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Use the $epsilon - delta$ definition to verify that $lim_{(x,y)to(0,0)}frac{x^3-y^3}{x^2+y^2} = 0$?...
This question already has an answer here:
Finding the limit of a 2-dimensional function
2 answers
I know that I have to find a $epsilon > 0$ s.t. |$frac{x^3-y^3}{x^2+y^2}$|<$epsilon$ whenever $0<sqrt{x^2+y^2}<delta$ but what are the steps to proving this?
I'm looking for an answer which shows the steps as well as explains the intuition.
Thanks
calculus limits multivariable-calculus
edited Nov 21 '18 at 17:37
StubbornAtom
5,38911138
asked Nov 21 '18 at 17:29
K.M.K.M.
293112
marked as duplicate by Decaf-Math, StubbornAtom, Trevor Gunn, GNUSupporter 8964民主女神 地下教會, amWhy calculus
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Nov 21 '18 at 19:40
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Finding the limit of a 2-dimensional function
2 answers
I know that I have to find a $epsilon > 0$ s.t. |$frac{x^3-y^3}{x^2+y^2}$|<$epsilon$ whenever $0<sqrt{x^2+y^2}<delta$ but what are the steps to proving this?
I'm looking for an answer which shows the steps as well as explains the intuition.
Thanks
calculus limits multivariable-calculus
edited Nov 21 '18 at 17:37
StubbornAtom
5,38911138
asked Nov 21 '18 at 17:29
K.M.K.M.
293112
marked as duplicate by Decaf-Math, StubbornAtom, Trevor Gunn, GNUSupporter 8964民主女神 地下教會, amWhy calculus
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Nov 21 '18 at 19:40
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
This one is asking for it "by definition" whereas the other answers use shortcuts.
– DanielV
Nov 21 '18 at 18:20
add a comment |
This question already has an answer here:
Finding the limit of a 2-dimensional function
2 answers
I know that I have to find a $epsilon > 0$ s.t. |$frac{x^3-y^3}{x^2+y^2}$|<$epsilon$ whenever $0<sqrt{x^2+y^2}<delta$ but what are the steps to proving this?
I'm looking for an answer which shows the steps as well as explains the intuition.
Thanks
calculus limits multivariable-calculus
edited Nov 21 '18 at 17:37
StubbornAtom
5,38911138
asked Nov 21 '18 at 17:29
K.M.K.M.
293112
This question already has an answer here:
Finding the limit of a 2-dimensional function
2 answers
I know that I have to find a $epsilon > 0$ s.t. |$frac{x^3-y^3}{x^2+y^2}$|<$epsilon$ whenever $0<sqrt{x^2+y^2}<delta$ but what are the steps to proving this?
I'm looking for an answer which shows the steps as well as explains the intuition.
Thanks
This question already has an answer here:
Finding the limit of a 2-dimensional function
2 answers
calculus limits multivariable-calculus
calculus limits multivariable-calculus
edited Nov 21 '18 at 17:37
StubbornAtom
5,38911138
asked Nov 21 '18 at 17:29
K.M.K.M.
293112
edited Nov 21 '18 at 17:37
StubbornAtom
5,38911138
asked Nov 21 '18 at 17:29
K.M.K.M.
293112
edited Nov 21 '18 at 17:37
StubbornAtom
5,38911138
edited Nov 21 '18 at 17:37
StubbornAtom
5,38911138
edited Nov 21 '18 at 17:37
StubbornAtom
5,38911138
5,38911138
asked Nov 21 '18 at 17:29
K.M.K.M.
293112
asked Nov 21 '18 at 17:29
K.M.K.M.
293112
asked Nov 21 '18 at 17:29
K.M.K.M.
293112
293112
marked as duplicate by Decaf-Math, StubbornAtom, Trevor Gunn, GNUSupporter 8964民主女神 地下教會, amWhy calculus
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Nov 21 '18 at 19:40
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Decaf-Math, StubbornAtom, Trevor Gunn, GNUSupporter 8964民主女神 地下教會, amWhy calculus
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Nov 21 '18 at 19:40
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
This one is asking for it "by definition" whereas the other answers use shortcuts.
– DanielV
Nov 21 '18 at 18:20
add a comment |
1
This one is asking for it "by definition" whereas the other answers use shortcuts.
– DanielV
Nov 21 '18 at 18:20
1
1
This one is asking for it "by definition" whereas the other answers use shortcuts.
– DanielV
Nov 21 '18 at 18:20
This one is asking for it "by definition" whereas the other answers use shortcuts.
