Proving a limit exists using delta and epsilon?
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First time posting, I did calculations, would someone be so kind to see if i'm right or wrong, i'm a part-time student so i do not have a lecturer etc.
f (x,y) = x2/ √(x2+ y2)
Prove from First principles
Lim f(x,y) = 0
(x,y) -> (0,0)
Answer:
The domain Df ∈ {(x,y) ∈ R}
0 < ll√[(x - 0)2 + (y - 0) 2]ll < 𝛿 ( from definition ll√[(x - a)2 + (y - b) 2]ll < 𝛿 )
0 < ll√[(x )2 + (y) 2]ll < 𝛿
0 < ll√[(x )2 + (y) 2]ll < 1 (where 1 < 𝛿)
lx-al = lxl = ll√[(x )2 + (y ) 2]ll = {(x,y) – (0,0)}
ly-bl = lyl = ll√[(x )2 + (y ) 2]ll = {(x,y) – (0,0)}
lf(x) - L l < ε
lx2/ √(x2+ y2) – 0 l < ε
lx2/ √(x2+ y2) – 0 l < ε
lx2l/l √(x2+ y2)l < ε
lxl 2/ √(x2+ y2) < ε
{√(x2+ y2)}2/ √(x2+ y2) < ε (substituted in lxl2 = ll√[(x )2 + (y ) 2]ll
√(x2+ y2) < ε
If 0 < ll√[(x )2 + (y) 2]ll < 1, where 1 < 𝛿, then lf(x) - L l < ε for Lim f(x,y) = 0
(x,y)->(0,0)
limits
asked Jan 23 at 6:10
Shaun WeinbergShaun Weinberg
62
$endgroup$
add a comment |
$begingroup$
First time posting, I did calculations, would someone be so kind to see if i'm right or wrong, i'm a part-time student so i do not have a lecturer etc.
f (x,y) = x2/ √(x2+ y2)
Prove from First principles
Lim f(x,y) = 0
(x,y) -> (0,0)
Answer:
The domain Df ∈ {(x,y) ∈ R}
0 < ll√[(x - 0)2 + (y - 0) 2]ll < 𝛿 ( from definition ll√[(x - a)2 + (y - b) 2]ll < 𝛿 )
0 < ll√[(x )2 + (y) 2]ll < 𝛿
0 < ll√[(x )2 + (y) 2]ll < 1 (where 1 < 𝛿)
lx-al = lxl = ll√[(x )2 + (y ) 2]ll = {(x,y) – (0,0)}
ly-bl = lyl = ll√[(x )2 + (y ) 2]ll = {(x,y) – (0,0)}
lf(x) - L l < ε
lx2/ √(x2+ y2) – 0 l < ε
lx2/ √(x2+ y2) – 0 l < ε
lx2l/l √(x2+ y2)l < ε
lxl 2/ √(x2+ y2) < ε
{√(x2+ y2)}2/ √(x2+ y2) < ε (substituted in lxl2 = ll√[(x )2 + (y ) 2]ll
√(x2+ y2) < ε
If 0 < ll√[(x )2 + (y) 2]ll < 1, where 1 < 𝛿, then lf(x) - L l < ε for Lim f(x,y) = 0
(x,y)->(0,0)
limits
asked Jan 23 at 6:10
Shaun WeinbergShaun Weinberg
62
$endgroup$
3
$begingroup$
Hi, here is a basic tutorial. Please type math using MathJax. Gate.
$endgroup$
– xbh
Jan 23 at 6:13
$begingroup$
Thanks going there now
$endgroup$
– Shaun Weinberg
Jan 23 at 6:24
add a comment |
$begingroup$
First time posting, I did calculations, would someone be so kind to see if i'm right or wrong, i'm a part-time student so i do not have a lecturer etc.
f (x,y) = x2/ √(x2+ y2)
Prove from First principles
Lim f(x,y) = 0
(x,y) -> (0,0)
Answer:
The domain Df ∈ {(x,y) ∈ R}
0 < ll√[(x - 0)2 + (y - 0) 2]ll < 𝛿 ( from definition ll√[(x - a)2 + (y - b) 2]ll < 𝛿 )
0 < ll√[(x )2 + (y) 2]ll < 𝛿
0 < ll√[(x )2 + (y) 2]ll < 1 (where 1 < 𝛿)
lx-al = lxl = ll√[(x )2 + (y ) 2]ll = {(x,y) – (0,0)}
ly-bl = lyl = ll√[(x )2 + (y ) 2]ll = {(x,y) – (0,0)}
lf(x) - L l < ε
lx2/ √(x2+ y2) – 0 l < ε
lx2/ √(x2+ y2) – 0 l < ε
lx2l/l √(x2+ y2)l < ε
lxl 2/ √(x2+ y2) < ε
{√(x2+ y2)}2/ √(x2+ y2) < ε (substituted in lxl2 = ll√[(x )2 + (y ) 2]ll
√(x2+ y2) < ε
If 0 < ll√[(x )2 + (y) 2]ll < 1, where 1 < 𝛿, then lf(x) - L l < ε for Lim f(x,y) = 0
(x,y)->(0,0)
limits
asked Jan 23 at 6:10
Shaun WeinbergShaun Weinberg
62
$endgroup$
First time posting, I did calculations, would someone be so kind to see if i'm right or wrong, i'm a part-time student so i do not have a lecturer etc.
