Notation: limit in two variables
I want to show that the function
$$
f: mathbb{R}^2 to mathbb{R},
(x,y) mapsto begin{cases}
x^2(x-1)(y-1)sin(xy), & (x,y) in [0,1]^2 \ 0, & text{elsewhere.}
end{cases}
$$
is continuous.
Obviously, both pieces are continuous because they are composed of elementary continuous functions.
Now I only need to show that the transition between both pieces is continuous and don't know how to notate it properly. My Idea was
begin{align*}
lim_{|x,y|_{infty} nearrow 1} f|_{[0,1]}
& = lim_{max(x,y) nearrow 1} x^2(x - 1)(y - 1)sin(xy) \
& = begin{cases}
lim_{x nearrow 1} x^2(x - 1)(y - 1)sin(xy) \
lim_{y nearrow 1} x^2(x - 1)(y - 1)sin(xy)
end{cases}
= begin{cases} 1^2(1 - 1)(y - 1)sin(y) \ x^2(x - 1)(1 - 1)sin(x) end{cases}
= 0
end{align*}
I've never seen it done that way but didn't have a better idea.
Second Attempt
For all $a in [0,1]$ we have
begin{equation*}
begin{cases}
limlimits_{(x,y)to(0,a)} f(x,y)
= 0^2(0 - 1)(a - 1)sin(0)
= 0
= f(0,a), \
limlimits_{(x,y)to(1,a)} f(x,y)
= 1^2(1 - 1)(a - 1)sin(y)
= 0
= f(1,a). \
limlimits_{(x,y)to(a,0)} f(x,y)
= a^2(a - 1)(0 - 1)sin(0)
= 0
= f(a,0), \
limlimits_{(x,y)to(a,1)} f(x,1)
= a^2(a - 1)(1 - 1)sin(x)
= 0
= f(a,1)
end{cases}
end{equation*}
real-analysis limits multivariable-calculus proof-verification continuity
asked Nov 25 '18 at 15:55
Viktor Glombik
578425
add a comment |
I want to show that the function
$$
f: mathbb{R}^2 to mathbb{R},
(x,y) mapsto begin{cases}
x^2(x-1)(y-1)sin(xy), & (x,y) in [0,1]^2 \ 0, & text{elsewhere.}
end{cases}
$$
is continuous.
Obviously, both pieces are continuous because they are composed of elementary continuous functions.
Now I only need to show that the transition between both pieces is continuous and don't know how to notate it properly. My Idea was
begin{align*}
lim_{|x,y|_{infty} nearrow 1} f|_{[0,1]}
& = lim_{max(x,y) nearrow 1} x^2(x - 1)(y - 1)sin(xy) \
& = begin{cases}
lim_{x nearrow 1} x^2(x - 1)(y - 1)sin(xy) \
lim_{y nearrow 1} x^2(x - 1)(y - 1)sin(xy)
end{cases}
= begin{cases} 1^2(1 - 1)(y - 1)sin(y) \ x^2(x - 1)(1 - 1)sin(x) end{cases}
= 0
end{align*}
I've never seen it done that way but didn't have a better idea.
Second Attempt
For all $a in [0,1]$ we have
begin{equation*}
begin{cases}
limlimits_{(x,y)to(0,a)} f(x,y)
= 0^2(0 - 1)(a - 1)sin(0)
= 0
= f(0,a), \
limlimits_{(x,y)to(1,a)} f(x,y)
= 1^2(1 - 1)(a - 1)sin(y)
= 0
= f(1,a). \
limlimits_{(x,y)to(a,0)} f(x,y)
= a^2(a - 1)(0 - 1)sin(0)
= 0
= f(a,0), \
limlimits_{(x,y)to(a,1)} f(x,1)
= a^2(a - 1)(1 - 1)sin(x)
= 0
= f(a,1)
end{cases}
end{equation*}
real-analysis limits multivariable-calculus proof-verification continuity
asked Nov 25 '18 at 15:55
Viktor Glombik
578425
add a comment |
I want to show that the function
$$
f: mathbb{R}^2 to mathbb{R},
(x,y) mapsto begin{cases}
x^2(x-1)(y-1)sin(xy), & (x,y) in [0,1]^2 \ 0, & text{elsewhere.}
end{cases}
$$
is continuous.
Obviously, both pieces are continuous because they are composed of elementary continuous functions.
