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Proving the limit of a multivariable function

1. Homework Statement
Find the limit if it exists, or show that the limit does not exist.
lim (x,y)-> (1,0) (xy-y)/((x-1)^2+y^2)

2. Homework Equations
lim (x,y)-> (a,b) f(x,y)
0<((x-a)^2+(y-b)^2)^1/2<[itex]\delta[/itex]
abs(f(x,y)-L)<[itex]\epsilon[/itex]

3. The Attempt at a Solution
I tried to prove that it does not exist by analyzing the limit coming in from the x & y axes, and along lines y=x, yatta yatta. I kept getting 0, so I then tried to prove the limit exists and equals zero using the delta epsilon method. There I ran into problems, I have a total of 3 calculus books, each only has one example for the method and they are all the same example, which is also the same and only example that was covered in my class. lim (x,y) -> (0,0) (3yx^2)/(x^2+y^2). I am just looking for a starting point.

So far I have
0<((x-1)^2+(y)^2)^1/2<[itex]\delta[/itex]
abs((xy-y)/((x-1)^2+y^2))<[itex]\epsilon[/itex]
I know I need to manipulate it so that I can relate delta to be some multiple of epsilon, but don't know how.
 

SammyS

Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,160
916
1. Homework Statement
Find the limit if it exists, or show that the limit does not exist.
lim (x,y)-> (1,0) (xy-y)/((x-1)^2+y^2)

2. Homework Equations
lim (x,y)-> (a,b) f(x,y)
0<((x-a)^2+(y-b)^2)^1/2<[itex]\delta[/itex]
abs(f(x,y)-L)<[itex]\epsilon[/itex]

3. The Attempt at a Solution
I tried to prove that it does not exist by analyzing the limit coming in from the x & y axes, and along lines y=x, yatta yatta. I kept getting 0, so I then tried to prove the limit exists and equals zero using the delta epsilon method. There I ran into problems, I have a total of 3 calculus books, each only has one example for the method and they are all the same example, which is also the same and only example that was covered in my class. lim (x,y) -> (0,0) (3yx^2)/(x^2+y^2). I am just looking for a starting point.

So far I have
0<((x-1)^2+(y)^2)^1/2<[itex]\delta[/itex]
abs((xy-y)/((x-1)^2+y^2))<[itex]\epsilon[/itex]
I know I need to manipulate it so that I can relate delta to be some multiple of epsilon, but don't know how.
Try another path.

y = m(x-1) approaches the point (1, 0) along a line of arbitrary slope.
 

Dick

Science Advisor
Homework Helper
26,258
618
1. Homework Statement
Find the limit if it exists, or show that the limit does not exist.
lim (x,y)-> (1,0) (xy-y)/((x-1)^2+y^2)

2. Homework Equations
lim (x,y)-> (a,b) f(x,y)
0<((x-a)^2+(y-b)^2)^1/2<[itex]\delta[/itex]
abs(f(x,y)-L)<[itex]\epsilon[/itex]

3. The Attempt at a Solution
I tried to prove that it does not exist by analyzing the limit coming in from the x & y axes, and along lines y=x, yatta yatta. I kept getting 0, so I then tried to prove the limit exists and equals zero using the delta epsilon method. There I ran into problems, I have a total of 3 calculus books, each only has one example for the method and they are all the same example, which is also the same and only example that was covered in my class. lim (x,y) -> (0,0) (3yx^2)/(x^2+y^2). I am just looking for a starting point.

So far I have
0<((x-1)^2+(y)^2)^1/2<[itex]\delta[/itex]
abs((xy-y)/((x-1)^2+y^2))<[itex]\epsilon[/itex]
I know I need to manipulate it so that I can relate delta to be some multiple of epsilon, but don't know how.
I'd first try a change of variables. Put u=x-1. So the limit is now (u,y)->(0,0). Denominator becomes u^2+y^2. What's the numerator? It might be easier to see that way.
 
Last edited:
Would either of you mind giving an example of using the delta epsilon method anyway?
 

Dick

Science Advisor
Homework Helper
26,258
618
Would either of you mind giving an example of using the delta epsilon method anyway?
You don't need it for this one. It has no limit.
 
Would either of you mind giving a different* example of using the delta epsilon method anyway?
 

Dick

Science Advisor
Homework Helper
26,258
618
Would either of you mind giving a different* example of using the delta epsilon method anyway?
Ok, take f(x,y)=4xy/sqrt(x^2+y^2). (x,y)->(0,0). Change it to polar coordinates. You get |4r*cos(θ)*r*sin(θ)/r|=|4r*cos(θ)*sin(θ)|<=4r. So the limit is 0. And r is the "((x-a)^2+(y-b)^2)^1/2" in your definition. Pick ε>0. You want |f(x,y)-0|<ε, |f(x,y)-0|<4r, so if you pick δ=ε/4 it works.
 

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