Mixing
Problem with multivariable inequality
Clash Royale CLAN TAG#URR8PPP
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Basically I'm trying to prove a limit with the delta epsilon definition and I'm reading about how to do it. In one example, the book does the following:
$lim_(x,y)to (1,-1) fracy+1(3(x-1)^2+2(y+1)^2)^1/3 leq lim_(x,y)to (1,-1) fraclVert[x-1,y+1]rVert(2lVert[x-1,y+1]rVert ^2+2lVert[x-1,y+1]rVert^2)^1/3$
I don't understand why this is true. I know for a fact that $(y+1) leq ||[x-1,y+1]||$ but I don't understand what's going on in the denominator. As far as I know the denominator of the first function should be greater than or equal to the one on the second function for this inequality to hold true, but I don't understand why it works here.
Edit: $lVert [x-1,y+1] rVert^2$ is equal to $sqrt (x-1)^2+(y+1)^2 $
analysis inequality
asked Jul 14 at 20:52
Direwolfox
62
 |Â
show 3 more comments
up vote
1
down vote
favorite
Basically I'm trying to prove a limit with the delta epsilon definition and I'm reading about how to do it. In one example, the book does the following:
$lim_(x,y)to (1,-1) fracy+1(3(x-1)^2+2(y+1)^2)^1/3 leq lim_(x,y)to (1,-1) fraclVert[x-1,y+1]rVert(2lVert[x-1,y+1]rVert ^2+2lVert[x-1,y+1]rVert^2)^1/3$
I don't understand why this is true. I know for a fact that $(y+1) leq ||[x-1,y+1]||$ but I don't understand what's going on in the denominator. As far as I know the denominator of the first function should be greater than or equal to the one on the second function for this inequality to hold true, but I don't understand why it works here.
Edit: $lVert [x-1,y+1] rVert^2$ is equal to $sqrt (x-1)^2+(y+1)^2 $
analysis inequality
asked Jul 14 at 20:52
Direwolfox
62
1
What is $||(x-1),(y+1)||$?
– Mostafa Ayaz
Jul 14 at 20:56
The right hand side is the sum of two identical terms?
– Bernard
Jul 14 at 21:04
@MostafaAyaz Oh sorry it is the norm of the vector. I already edited my notation to the one in MathJax.
– Direwolfox
Jul 14 at 21:12
@Bernard Yes it is, I guess it is written that way to make it more understandable, as in one of those is related to 3(x-1)^2 and the other one to 2(y+1)^2. In the next step he adds them up and ends up with 4||(x-1),(y+1)||^2
– Direwolfox
Jul 14 at 21:14
@Direwolfox do you mean a vector with entries $[x-1,y+1]$?
– mathreadler
Jul 14 at 21:14
 |Â
show 3 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Basically I'm trying to prove a limit with the delta epsilon definition and I'm reading about how to do it. In one example, the book does the following:
$lim_(x,y)to (1,-1) fracy+1(3(x-1)^2+2(y+1)^2)^1/3 leq lim_(x,y)to (1,-1) fraclVert[x-1,y+1]rVert(2lVert[x-1,y+1]rVert ^2+2lVert[x-1,y+1]rVert^2)^1/3$
I don't understand why this is true. I know for a fact that $(y+1) leq ||[x-1,y+1]||$ but I don't understand what's going on in the denominator. As far as I know the denominator of the first function should be greater than or equal to the one on the second function for this inequality to hold true, but I don't understand why it works here.
Edit: $lVert [x-1,y+1] rVert^2$ is equal to $sqrt (x-1)^2+(y+1)^2 $
analysis inequality
asked Jul 14 at 20:52
Direwolfox
62
Basically I'm trying to prove a limit with the delta epsilon definition and I'm reading about how to do it. In one example, the book does the following:
$lim_(x,y)to (1,-1) fracy+1(3(x-1)^2+2(y+1)^2)^1/3 leq lim_(x,y)to (1,-1) fraclVert[x-1,y+1]rVert(2lVert[x-1,y+1]rVert ^2+2lVert[x-1,y+1]rVert^2)^1/3$
I don't understand why this is true. I know for a fact that $(y+1) leq ||[x-1,y+1]||$ but I don't understand what's going on in the denominator. As far as I know the denominator of the first function should be greater than or equal to the one on the second function for this inequality to hold true, but I don't understand why it works here.
