0.$
Choose $delta =epsilon/3$.
answered 40 mins ago
Peter Szilas
8,9532719
add a comment |Â
up vote
3
down vote
accepted
$|dfracx(x^2-y^2)x^2+y^2|le$
$|x| + |dfracx(-2y^2)x^2+y^2| le $
$|x| + |2x| = 3|x| =$
$3sqrtx^2 le 3sqrtx^2+y^2$.
Given $epsilon >0.$
Choose $delta =epsilon/3$.
answered 40 mins ago
Peter Szilas
8,9532719
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$|dfracx(x^2-y^2)x^2+y^2|le$
$|x| + |dfracx(-2y^2)x^2+y^2| le $
$|x| + |2x| = 3|x| =$
$3sqrtx^2 le 3sqrtx^2+y^2$.
Given $epsilon >0.$
Choose $delta =epsilon/3$.
answered 40 mins ago
Peter Szilas
8,9532719
$|dfracx(x^2-y^2)x^2+y^2|le$
$|x| + |dfracx(-2y^2)x^2+y^2| le $
$|x| + |2x| = 3|x| =$
$3sqrtx^2 le 3sqrtx^2+y^2$.
Given $epsilon >0.$
Choose $delta =epsilon/3$.
answered 40 mins ago
Peter Szilas
8,9532719
answered 40 mins ago
Peter Szilas
8,9532719
answered 40 mins ago
Peter Szilas
8,9532719
answered 40 mins ago
Peter Szilas
8,9532719
8,9532719
add a comment |Â
add a comment |Â
up vote
5
down vote
Showing that on y=mx curve your limit is the same for every m isn't sufficient, because u can approach the point (0,0) through, let's say, y=sin(x) curve. (sin(x) tends to 0, as x tends to 0, everything's still okay).
But not to stray away from the topic, let's see:
$$fracx^3-xy^2x^2+y^2 = frac(x+y)(x-y)x^2+y^2x $$
Let's forget for a while about the last "x" that I've left from fraction.
Look at phrase:
$$ fracx^2-y^2x^2+y^2 = 1 - 2 fracy^2x^2+y^2 $$
Clearly, that cannot be either greater than 1 ( cause $fracy^2x^2+y^2 $ isn't negative, or be less than -1 ( cause $fracy^2x^2+y^2$ cannot be greater than 1)
So that we get VERY important inequality:
$$ 0 leq |fracx^3-xy^2x^2+y^2| leq |x| $$
Finally, let's conclude that x tends to 0 as (x,y) tends to (0,0).
By squeeze theorem ( I believe it is called so in English) you have what you wanted.
answered 42 mins ago
Dominik Kutek
513
New contributor
Dominik Kutek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Dominik.Very nice +.
– Peter Szilas
27 mins ago
add a comment |Â
up vote
5
down vote
Showing that on y=mx curve your limit is the same for every m isn't sufficient, because u can approach the point (0,0) through, let's say, y=sin(x) curve. (sin(x) tends to 0, as x tends to 0, everything's still okay).
But not to stray away from the topic, let's see:
$$fracx^3-xy^2x^2+y^2 = frac(x+y)(x-y)x^2+y^2x $$
Let's forget for a while about the last "x" that I've left from fraction.
Look at phrase:
$$ fracx^2-y^2x^2+y^2 = 1 - 2 fracy^2x^2+y^2 $$
Clearly, that cannot be either greater than 1 ( cause $fracy^2x^2+y^2 $ isn't negative, or be less than -1 ( cause $fracy^2x^2+y^2$ cannot be greater than 1)
So that we get VERY important inequality:
$$ 0 leq |fracx^3-xy^2x^2+y^2| leq |x| $$
Finally, let's conclude that x tends to 0 as (x,y) tends to (0,0).
By squeeze theorem ( I believe it is called so in English) you have what you wanted.
answered 42 mins ago
Dominik Kutek
513
New contributor
Dominik Kutek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Dominik.Very nice +.
– Peter Szilas
27 mins ago
add a comment |Â
up vote
5
down vote
up vote
5
down vote
Showing that on y=mx curve your limit is the same for every m isn't sufficient, because u can approach the point (0,0) through, let's say, y=sin(x) curve. (sin(x) tends to 0, as x tends to 0, everything's still okay).
But not to stray away from the topic, let's see:
$$fracx^3-xy^2x^2+y^2 = frac(x+y)(x-y)x^2+y^2x $$
Let's forget for a while about the last "x" that I've left from fraction.
Look at phrase:
$$ fracx^2-y^2x^2+y^2 = 1 - 2 fracy^2x^2+y^2 $$
Clearly, that cannot be either greater than 1 ( cause $fracy^2x^2+y^2 $ isn't negative, or be less than -1 ( cause $fracy^2x^2+y^2$ cannot be greater than 1)
So that we get VERY important inequality:
$$ 0 leq |fracx^3-xy^2x^2+y^2| leq |x| $$
Finally, let's conclude that x tends to 0 as (x,y) tends to (0,0).
By squeeze theorem ( I believe it is called so in English) you have what you wanted.
answered 42 mins ago
Dominik Kutek
513
New contributor
Dominik Kutek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Showing that on y=mx curve your limit is the same for every m isn't sufficient, because u can approach the point (0,0) through, let's say, y=sin(x) curve. (sin(x) tends to 0, as x tends to 0, everything's still okay).
But not to stray away from the topic, let's see:
$$fracx^3-xy^2x^2+y^2 = frac(x+y)(x-y)x^2+y^2x $$
Let's forget for a while about the last "x" that I've left from fraction.
