Proving a limit exists using delta and epsilon?












0












$begingroup$


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







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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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 23 at 7:46







Shaun Weinberg

















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














  • 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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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 $|f(x,y)| <epsilon$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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













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1 Answer
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active

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1 Answer
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active

oldest

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oldest

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1












$begingroup$

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 $|f(x,y)| <epsilon$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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


















1












$begingroup$

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 $|f(x,y)| <epsilon$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















1












1








1





$begingroup$

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 $|f(x,y)| <epsilon$.






share|cite|improve this answer











$endgroup$



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 $|f(x,y)| <epsilon$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 23 at 6:44

























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




















  • $begingroup$
    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




















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