Edit: $lVert [x-1,y+1] rVert^2$ is equal to $sqrt (x-1)^2+(y+1)^2 $
analysis inequality
asked Jul 14 at 20:52
Direwolfox
62
edited Jul 14 at 21:59
asked Jul 14 at 20:52
Direwolfox
62
asked Jul 14 at 20:52
Direwolfox
62
asked Jul 14 at 20:52
Direwolfox
62
62
1
What is $||(x-1),(y+1)||$?
– Mostafa Ayaz
Jul 14 at 20:56
The right hand side is the sum of two identical terms?
– Bernard
Jul 14 at 21:04
@MostafaAyaz Oh sorry it is the norm of the vector. I already edited my notation to the one in MathJax.
– Direwolfox
Jul 14 at 21:12
@Bernard Yes it is, I guess it is written that way to make it more understandable, as in one of those is related to 3(x-1)^2 and the other one to 2(y+1)^2. In the next step he adds them up and ends up with 4||(x-1),(y+1)||^2
– Direwolfox
Jul 14 at 21:14
@Direwolfox do you mean a vector with entries $[x-1,y+1]$?
– mathreadler
Jul 14 at 21:14
 |Â
show 3 more comments
1
What is $||(x-1),(y+1)||$?
– Mostafa Ayaz
Jul 14 at 20:56
The right hand side is the sum of two identical terms?
– Bernard
Jul 14 at 21:04
@MostafaAyaz Oh sorry it is the norm of the vector. I already edited my notation to the one in MathJax.
– Direwolfox
Jul 14 at 21:12
@Bernard Yes it is, I guess it is written that way to make it more understandable, as in one of those is related to 3(x-1)^2 and the other one to 2(y+1)^2. In the next step he adds them up and ends up with 4||(x-1),(y+1)||^2
– Direwolfox
Jul 14 at 21:14
@Direwolfox do you mean a vector with entries $[x-1,y+1]$?
– mathreadler
Jul 14 at 21:14
1
1
What is $||(x-1),(y+1)||$?
– Mostafa Ayaz
Jul 14 at 20:56
What is $||(x-1),(y+1)||$?
– Mostafa Ayaz
Jul 14 at 20:56
The right hand side is the sum of two identical terms?
– Bernard
Jul 14 at 21:04
The right hand side is the sum of two identical terms?
– Bernard
Jul 14 at 21:04
@MostafaAyaz Oh sorry it is the norm of the vector. I already edited my notation to the one in MathJax.
– Direwolfox
Jul 14 at 21:12
@MostafaAyaz Oh sorry it is the norm of the vector. I already edited my notation to the one in MathJax.
– Direwolfox
Jul 14 at 21:12
@Bernard Yes it is, I guess it is written that way to make it more understandable, as in one of those is related to 3(x-1)^2 and the other one to 2(y+1)^2. In the next step he adds them up and ends up with 4||(x-1),(y+1)||^2
– Direwolfox
Jul 14 at 21:14
@Bernard Yes it is, I guess it is written that way to make it more understandable, as in one of those is related to 3(x-1)^2 and the other one to 2(y+1)^2. In the next step he adds them up and ends up with 4||(x-1),(y+1)||^2
– Direwolfox
Jul 14 at 21:14
@Direwolfox do you mean a vector with entries $[x-1,y+1]$?
– mathreadler
Jul 14 at 21:14
@Direwolfox do you mean a vector with entries $[x-1,y+1]$?
– mathreadler
Jul 14 at 21:14
 |Â
show 3 more comments
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1
What is $||(x-1),(y+1)||$?
– Mostafa Ayaz
Jul 14 at 20:56
The right hand side is the sum of two identical terms?
– Bernard
Jul 14 at 21:04
@MostafaAyaz Oh sorry it is the norm of the vector. I already edited my notation to the one in MathJax.
– Direwolfox
Jul 14 at 21:12
@Bernard Yes it is, I guess it is written that way to make it more understandable, as in one of those is related to 3(x-1)^2 and the other one to 2(y+1)^2. In the next step he adds them up and ends up with 4||(x-1),(y+1)||^2
– Direwolfox
Jul 14 at 21:14
@Direwolfox do you mean a vector with entries $[x-1,y+1]$?
– mathreadler
Jul 14 at 21:14