f (x,y) = x2/ √(x2+ y2)
Prove from First principles
Lim f(x,y) = 0
(x,y) -> (0,0)
Answer:
The domain Df ∈ {(x,y) ∈ R}
0 < ll√[(x - 0)2 + (y - 0) 2]ll < 𝛿 ( from definition ll√[(x - a)2 + (y - b) 2]ll < 𝛿 )
0 < ll√[(x )2 + (y) 2]ll < 𝛿
0 < ll√[(x )2 + (y) 2]ll < 1 (where 1 < 𝛿)
lx-al = lxl = ll√[(x )2 + (y ) 2]ll = {(x,y) – (0,0)}
ly-bl = lyl = ll√[(x )2 + (y ) 2]ll = {(x,y) – (0,0)}
lf(x) - L l < ε
lx2/ √(x2+ y2) – 0 l < ε
lx2/ √(x2+ y2) – 0 l < ε
lx2l/l √(x2+ y2)l < ε
lxl 2/ √(x2+ y2) < ε
{√(x2+ y2)}2/ √(x2+ y2) < ε (substituted in lxl2 = ll√[(x )2 + (y ) 2]ll
√(x2+ y2) < ε
If 0 < ll√[(x )2 + (y) 2]ll < 1, where 1 < 𝛿, then lf(x) - L l < ε for Lim f(x,y) = 0
(x,y)->(0,0)
limits
limits
asked Jan 23 at 6:10
Shaun WeinbergShaun Weinberg
62
asked Jan 23 at 6:10
Shaun WeinbergShaun Weinberg
62
edited Jan 23 at 7:46
Shaun Weinberg
asked Jan 23 at 6:10
Shaun WeinbergShaun Weinberg
62
asked Jan 23 at 6:10
Shaun WeinbergShaun Weinberg
62
asked Jan 23 at 6:10
Shaun WeinbergShaun Weinberg
62
62
3
$begingroup$
Hi, here is a basic tutorial. Please type math using MathJax. Gate.
$endgroup$
– xbh
Jan 23 at 6:13
$begingroup$
Thanks going there now
$endgroup$
– Shaun Weinberg
Jan 23 at 6:24
add a comment |
3
$begingroup$
Hi, here is a basic tutorial. Please type math using MathJax. Gate.
$endgroup$
– xbh
Jan 23 at 6:13
$begingroup$
Thanks going there now
$endgroup$
– Shaun Weinberg
Jan 23 at 6:24
3
3
$begingroup$
Hi, here is a basic tutorial. Please type math using MathJax. Gate.
$endgroup$
– xbh
Jan 23 at 6:13
$begingroup$
Hi, here is a basic tutorial. Please type math using MathJax. Gate.
$endgroup$
– xbh
Jan 23 at 6:13
$begingroup$
Thanks going there now
$endgroup$
– Shaun Weinberg
Jan 23 at 6:24
$begingroup$
Thanks going there now
$endgroup$
– Shaun Weinberg
Jan 23 at 6:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I will suggest a much simpler argument. $delta =epsilon$ works because $x^{2} leq {x^{2}+y^{2}}$, so $|f(x,y)| leq sqrt {x^{2}+y^{2}}$. Hence $sqrt {x^{2}+y^{2}} <epsilon$ implies $|f(x,y)| <epsilon$.
answered Jan 23 at 6:39
Kavi Rama MurthyKavi Rama Murthy
63.9k42464
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$begingroup$
Thank you, i have updated the question i am still struggling with this sites way of doing things. Could you kindly explain?
$endgroup$
– Shaun Weinberg
Jan 23 at 7:47
$begingroup$
@ShaunWeinberg Sorry, I find it very hard to read what you have written. There is nothing wrong in doing some simple manipulations before coming up with a $delta$, so my answer is usually accepted by any teacher.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 7:50
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
$begingroup$
I will suggest a much simpler argument. $delta =epsilon$ works because $x^{2} leq {x^{2}+y^{2}}$, so $|f(x,y)| leq sqrt {x^{2}+y^{2}}$. Hence $sqrt {x^{2}+y^{2}} <epsilon$ implies $|f(x,y)| <epsilon$.
answered Jan 23 at 6:39
Kavi Rama MurthyKavi Rama Murthy
63.9k42464
$endgroup$
$begingroup$
Thank you, i have updated the question i am still struggling with this sites way of doing things. Could you kindly explain?