Now I only need to show that the transition between both pieces is continuous and don't know how to notate it properly. My Idea was
begin{align*}
lim_{|x,y|_{infty} nearrow 1} f|_{[0,1]}
& = lim_{max(x,y) nearrow 1} x^2(x - 1)(y - 1)sin(xy) \
& = begin{cases}
lim_{x nearrow 1} x^2(x - 1)(y - 1)sin(xy) \
lim_{y nearrow 1} x^2(x - 1)(y - 1)sin(xy)
end{cases}
= begin{cases} 1^2(1 - 1)(y - 1)sin(y) \ x^2(x - 1)(1 - 1)sin(x) end{cases}
= 0
end{align*}
I've never seen it done that way but didn't have a better idea.
Second Attempt
For all $a in [0,1]$ we have
begin{equation*}
begin{cases}
limlimits_{(x,y)to(0,a)} f(x,y)
= 0^2(0 - 1)(a - 1)sin(0)
= 0
= f(0,a), \
limlimits_{(x,y)to(1,a)} f(x,y)
= 1^2(1 - 1)(a - 1)sin(y)
= 0
= f(1,a). \
limlimits_{(x,y)to(a,0)} f(x,y)
= a^2(a - 1)(0 - 1)sin(0)
= 0
= f(a,0), \
limlimits_{(x,y)to(a,1)} f(x,1)
= a^2(a - 1)(1 - 1)sin(x)
= 0
= f(a,1)
end{cases}
end{equation*}
real-analysis limits multivariable-calculus proof-verification continuity
asked Nov 25 '18 at 15:55
Viktor Glombik
578425
I want to show that the function
$$
f: mathbb{R}^2 to mathbb{R},
(x,y) mapsto begin{cases}
x^2(x-1)(y-1)sin(xy), & (x,y) in [0,1]^2 \ 0, & text{elsewhere.}
end{cases}
$$
is continuous.
Obviously, both pieces are continuous because they are composed of elementary continuous functions.
Now I only need to show that the transition between both pieces is continuous and don't know how to notate it properly. My Idea was
begin{align*}
lim_{|x,y|_{infty} nearrow 1} f|_{[0,1]}
& = lim_{max(x,y) nearrow 1} x^2(x - 1)(y - 1)sin(xy) \
& = begin{cases}
lim_{x nearrow 1} x^2(x - 1)(y - 1)sin(xy) \
lim_{y nearrow 1} x^2(x - 1)(y - 1)sin(xy)
end{cases}
= begin{cases} 1^2(1 - 1)(y - 1)sin(y) \ x^2(x - 1)(1 - 1)sin(x) end{cases}
= 0
end{align*}
I've never seen it done that way but didn't have a better idea.
Second Attempt
For all $a in [0,1]$ we have
begin{equation*}
begin{cases}
limlimits_{(x,y)to(0,a)} f(x,y)
= 0^2(0 - 1)(a - 1)sin(0)
= 0
= f(0,a), \
limlimits_{(x,y)to(1,a)} f(x,y)
= 1^2(1 - 1)(a - 1)sin(y)
= 0
= f(1,a). \
limlimits_{(x,y)to(a,0)} f(x,y)
= a^2(a - 1)(0 - 1)sin(0)
= 0
= f(a,0), \
limlimits_{(x,y)to(a,1)} f(x,1)
= a^2(a - 1)(1 - 1)sin(x)
= 0
= f(a,1)
end{cases}
end{equation*}
real-analysis limits multivariable-calculus proof-verification continuity
real-analysis limits multivariable-calculus proof-verification continuity
asked Nov 25 '18 at 15:55
Viktor Glombik
578425
asked Nov 25 '18 at 15:55
Viktor Glombik
578425
edited Nov 25 '18 at 17:05
asked Nov 25 '18 at 15:55
Viktor Glombik
578425
asked Nov 25 '18 at 15:55
Viktor Glombik
578425
asked Nov 25 '18 at 15:55
Viktor Glombik
578425
578425
add a comment |
add a comment |
1 Answer
1
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oldest
votes
To prove continuity we need to show that for $0le ale 1$
$lim_{(x,y)to(0,a)}f(x,y)=f(0,a)=0$
$lim_{(x,y)to(1,a)}f(x,y)=f(1,a)=0$
$lim_{(x,y)to(a,0)}f(x,y)=f(a,0)=0$
$lim_{(x,y)to(a,1)}f(x,y)=f(a,1)=0$
answered Nov 25 '18 at 16:04
gimusi
1
1
The variables $x, y$ in an expression of the form $lim_{(x, y)to(a, b)}f(x, y)$ are bound to their occurrences in $f(x, y)$. You cannot simultaneously use them as free variables occurring in $(a, b)$.