Look at phrase:
$$ fracx^2-y^2x^2+y^2 = 1 - 2 fracy^2x^2+y^2 $$
Clearly, that cannot be either greater than 1 ( cause $fracy^2x^2+y^2 $ isn't negative, or be less than -1 ( cause $fracy^2x^2+y^2$ cannot be greater than 1)
So that we get VERY important inequality:
$$ 0 leq |fracx^3-xy^2x^2+y^2| leq |x| $$
Finally, let's conclude that x tends to 0 as (x,y) tends to (0,0).
By squeeze theorem ( I believe it is called so in English) you have what you wanted.
answered 42 mins ago
Dominik Kutek
513
New contributor
Dominik Kutek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 42 mins ago
Dominik Kutek
513
New contributor
Dominik Kutek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 42 mins ago
Dominik Kutek
513
answered 42 mins ago
Dominik Kutek
513
513
New contributor
Dominik Kutek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Dominik Kutek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Dominik Kutek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Dominik.Very nice +.
– Peter Szilas
27 mins ago
add a comment |Â
Dominik.Very nice +.
– Peter Szilas
27 mins ago
Dominik.Very nice +.
– Peter Szilas
27 mins ago
Dominik.Very nice +.
– Peter Szilas
27 mins ago
add a comment |Â
up vote
1
down vote
I would write this equation in polar coordinates. $$x=rcostheta\y=rsintheta$$
Then $$lim_(x,y)to(0,0)f(x,y)=lim_rto 0fracr^3(cos^3theta-costhetasin^2theta)r^2$$
The absolute value of the expression in parenthesis is always less that $2$, so it's easy to write the $delta-varepsilon$ solution.
answered 54 mins ago
Andrei
8,6032923
You are right it's easy to show in polar coordinates but can we show in Cartesian coordinates?
– Noor Aslam
50 mins ago
add a comment |Â
up vote
1
down vote
I would write this equation in polar coordinates. $$x=rcostheta\y=rsintheta$$
Then $$lim_(x,y)to(0,0)f(x,y)=lim_rto 0fracr^3(cos^3theta-costhetasin^2theta)r^2$$
The absolute value of the expression in parenthesis is always less that $2$, so it's easy to write the $delta-varepsilon$ solution.
answered 54 mins ago
Andrei
8,6032923
You are right it's easy to show in polar coordinates but can we show in Cartesian coordinates?
– Noor Aslam
50 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I would write this equation in polar coordinates. $$x=rcostheta\y=rsintheta$$
Then $$lim_(x,y)to(0,0)f(x,y)=lim_rto 0fracr^3(cos^3theta-costhetasin^2theta)r^2$$
The absolute value of the expression in parenthesis is always less that $2$, so it's easy to write the $delta-varepsilon$ solution.
answered 54 mins ago
Andrei
8,6032923
I would write this equation in polar coordinates. $$x=rcostheta\y=rsintheta$$
Then $$lim_(x,y)to(0,0)f(x,y)=lim_rto 0fracr^3(cos^3theta-costhetasin^2theta)r^2$$
The absolute value of the expression in parenthesis is always less that $2$, so it's easy to write the $delta-varepsilon$ solution.
answered 54 mins ago
Andrei
8,6032923
answered 54 mins ago
Andrei
8,6032923
answered 54 mins ago
Andrei
8,6032923
answered 54 mins ago
Andrei
8,6032923
8,6032923
You are right it's easy to show in polar coordinates but can we show in Cartesian coordinates?
– Noor Aslam
50 mins ago
add a comment |Â
You are right it's easy to show in polar coordinates but can we show in Cartesian coordinates?
– Noor Aslam
50 mins ago
You are right it's easy to show in polar coordinates but can we show in Cartesian coordinates?
– Noor Aslam
50 mins ago
You are right it's easy to show in polar coordinates but can we show in Cartesian coordinates?
– Noor Aslam
50 mins ago
add a comment |Â
up vote
1
down vote
For $x,yinmathbbR$ is $$0leq|x^2-y^2|leq x^2+y^2,$$ thus if $xy neq 0$ we get $$0leqfracx^2+y^2leq 1.$$ Then $$0<|x|fracx^2+y^2leq |x|$$ and the value $0$ of the limit follows immediately.
answered 26 mins ago
user376343
1,203614
add a comment |Â
up vote
1
down vote
For $x,yinmathbbR$ is $$0leq|x^2-y^2|leq x^2+y^2,$$ thus if $xy neq 0$ we get $$0leqfracx^2+y^2leq 1.$$ Then $$0<|x|fracx^2+y^2leq |x|$$ and the value $0$ of the limit follows immediately.
answered 26 mins ago
user376343
1,203614
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For $x,yinmathbbR$ is $$0leq|x^2-y^2|leq x^2+y^2,$$ thus if $xy neq 0$ we get $$0leqfracx^2+y^2leq 1.$$ Then $$0<|x|fracx^2+y^2leq |x|$$ and the value $0$ of the limit follows immediately.
answered 26 mins ago
user376343
1,203614
For $x,yinmathbbR$ is $$0leq|x^2-y^2|leq x^2+y^2,$$ thus if $xy neq 0$ we get $$0leqfracx^2+y^2leq 1.$$ Then $$0<|x|fracx^2+y^2leq |x|$$ and the value $0$ of the limit follows immediately.
answered 26 mins ago
user376343
1,203614
answered 26 mins ago
user376343
1,203614
answered 26 mins ago
user376343
1,203614
answered 26 mins ago
user376343
1,203614
1,203614
add a comment |Â
add a comment |Â
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1
This approach is not valid if you want to prove the existence of the limit. To obtain equal limit along some (even if they are numerous) paths does not suffice.
– user376343
23 mins ago