Mixing
Finding an $epsilon-delta$ proof for a multivariable limit.
$begingroup$
Suppose $f:mathbb{R}^2rightarrow mathbb{R}$ is defined as $$(x,y) longmapsto
left{
begin{array}{cl}
dfrac{4x^2y^3 +x^4y - y^5}{(x^2+y^2)^2} & mbox {if } (x,y) neq (0,0) \
\
0 & mbox {if } (x,y) = (0,0)
end{array}
right.
$$
I have shown that $f$ is continuous at $(0,0)$ using polar coordinates, but I am really trying to improve my $epsilon-delta$ proofs for such limits. I always end up getting confused with the inequalities. (Sorry if the question is a bit repetitive here)
So for $epsilon > 0$ I need to find a $delta$ such that $|(x,y)| =sqrt{x^2 + y^2} < delta$ implies $|f(x,y)| < epsilon$
I have tried to work backwards using the inequality $(x^2+y^2)^2 geq 4x^2y^2$
$$begin{align*}
left|frac{4x^2y^3 +x^4y - y^5}{(x^2+y^2)^2}right|&=left|frac{4x^2y^3}{(x^2+y^2)^2} + frac{x^4 y}{(x^2+y^2)^2} - frac{y^5}{(x^2+y^2)^2}right|\
&leqleft|frac{4x^2y^3}{(x^2+y^2)^2}right| + left|frac{x^4 y}{(x^2+y^2)^2}right| + left|frac{y^5}{(x^2+y^2)^2}right|\
&leqleft|frac{4x^2y^3}{4x^2y^2}right| + left|frac{x^4 y}{4x^2y^2}right| + left|frac{y^5}{4x^2y^2}right|\
&= left|yright| + left|frac{x^2}{4y}right| + left|frac{y^3}{4x^2}right|
end{align*}$$
Got stuck here and not sure if my approach is correct.
real-analysis limits multivariable-calculus epsilon-delta
edited Jan 28 at 17:57
MathOverview
8,96243164
asked Jan 28 at 17:49
mikemike
564
$endgroup$
add a comment |
$begingroup$
Suppose $f:mathbb{R}^2rightarrow mathbb{R}$ is defined as $$(x,y) longmapsto
left{
begin{array}{cl}
dfrac{4x^2y^3 +x^4y - y^5}{(x^2+y^2)^2} & mbox {if } (x,y) neq (0,0) \
\
0 & mbox {if } (x,y) = (0,0)
end{array}
right.
$$
I have shown that $f$ is continuous at $(0,0)$ using polar coordinates, but I am really trying to improve my $epsilon-delta$ proofs for such limits. I always end up getting confused with the inequalities. (Sorry if the question is a bit repetitive here)
So for $epsilon > 0$ I need to find a $delta$ such that $|(x,y)| =sqrt{x^2 + y^2} < delta$ implies $|f(x,y)| < epsilon$
I have tried to work backwards using the inequality $(x^2+y^2)^2 geq 4x^2y^2$
$$begin{align*}
left|frac{4x^2y^3 +x^4y - y^5}{(x^2+y^2)^2}right|&=left|frac{4x^2y^3}{(x^2+y^2)^2} + frac{x^4 y}{(x^2+y^2)^2} - frac{y^5}{(x^2+y^2)^2}right|\
&leqleft|frac{4x^2y^3}{(x^2+y^2)^2}right| + left|frac{x^4 y}{(x^2+y^2)^2}right| + left|frac{y^5}{(x^2+y^2)^2}right|\
&leqleft|frac{4x^2y^3}{4x^2y^2}right| + left|frac{x^4 y}{4x^2y^2}right| + left|frac{y^5}{4x^2y^2}right|\
&= left|yright| + left|frac{x^2}{4y}right| + left|frac{y^3}{4x^2}right|
end{align*}$$
Got stuck here and not sure if my approach is correct.
real-analysis limits multivariable-calculus epsilon-delta
edited Jan 28 at 17:57
MathOverview
8,96243164
asked Jan 28 at 17:49
mikemike
564
$endgroup$
$begingroup$
Having $x$ and $y$ in the denominator is going to cause you horrible problems with things blowing up. Just use the fact that $|x|$ and $|y|$ are both $le (x^2+y^2)^{1/2}$.
$endgroup$
– Ted Shifrin
Jan 28 at 19:16
add a comment |
$begingroup$
Suppose $f:mathbb{R}^2rightarrow mathbb{R}$ is defined as $$(x,y) longmapsto
left{
begin{array}{cl}
dfrac{4x^2y^3 +x^4y - y^5}{(x^2+y^2)^2} & mbox {if } (x,y) neq (0,0) \
\
0 & mbox {if } (x,y) = (0,0)
end{array}
right.