– Calum Gilhooley
Nov 25 '18 at 16:21
@CalumGilhooley Ok that’s a good point! I fix that, thanks.
– gimusi
Nov 25 '18 at 16:24
So my reasoning for the first would be $$limlimits_{(x,y)to(0,a)} f(x,y) = 0^2(0 - 1)(a - 1)sin(0) = 0 = f(0,a)$$?
– Viktor Glombik
Nov 25 '18 at 16:34
@ViktorGlombik For the first for we have $|x^2(x-1)(y-1)sin(xy) to 0cdot (-1) cdot (a-1)cdot sin (0cdot a)=0$.
– gimusi
Nov 25 '18 at 16:43
Well that's exactly what I wrote, right?
– Viktor Glombik
Nov 25 '18 at 16:54
|
show 3 more comments
Your Answer
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1 Answer
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1 Answer
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active
oldest
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active
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active
oldest
votes
To prove continuity we need to show that for $0le ale 1$
$lim_{(x,y)to(0,a)}f(x,y)=f(0,a)=0$
$lim_{(x,y)to(1,a)}f(x,y)=f(1,a)=0$
$lim_{(x,y)to(a,0)}f(x,y)=f(a,0)=0$
$lim_{(x,y)to(a,1)}f(x,y)=f(a,1)=0$
answered Nov 25 '18 at 16:04
gimusi
1
1
The variables $x, y$ in an expression of the form $lim_{(x, y)to(a, b)}f(x, y)$ are bound to their occurrences in $f(x, y)$. You cannot simultaneously use them as free variables occurring in $(a, b)$.
– Calum Gilhooley
Nov 25 '18 at 16:21
@CalumGilhooley Ok that’s a good point! I fix that, thanks.
– gimusi
Nov 25 '18 at 16:24
So my reasoning for the first would be $$limlimits_{(x,y)to(0,a)} f(x,y) = 0^2(0 - 1)(a - 1)sin(0) = 0 = f(0,a)$$?
– Viktor Glombik
Nov 25 '18 at 16:34
@ViktorGlombik For the first for we have $|x^2(x-1)(y-1)sin(xy) to 0cdot (-1) cdot (a-1)cdot sin (0cdot a)=0$.
– gimusi
Nov 25 '18 at 16:43
Well that's exactly what I wrote, right?
– Viktor Glombik
Nov 25 '18 at 16:54
|
show 3 more comments
To prove continuity we need to show that for $0le ale 1$
$lim_{(x,y)to(0,a)}f(x,y)=f(0,a)=0$
$lim_{(x,y)to(1,a)}f(x,y)=f(1,a)=0$
$lim_{(x,y)to(a,0)}f(x,y)=f(a,0)=0$
$lim_{(x,y)to(a,1)}f(x,y)=f(a,1)=0$
answered Nov 25 '18 at 16:04
gimusi
1
1
The variables $x, y$ in an expression of the form $lim_{(x, y)to(a, b)}f(x, y)$ are bound to their occurrences in $f(x, y)$. You cannot simultaneously use them as free variables occurring in $(a, b)$.
– Calum Gilhooley
Nov 25 '18 at 16:21
@CalumGilhooley Ok that’s a good point! I fix that, thanks.
– gimusi
Nov 25 '18 at 16:24
So my reasoning for the first would be $$limlimits_{(x,y)to(0,a)} f(x,y) = 0^2(0 - 1)(a - 1)sin(0) = 0 = f(0,a)$$?
– Viktor Glombik
Nov 25 '18 at 16:34
@ViktorGlombik For the first for we have $|x^2(x-1)(y-1)sin(xy) to 0cdot (-1) cdot (a-1)cdot sin (0cdot a)=0$.
– gimusi
Nov 25 '18 at 16:43
Well that's exactly what I wrote, right?
– Viktor Glombik
Nov 25 '18 at 16:54
|
show 3 more comments
To prove continuity we need to show that for $0le ale 1$
$lim_{(x,y)to(0,a)}f(x,y)=f(0,a)=0$
$lim_{(x,y)to(1,a)}f(x,y)=f(1,a)=0$
$lim_{(x,y)to(a,0)}f(x,y)=f(a,0)=0$
$lim_{(x,y)to(a,1)}f(x,y)=f(a,1)=0$
answered Nov 25 '18 at 16:04
gimusi
1
To prove continuity we need to show that for $0le ale 1$
$lim_{(x,y)to(0,a)}f(x,y)=f(0,a)=0$
$lim_{(x,y)to(1,a)}f(x,y)=f(1,a)=0$
$lim_{(x,y)to(a,0)}f(x,y)=f(a,0)=0$
$lim_{(x,y)to(a,1)}f(x,y)=f(a,1)=0$
answered Nov 25 '18 at 16:04
gimusi
1
edited Nov 25 '18 at 16:27
answered Nov 25 '18 at 16:04
gimusi
1
answered Nov 25 '18 at 16:04
gimusi
1
answered Nov 25 '18 at 16:04
gimusi
1
1
1
The variables $x, y$ in an expression of the form $lim_{(x, y)to(a, b)}f(x, y)$ are bound to their occurrences in $f(x, y)$. You cannot simultaneously use them as free variables occurring in $(a, b)$.