– DanielV
Nov 21 '18 at 18:20
add a comment |
2 Answers
2
active
oldest
votes
Numerator:
$|(x^3-y^3)|=$
$|(x-y)|(x^2+xy+y^2)|le$
$|x-y||x^2+y^2| +|xy|| le$
$|x-y||x^2+y^2| +|x|^2+|y|^2) le$
$2|x-y||x^2+y^2| le$
$ 2(|x|+|y|) (x^2+y^2)le$
$4 sqrt{x^2+y^2}(x^2+y^2)$.
Finally
$dfrac{|x^3-y^3|}{x^2+y^2} le 4sqrt{x^2+y^2}.$
Choose $delta =epsilon/4.$
Used: $x^2+y^2 ge 2|xy|;$
$|x| =sqrt{x^2} le sqrt{x^2+y^2}$,
$|y| =sqrt{y^2} le sqrt{x^2+y^2}$.
answered Nov 21 '18 at 19:05
Peter SzilasPeter Szilas
10.8k2720
add a comment |
$$(forall y > 0 ~ exists r > 0 ~ forall bar x )~ 0 < | bar x - bar c | < r implies |f bar x - L| < y$$
$(forall y > 0 ~ exists r > 0 ~ forall bar x )$ is equivalent to asking for the existence of a function $g$ that, given a maximum deviation from $L$, outputs the neighborhood $r$ that satisfies that maximum deviation. In other words, find the disc centered at $bar c$ with radius $g(y)$ such that every value of $f$ within that region is within $y$ of $L$. Symbolically:
$$(exists g : mathbb R to mathbb R ~forall y > 0 ~ forall bar x) ~ 0 < | bar x - bar c | < g(y) implies |f bar x - L| < y$$
Putting in the values for your problem:
$$(exists g : mathbb R to mathbb R ~forall y > 0 ~ forall bar x) ~ 0 < | bar x | < g(y) implies left|frac{x_1{}^3 - x_2{}^3}{x_1{}^2 + x_2{}^2}right| < y$$
or to put more tersely, find $g : mathbb R to mathbb R$ such that for $bar x ne (0, 0)$
$$|bar x| < g(y) implies left|frac{x_1{}^3 - x_2{}^3}{x_1{}^2 + x_2{}^2}right| < y$$
A polar substitution helps:
$$r < g(y) implies left|frac{(rcos t)^3 - (rsin t)^3}{(rcos t)^2 + (rsin t)^2}right| < y$$
Can you find the $g$ now? One more hint if you need it:
$-1 le cos t, sin t le 1$
answered Nov 21 '18 at 18:19
DanielVDanielV
17.8k42754
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Numerator:
$|(x^3-y^3)|=$
$|(x-y)|(x^2+xy+y^2)|le$
$|x-y||x^2+y^2| +|xy|| le$
$|x-y||x^2+y^2| +|x|^2+|y|^2) le$
$2|x-y||x^2+y^2| le$
$ 2(|x|+|y|) (x^2+y^2)le$
$4 sqrt{x^2+y^2}(x^2+y^2)$.
Finally
$dfrac{|x^3-y^3|}{x^2+y^2} le 4sqrt{x^2+y^2}.$
Choose $delta =epsilon/4.$
Used: $x^2+y^2 ge 2|xy|;$
$|x| =sqrt{x^2} le sqrt{x^2+y^2}$,
$|y| =sqrt{y^2} le sqrt{x^2+y^2}$.
answered Nov 21 '18 at 19:05
Peter SzilasPeter Szilas
10.8k2720
add a comment |
Numerator:
$|(x^3-y^3)|=$
$|(x-y)|(x^2+xy+y^2)|le$
$|x-y||x^2+y^2| +|xy|| le$
$|x-y||x^2+y^2| +|x|^2+|y|^2) le$
$2|x-y||x^2+y^2| le$
$ 2(|x|+|y|) (x^2+y^2)le$
$4 sqrt{x^2+y^2}(x^2+y^2)$.
Finally
$dfrac{|x^3-y^3|}{x^2+y^2} le 4sqrt{x^2+y^2}.$
Choose $delta =epsilon/4.$
Used: $x^2+y^2 ge 2|xy|;$
$|x| =sqrt{x^2} le sqrt{x^2+y^2}$,
$|y| =sqrt{y^2} le sqrt{x^2+y^2}$.
answered Nov 21 '18 at 19:05
Peter SzilasPeter Szilas
10.8k2720
add a comment |
Numerator:
$|(x^3-y^3)|=$
$|(x-y)|(x^2+xy+y^2)|le$
$|x-y||x^2+y^2| +|xy|| le$
$|x-y||x^2+y^2| +|x|^2+|y|^2) le$
$2|x-y||x^2+y^2| le$
$ 2(|x|+|y|) (x^2+y^2)le$
$4 sqrt{x^2+y^2}(x^2+y^2)$.