$$
I have shown that $f$ is continuous at $(0,0)$ using polar coordinates, but I am really trying to improve my $epsilon-delta$ proofs for such limits. I always end up getting confused with the inequalities. (Sorry if the question is a bit repetitive here)
So for $epsilon > 0$ I need to find a $delta$ such that $|(x,y)| =sqrt{x^2 + y^2} < delta$ implies $|f(x,y)| < epsilon$
I have tried to work backwards using the inequality $(x^2+y^2)^2 geq 4x^2y^2$
$$begin{align*}
left|frac{4x^2y^3 +x^4y - y^5}{(x^2+y^2)^2}right|&=left|frac{4x^2y^3}{(x^2+y^2)^2} + frac{x^4 y}{(x^2+y^2)^2} - frac{y^5}{(x^2+y^2)^2}right|\
&leqleft|frac{4x^2y^3}{(x^2+y^2)^2}right| + left|frac{x^4 y}{(x^2+y^2)^2}right| + left|frac{y^5}{(x^2+y^2)^2}right|\
&leqleft|frac{4x^2y^3}{4x^2y^2}right| + left|frac{x^4 y}{4x^2y^2}right| + left|frac{y^5}{4x^2y^2}right|\
&= left|yright| + left|frac{x^2}{4y}right| + left|frac{y^3}{4x^2}right|
end{align*}$$
Got stuck here and not sure if my approach is correct.
real-analysis limits multivariable-calculus epsilon-delta
edited Jan 28 at 17:57
MathOverview
8,96243164
asked Jan 28 at 17:49
mikemike
564
$endgroup$
Suppose $f:mathbb{R}^2rightarrow mathbb{R}$ is defined as $$(x,y) longmapsto
left{
begin{array}{cl}
dfrac{4x^2y^3 +x^4y - y^5}{(x^2+y^2)^2} & mbox {if } (x,y) neq (0,0) \
\
0 & mbox {if } (x,y) = (0,0)
end{array}
right.
$$
I have shown that $f$ is continuous at $(0,0)$ using polar coordinates, but I am really trying to improve my $epsilon-delta$ proofs for such limits. I always end up getting confused with the inequalities. (Sorry if the question is a bit repetitive here)
So for $epsilon > 0$ I need to find a $delta$ such that $|(x,y)| =sqrt{x^2 + y^2} < delta$ implies $|f(x,y)| < epsilon$
I have tried to work backwards using the inequality $(x^2+y^2)^2 geq 4x^2y^2$
$$begin{align*}
left|frac{4x^2y^3 +x^4y - y^5}{(x^2+y^2)^2}right|&=left|frac{4x^2y^3}{(x^2+y^2)^2} + frac{x^4 y}{(x^2+y^2)^2} - frac{y^5}{(x^2+y^2)^2}right|\
&leqleft|frac{4x^2y^3}{(x^2+y^2)^2}right| + left|frac{x^4 y}{(x^2+y^2)^2}right| + left|frac{y^5}{(x^2+y^2)^2}right|\
&leqleft|frac{4x^2y^3}{4x^2y^2}right| + left|frac{x^4 y}{4x^2y^2}right| + left|frac{y^5}{4x^2y^2}right|\
&= left|yright| + left|frac{x^2}{4y}right| + left|frac{y^3}{4x^2}right|
end{align*}$$
Got stuck here and not sure if my approach is correct.
real-analysis limits multivariable-calculus epsilon-delta
real-analysis limits multivariable-calculus epsilon-delta
edited Jan 28 at 17:57
MathOverview
8,96243164
asked Jan 28 at 17:49
mikemike
564
edited Jan 28 at 17:57
MathOverview
8,96243164
asked Jan 28 at 17:49
mikemike
564
edited Jan 28 at 17:57
MathOverview
8,96243164
edited Jan 28 at 17:57
MathOverview
8,96243164
edited Jan 28 at 17:57
MathOverview
8,96243164
8,96243164
asked Jan 28 at 17:49
mikemike
564
asked Jan 28 at 17:49
mikemike
564
asked Jan 28 at 17:49
mikemike
564
564
$begingroup$
Having $x$ and $y$ in the denominator is going to cause you horrible problems with things blowing up. Just use the fact that $|x|$ and $|y|$ are both $le (x^2+y^2)^{1/2}$.
$endgroup$
– Ted Shifrin
Jan 28 at 19:16
add a comment |
$begingroup$
Having $x$ and $y$ in the denominator is going to cause you horrible problems with things blowing up. Just use the fact that $|x|$ and $|y|$ are both $le (x^2+y^2)^{1/2}$.
$endgroup$
– Ted Shifrin
Jan 28 at 19:16
$begingroup$
Having $x$ and $y$ in the denominator is going to cause you horrible problems with things blowing up. Just use the fact that $|x|$ and $|y|$ are both $le (x^2+y^2)^{1/2}$.
$endgroup$
– Ted Shifrin
Jan 28 at 19:16
$begingroup$
Having $x$ and $y$ in the denominator is going to cause you horrible problems with things blowing up. Just use the fact that $|x|$ and $|y|$ are both $le (x^2+y^2)^{1/2}$.
