Proving a Limit with Delta-Epsilon

[ahmetbaba]

Delta epsilon definition of a limit and its proof

Homework Statement

Prove; lim x²+y²= a²+b²
(x,y) ---->(a,b)

Homework Equations

Not much equations are present, just some algebra, and the definition of a limit.

The Attempt at a Solution

Well, this is what I did

(x²+y²)-(a²+b²) < \sqrt{(x-a)²+(y-b)²} = or < than \delta.

so i opened the right hand, and left hand of the equation with the square root out.

x²+y²-a²-b²=(x²+y²+a²+b²-2xa-2yb)-2a²-2b²+2xa+2yb

The parenthesis in the right hand side is /delta². and all the other stuff on the right hand side, I'm going to try to write it in terms of \delta .

-2a²-2b²+2xa+2yb= (x-a)+(y-b)

so, I fix these up a bit;

(x-a)+(y-b)=2a(x-a) + 2b(y-b) < \delta

so here, 2a(x-a) is 2a\delta and 2b(y-b) is 2b\delta

and we know that x-a<\delta and y-a<\delta

so in the worst case, let's equal these pieced to \delta, we will have

2a\delta + 2b\delta + \delta² <\epsilon


And from here, I just solve for \delta.

Is this the correct way to solve this? Lend me a helping hand the masters of math.

 

[lanedance]

Homework Statement

Prove; lim x²+y²= a²+b²
(x,y) ---->(a,b)

Homework Equations

Not much equations are present, just some algebra, and the definition of a limit.

The Attempt at a Solution

Well, this is what I did

(x²+y²)-(a²+b²) < \sqrt{(x-a)²+(y-b)²} = or < than \delta.

this statement isn't true

 

[ahmetbaba]

well that first statement is from the epsilon-delta definition of a limit. I know they don't equal each other, but they do have a relationship, and that's how you approach the problem, that's what I think.

 

[Susanne217]

Homework Statement

Prove; lim x²+y²= a²+b²
(x,y) ---->(a,b)

Homework Equations

Not much equations are present, just some algebra, and the definition of a limit.

The Attempt at a Solution

Well, this is what I did

(x²+y²)-(a²+b²) < \sqrt{(x-a)²+(y-b)²} = or < than \delta.

so i opened the right hand, and left hand of the equation with the square root out.

x²+y²-a²-b²=(x²+y²+a²+b²-2xa-2yb)-2a²-2b²+2xa+2yb

The parenthesis in the right hand side is /delta². and all the other stuff on the right hand side, I'm going to try to write it in terms of \delta .

-2a²-2b²+2xa+2yb= (x-a)+(y-b)

so, I fix these up a bit;

(x-a)+(y-b)=2a(x-a) + 2b(y-b) < \delta

so here, 2a(x-a) is 2a\delta and 2b(y-b) is 2b\delta

and we know that x-a<\delta and y-a<\delta

so in the worst case, let's equal these pieced to \delta, we will have

2a\delta + 2b\delta + \delta² <\epsilon


And from here, I just solve for \delta.

Is this the correct way to solve this? Lend me a helping hand the masters of math.

To better understand the definition of a multiviariable func

here is a step-by-step guide http://www.math.utah.edu/~arcara/teaching/0308/handout.pdf

 

[ahmetbaba]

Suzanne, the link that you posted doesn't say how to prove that a limit exists, it basically says that we did such things and now do these for practice.

Is there no one here that knows about these things in detail?

 

[Mark44]

You need to show that for any positive epsilon, there is a positive number delta such that if |(x, y) - (a, b)| < delta, then |x^2 + y^2 - a^2 - b^2)| < epsilon.

What the first inequality says is that (x, y) can be any point within delta of the point (a, b). I'm having a hard time following your work, especially the part just before you said you "opened" the right side.

 

[ahmetbaba]

well the right hand side is the length. [(x-a)^2+(y-b)^2]^.5, When I say I open this up, I mean, i take its square and distribute the squares.

 

[Gib Z]

It's extremely hard to read your post since you don't use latex, so I'm going to try lead you down a different way.

You want to show that for every ε>0 there exists δ>0 so that (x-a)^2 + (y-b)^2 &lt; \delta^2 implies ( x^2-a^2 + y^2-b^2) = (x-a)(x+a) + (y-b)(y+b) &lt; \epsilon.

So a method you could use is to start with (x-a)^2 + (y-b)^2 &lt; \delta^2, and use it to show that it implies some other inequality (x-a)(x+a) + (y-b)(y+b) &lt; f(\delta). Then if you can show that f is invertible, so that for every ε>0 there exists δ>0 so that f(δ) = ε, then we have found a way to satisfy our definition. The first step in this method is to use the starting inequality and get some basic, weaker inequalities from it, so do that first and play around with it.

 

[HallsofIvy]

With functions of two variables, I find it easiest to "translate" so that the limit is as (x, y) goes to (0, 0) and then change to polar coordinates. That way the distance to (0, 0) is measured by the single variable, r.

Let x'= x- a, y'= y- b. Then x= x'+ a, y= y'+ b, and x^2+ y^2= (x&#039;+a)^2+ (y&#039;+ b)^2=x&#039;^2+ y&#039;^2+ 2ax&#039;+ 2by&#039;+ a^2+ b^2. In polar coordinates, x&#039;^2+ y&#039;^2= r^2 2ax&#039;= 2ar cos(\theta) and 2by&#039;= 2br sin(\theta) so that x^2+ y^2= r^2+ 2r(a cos(\theta)+ b sin(\theta)+ a^2+ b^2.

That will have limit, as (x, y) goes to (a, b), a^2+ b^2 if and only if r^2+ 2r(a cos(\theta)+ b sin(\theta)= r(r+ 2(acos(\theta)+ b sin(\theta)) goes to 0 as r goes to 0 (for any value of \theta).