– Calum Gilhooley
Nov 25 '18 at 16:21
@CalumGilhooley Ok that’s a good point! I fix that, thanks.
– gimusi
Nov 25 '18 at 16:24
So my reasoning for the first would be $$limlimits_{(x,y)to(0,a)} f(x,y) = 0^2(0 - 1)(a - 1)sin(0) = 0 = f(0,a)$$?
– Viktor Glombik
Nov 25 '18 at 16:34
@ViktorGlombik For the first for we have $|x^2(x-1)(y-1)sin(xy) to 0cdot (-1) cdot (a-1)cdot sin (0cdot a)=0$.
– gimusi
Nov 25 '18 at 16:43
Well that's exactly what I wrote, right?
– Viktor Glombik
Nov 25 '18 at 16:54
|
show 3 more comments
1
The variables $x, y$ in an expression of the form $lim_{(x, y)to(a, b)}f(x, y)$ are bound to their occurrences in $f(x, y)$. You cannot simultaneously use them as free variables occurring in $(a, b)$.
– Calum Gilhooley
Nov 25 '18 at 16:21
@CalumGilhooley Ok that’s a good point! I fix that, thanks.
– gimusi
Nov 25 '18 at 16:24
So my reasoning for the first would be $$limlimits_{(x,y)to(0,a)} f(x,y) = 0^2(0 - 1)(a - 1)sin(0) = 0 = f(0,a)$$?
– Viktor Glombik
Nov 25 '18 at 16:34
@ViktorGlombik For the first for we have $|x^2(x-1)(y-1)sin(xy) to 0cdot (-1) cdot (a-1)cdot sin (0cdot a)=0$.
– gimusi
Nov 25 '18 at 16:43
Well that's exactly what I wrote, right?
– Viktor Glombik
Nov 25 '18 at 16:54
1
1
The variables $x, y$ in an expression of the form $lim_{(x, y)to(a, b)}f(x, y)$ are bound to their occurrences in $f(x, y)$. You cannot simultaneously use them as free variables occurring in $(a, b)$.
– Calum Gilhooley
Nov 25 '18 at 16:21
The variables $x, y$ in an expression of the form $lim_{(x, y)to(a, b)}f(x, y)$ are bound to their occurrences in $f(x, y)$. You cannot simultaneously use them as free variables occurring in $(a, b)$.
– Calum Gilhooley
Nov 25 '18 at 16:21
@CalumGilhooley Ok that’s a good point! I fix that, thanks.
– gimusi
Nov 25 '18 at 16:24
@CalumGilhooley Ok that’s a good point! I fix that, thanks.
– gimusi
Nov 25 '18 at 16:24
So my reasoning for the first would be $$limlimits_{(x,y)to(0,a)} f(x,y) = 0^2(0 - 1)(a - 1)sin(0) = 0 = f(0,a)$$?
– Viktor Glombik
Nov 25 '18 at 16:34
So my reasoning for the first would be $$limlimits_{(x,y)to(0,a)} f(x,y) = 0^2(0 - 1)(a - 1)sin(0) = 0 = f(0,a)$$?
– Viktor Glombik
Nov 25 '18 at 16:34
@ViktorGlombik For the first for we have $|x^2(x-1)(y-1)sin(xy) to 0cdot (-1) cdot (a-1)cdot sin (0cdot a)=0$.
– gimusi
Nov 25 '18 at 16:43
@ViktorGlombik For the first for we have $|x^2(x-1)(y-1)sin(xy) to 0cdot (-1) cdot (a-1)cdot sin (0cdot a)=0$.
– gimusi
Nov 25 '18 at 16:43
Well that's exactly what I wrote, right?
– Viktor Glombik
Nov 25 '18 at 16:54
Well that's exactly what I wrote, right?
– Viktor Glombik
Nov 25 '18 at 16:54
|
show 3 more comments
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