Finally
$dfrac{|x^3-y^3|}{x^2+y^2} le 4sqrt{x^2+y^2}.$
Choose $delta =epsilon/4.$
Used: $x^2+y^2 ge 2|xy|;$
$|x| =sqrt{x^2} le sqrt{x^2+y^2}$,
$|y| =sqrt{y^2} le sqrt{x^2+y^2}$.
answered Nov 21 '18 at 19:05
Peter SzilasPeter Szilas
10.8k2720
Numerator:
$|(x^3-y^3)|=$
$|(x-y)|(x^2+xy+y^2)|le$
$|x-y||x^2+y^2| +|xy|| le$
$|x-y||x^2+y^2| +|x|^2+|y|^2) le$
$2|x-y||x^2+y^2| le$
$ 2(|x|+|y|) (x^2+y^2)le$
$4 sqrt{x^2+y^2}(x^2+y^2)$.
Finally
$dfrac{|x^3-y^3|}{x^2+y^2} le 4sqrt{x^2+y^2}.$
Choose $delta =epsilon/4.$
Used: $x^2+y^2 ge 2|xy|;$
$|x| =sqrt{x^2} le sqrt{x^2+y^2}$,
$|y| =sqrt{y^2} le sqrt{x^2+y^2}$.
answered Nov 21 '18 at 19:05
Peter SzilasPeter Szilas
10.8k2720
edited Nov 21 '18 at 19:10
answered Nov 21 '18 at 19:05
Peter SzilasPeter Szilas
10.8k2720
answered Nov 21 '18 at 19:05
Peter SzilasPeter Szilas
10.8k2720
answered Nov 21 '18 at 19:05
Peter SzilasPeter Szilas
10.8k2720
10.8k2720
add a comment |
add a comment |
$$(forall y > 0 ~ exists r > 0 ~ forall bar x )~ 0 < | bar x - bar c | < r implies |f bar x - L| < y$$
$(forall y > 0 ~ exists r > 0 ~ forall bar x )$ is equivalent to asking for the existence of a function $g$ that, given a maximum deviation from $L$, outputs the neighborhood $r$ that satisfies that maximum deviation. In other words, find the disc centered at $bar c$ with radius $g(y)$ such that every value of $f$ within that region is within $y$ of $L$. Symbolically:
$$(exists g : mathbb R to mathbb R ~forall y > 0 ~ forall bar x) ~ 0 < | bar x - bar c | < g(y) implies |f bar x - L| < y$$
Putting in the values for your problem:
$$(exists g : mathbb R to mathbb R ~forall y > 0 ~ forall bar x) ~ 0 < | bar x | < g(y) implies left|frac{x_1{}^3 - x_2{}^3}{x_1{}^2 + x_2{}^2}right| < y$$
or to put more tersely, find $g : mathbb R to mathbb R$ such that for $bar x ne (0, 0)$
$$|bar x| < g(y) implies left|frac{x_1{}^3 - x_2{}^3}{x_1{}^2 + x_2{}^2}right| < y$$
A polar substitution helps:
$$r < g(y) implies left|frac{(rcos t)^3 - (rsin t)^3}{(rcos t)^2 + (rsin t)^2}right| < y$$
Can you find the $g$ now? One more hint if you need it:
$-1 le cos t, sin t le 1$
answered Nov 21 '18 at 18:19
DanielVDanielV
17.8k42754
add a comment |
$$(forall y > 0 ~ exists r > 0 ~ forall bar x )~ 0 < | bar x - bar c | < r implies |f bar x - L| < y$$
$(forall y > 0 ~ exists r > 0 ~ forall bar x )$ is equivalent to asking for the existence of a function $g$ that, given a maximum deviation from $L$, outputs the neighborhood $r$ that satisfies that maximum deviation. In other words, find the disc centered at $bar c$ with radius $g(y)$ such that every value of $f$ within that region is within $y$ of $L$. Symbolically:
$$(exists g : mathbb R to mathbb R ~forall y > 0 ~ forall bar x) ~ 0 < | bar x - bar c | < g(y) implies |f bar x - L| < y$$
Putting in the values for your problem:
$$(exists g : mathbb R to mathbb R ~forall y > 0 ~ forall bar x) ~ 0 < | bar x | < g(y) implies left|frac{x_1{}^3 - x_2{}^3}{x_1{}^2 + x_2{}^2}right| < y$$
or to put more tersely, find $g : mathbb R to mathbb R$ such that for $bar x ne (0, 0)$