$endgroup$
– Ted Shifrin
Jan 28 at 19:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The essential point is that $r:=sqrt{x^2+y^2}$ governs the distance from some point $(x,y)$ to $(0,0)$. In particular $|x|leq sqrt{x^2+y^2}=r$, and similarly $|y|leq r$. It follows that
$$left|{4x^2y^3+x^4y-y^5over(x^2+y^2)^2}right|leq(4+1+1){r^5over r^4}=6r .tag{1}$$
Given an $epsilon>0$ choose $delta:={epsilonover6}$. If $|(x,y)|=r<delta$ then, according to $(1)$, we have
$$left|{4x^2y^3+x^4y-y^5over(x^2+y^2)^2}right|leq6 r<6delta=epsilon .$$
answered Jan 28 at 20:04
Christian BlatterChristian Blatter
176k8115327
$endgroup$
$begingroup$
Thank a lot for your answer. It made it very simple :)
$endgroup$
– mike
Jan 28 at 22:05
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
The essential point is that $r:=sqrt{x^2+y^2}$ governs the distance from some point $(x,y)$ to $(0,0)$. In particular $|x|leq sqrt{x^2+y^2}=r$, and similarly $|y|leq r$. It follows that
$$left|{4x^2y^3+x^4y-y^5over(x^2+y^2)^2}right|leq(4+1+1){r^5over r^4}=6r .tag{1}$$
Given an $epsilon>0$ choose $delta:={epsilonover6}$. If $|(x,y)|=r<delta$ then, according to $(1)$, we have
$$left|{4x^2y^3+x^4y-y^5over(x^2+y^2)^2}right|leq6 r<6delta=epsilon .$$
answered Jan 28 at 20:04
Christian BlatterChristian Blatter
176k8115327
$endgroup$
$begingroup$
Thank a lot for your answer. It made it very simple :)
$endgroup$
– mike
Jan 28 at 22:05
add a comment |
$begingroup$
The essential point is that $r:=sqrt{x^2+y^2}$ governs the distance from some point $(x,y)$ to $(0,0)$. In particular $|x|leq sqrt{x^2+y^2}=r$, and similarly $|y|leq r$. It follows that
$$left|{4x^2y^3+x^4y-y^5over(x^2+y^2)^2}right|leq(4+1+1){r^5over r^4}=6r .tag{1}$$
Given an $epsilon>0$ choose $delta:={epsilonover6}$. If $|(x,y)|=r<delta$ then, according to $(1)$, we have
$$left|{4x^2y^3+x^4y-y^5over(x^2+y^2)^2}right|leq6 r<6delta=epsilon .$$
answered Jan 28 at 20:04
Christian BlatterChristian Blatter
176k8115327
$endgroup$
$begingroup$
Thank a lot for your answer. It made it very simple :)
$endgroup$
– mike
Jan 28 at 22:05
add a comment |
$begingroup$
The essential point is that $r:=sqrt{x^2+y^2}$ governs the distance from some point $(x,y)$ to $(0,0)$. In particular $|x|leq sqrt{x^2+y^2}=r$, and similarly $|y|leq r$. It follows that
$$left|{4x^2y^3+x^4y-y^5over(x^2+y^2)^2}right|leq(4+1+1){r^5over r^4}=6r .tag{1}$$
Given an $epsilon>0$ choose $delta:={epsilonover6}$. If $|(x,y)|=r<delta$ then, according to $(1)$, we have
$$left|{4x^2y^3+x^4y-y^5over(x^2+y^2)^2}right|leq6 r<6delta=epsilon .$$
answered Jan 28 at 20:04
Christian BlatterChristian Blatter
176k8115327
$endgroup$
The essential point is that $r:=sqrt{x^2+y^2}$ governs the distance from some point $(x,y)$ to $(0,0)$. In particular $|x|leq sqrt{x^2+y^2}=r$, and similarly $|y|leq r$. It follows that
$$left|{4x^2y^3+x^4y-y^5over(x^2+y^2)^2}right|leq(4+1+1){r^5over r^4}=6r .tag{1}$$
Given an $epsilon>0$ choose $delta:={epsilonover6}$. If $|(x,y)|=r<delta$ then, according to $(1)$, we have
$$left|{4x^2y^3+x^4y-y^5over(x^2+y^2)^2}right|leq6 r<6delta=epsilon .$$
answered Jan 28 at 20:04
Christian BlatterChristian Blatter
176k8115327
answered Jan 28 at 20:04
Christian BlatterChristian Blatter
176k8115327
answered Jan 28 at 20:04
Christian BlatterChristian Blatter
176k8115327
answered Jan 28 at 20:04
Christian BlatterChristian Blatter
176k8115327
176k8115327
$begingroup$
Thank a lot for your answer. It made it very simple :)
$endgroup$
– mike
Jan 28 at 22:05
add a comment |
$begingroup$
Thank a lot for your answer. It made it very simple :)
$endgroup$
– mike
Jan 28 at 22:05
$begingroup$
Thank a lot for your answer. It made it very simple :)
$endgroup$
– mike
Jan 28 at 22:05
$begingroup$
Thank a lot for your answer. It made it very simple :)
$endgroup$
– mike
Jan 28 at 22:05
add a comment |
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$begingroup$
Having $x$ and $y$ in the denominator is going to cause you horrible problems with things blowing up. Just use the fact that $|x|$ and $|y|$ are both $le (x^2+y^2)^{1/2}$.
$endgroup$
– Ted Shifrin
Jan 28 at 19:16