$endgroup$
– Shaun Weinberg
Jan 23 at 7:47
$begingroup$
@ShaunWeinberg Sorry, I find it very hard to read what you have written. There is nothing wrong in doing some simple manipulations before coming up with a $delta$, so my answer is usually accepted by any teacher.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 7:50
add a comment |
$begingroup$
I will suggest a much simpler argument. $delta =epsilon$ works because $x^{2} leq {x^{2}+y^{2}}$, so $|f(x,y)| leq sqrt {x^{2}+y^{2}}$. Hence $sqrt {x^{2}+y^{2}} <epsilon$ implies $|f(x,y)| <epsilon$.
answered Jan 23 at 6:39
Kavi Rama MurthyKavi Rama Murthy
63.9k42464
$endgroup$
$begingroup$
Thank you, i have updated the question i am still struggling with this sites way of doing things. Could you kindly explain?
$endgroup$
– Shaun Weinberg
Jan 23 at 7:47
$begingroup$
@ShaunWeinberg Sorry, I find it very hard to read what you have written. There is nothing wrong in doing some simple manipulations before coming up with a $delta$, so my answer is usually accepted by any teacher.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 7:50
add a comment |
$begingroup$
I will suggest a much simpler argument. $delta =epsilon$ works because $x^{2} leq {x^{2}+y^{2}}$, so $|f(x,y)| leq sqrt {x^{2}+y^{2}}$. Hence $sqrt {x^{2}+y^{2}} <epsilon$ implies $|f(x,y)| <epsilon$.
answered Jan 23 at 6:39
Kavi Rama MurthyKavi Rama Murthy
63.9k42464
$endgroup$
I will suggest a much simpler argument. $delta =epsilon$ works because $x^{2} leq {x^{2}+y^{2}}$, so $|f(x,y)| leq sqrt {x^{2}+y^{2}}$. Hence $sqrt {x^{2}+y^{2}} <epsilon$ implies $|f(x,y)| <epsilon$.
answered Jan 23 at 6:39
Kavi Rama MurthyKavi Rama Murthy
63.9k42464
edited Jan 23 at 6:44
answered Jan 23 at 6:39
Kavi Rama MurthyKavi Rama Murthy
63.9k42464
answered Jan 23 at 6:39
Kavi Rama MurthyKavi Rama Murthy
63.9k42464
answered Jan 23 at 6:39
Kavi Rama MurthyKavi Rama Murthy
63.9k42464
63.9k42464
$begingroup$
Thank you, i have updated the question i am still struggling with this sites way of doing things. Could you kindly explain?
$endgroup$
– Shaun Weinberg
Jan 23 at 7:47
$begingroup$
@ShaunWeinberg Sorry, I find it very hard to read what you have written. There is nothing wrong in doing some simple manipulations before coming up with a $delta$, so my answer is usually accepted by any teacher.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 7:50
add a comment |
$begingroup$
Thank you, i have updated the question i am still struggling with this sites way of doing things. Could you kindly explain?
$endgroup$
– Shaun Weinberg
Jan 23 at 7:47
$begingroup$
@ShaunWeinberg Sorry, I find it very hard to read what you have written. There is nothing wrong in doing some simple manipulations before coming up with a $delta$, so my answer is usually accepted by any teacher.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 7:50
$begingroup$
Thank you, i have updated the question i am still struggling with this sites way of doing things. Could you kindly explain?
$endgroup$
– Shaun Weinberg
Jan 23 at 7:47
$begingroup$
Thank you, i have updated the question i am still struggling with this sites way of doing things. Could you kindly explain?
$endgroup$
– Shaun Weinberg
Jan 23 at 7:47
$begingroup$
@ShaunWeinberg Sorry, I find it very hard to read what you have written. There is nothing wrong in doing some simple manipulations before coming up with a $delta$, so my answer is usually accepted by any teacher.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 7:50
$begingroup$
@ShaunWeinberg Sorry, I find it very hard to read what you have written. There is nothing wrong in doing some simple manipulations before coming up with a $delta$, so my answer is usually accepted by any teacher.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 7:50
add a comment |
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3
$begingroup$
Hi, here is a basic tutorial. Please type math using MathJax. Gate.
$endgroup$
– xbh
Jan 23 at 6:13
$begingroup$
Thanks going there now
$endgroup$
– Shaun Weinberg
Jan 23 at 6:24