$$|bar x| < g(y) implies left|frac{x_1{}^3 - x_2{}^3}{x_1{}^2 + x_2{}^2}right| < y$$
A polar substitution helps:
$$r < g(y) implies left|frac{(rcos t)^3 - (rsin t)^3}{(rcos t)^2 + (rsin t)^2}right| < y$$
Can you find the $g$ now? One more hint if you need it:
$-1 le cos t, sin t le 1$
answered Nov 21 '18 at 18:19
DanielVDanielV
17.8k42754
add a comment |
$$(forall y > 0 ~ exists r > 0 ~ forall bar x )~ 0 < | bar x - bar c | < r implies |f bar x - L| < y$$
$(forall y > 0 ~ exists r > 0 ~ forall bar x )$ is equivalent to asking for the existence of a function $g$ that, given a maximum deviation from $L$, outputs the neighborhood $r$ that satisfies that maximum deviation. In other words, find the disc centered at $bar c$ with radius $g(y)$ such that every value of $f$ within that region is within $y$ of $L$. Symbolically:
$$(exists g : mathbb R to mathbb R ~forall y > 0 ~ forall bar x) ~ 0 < | bar x - bar c | < g(y) implies |f bar x - L| < y$$
Putting in the values for your problem:
$$(exists g : mathbb R to mathbb R ~forall y > 0 ~ forall bar x) ~ 0 < | bar x | < g(y) implies left|frac{x_1{}^3 - x_2{}^3}{x_1{}^2 + x_2{}^2}right| < y$$
or to put more tersely, find $g : mathbb R to mathbb R$ such that for $bar x ne (0, 0)$
$$|bar x| < g(y) implies left|frac{x_1{}^3 - x_2{}^3}{x_1{}^2 + x_2{}^2}right| < y$$
A polar substitution helps:
$$r < g(y) implies left|frac{(rcos t)^3 - (rsin t)^3}{(rcos t)^2 + (rsin t)^2}right| < y$$
Can you find the $g$ now? One more hint if you need it:
$-1 le cos t, sin t le 1$
answered Nov 21 '18 at 18:19
DanielVDanielV
17.8k42754
$$(forall y > 0 ~ exists r > 0 ~ forall bar x )~ 0 < | bar x - bar c | < r implies |f bar x - L| < y$$
$(forall y > 0 ~ exists r > 0 ~ forall bar x )$ is equivalent to asking for the existence of a function $g$ that, given a maximum deviation from $L$, outputs the neighborhood $r$ that satisfies that maximum deviation. In other words, find the disc centered at $bar c$ with radius $g(y)$ such that every value of $f$ within that region is within $y$ of $L$. Symbolically:
$$(exists g : mathbb R to mathbb R ~forall y > 0 ~ forall bar x) ~ 0 < | bar x - bar c | < g(y) implies |f bar x - L| < y$$
Putting in the values for your problem:
$$(exists g : mathbb R to mathbb R ~forall y > 0 ~ forall bar x) ~ 0 < | bar x | < g(y) implies left|frac{x_1{}^3 - x_2{}^3}{x_1{}^2 + x_2{}^2}right| < y$$
or to put more tersely, find $g : mathbb R to mathbb R$ such that for $bar x ne (0, 0)$
$$|bar x| < g(y) implies left|frac{x_1{}^3 - x_2{}^3}{x_1{}^2 + x_2{}^2}right| < y$$
A polar substitution helps:
$$r < g(y) implies left|frac{(rcos t)^3 - (rsin t)^3}{(rcos t)^2 + (rsin t)^2}right| < y$$
Can you find the $g$ now? One more hint if you need it:
$-1 le cos t, sin t le 1$
answered Nov 21 '18 at 18:19
DanielVDanielV
17.8k42754
answered Nov 21 '18 at 18:19
DanielVDanielV
17.8k42754
answered Nov 21 '18 at 18:19
DanielVDanielV
17.8k42754
answered Nov 21 '18 at 18:19
DanielVDanielV
17.8k42754
17.8k42754
add a comment |
add a comment |
1
This one is asking for it "by definition" whereas the other answers use shortcuts.
– DanielV
Nov 21 '18 at 18:20