Limits of more than one function

[Mdhiggenz]

Homework Statement

For some reason I'm not grasping the concept of limits when another function is included.

For example,

lim (x²-y²)/(x²+y²)
(x,y)---->(0,0)


So pretty much what I did was took the Limit as (x,y=x)------>0 and got 0. Then

I took the limit as (x,y=0)------>0 and got 1.

They are not equal so the limit does not exist.

However the person who solved the question instead of taking the Limit as (x,y=x) they took the limit as (x=0,y) and got -1 and then from there stated that the limit did not exist.

Was my method wrong?

Homework Equations

The Attempt at a Solution

 

[STEMucator]

Homework Statement

For some reason I'm not grasping the concept of limits when another function is included.

For example,

lim (x²-y²)/(x²+y²)
(x,y)---->(0,0)


So pretty much what I did was took the Limit as (x,y=x)------>0 and got 0. Then

I took the limit as (x,y=0)------>0 and got 1.

They are not equal so the limit does not exist.

However the person who solved the question instead of taking the Limit as (x,y=x) they took the limit as (x=0,y) and got -1 and then from there stated that the limit did not exist.

Was my method wrong?

Homework Equations

The Attempt at a Solution

The limit of a function f(x,y) exists only if all of the limits match no matter what particular choice of (x,y) you choose.

The limit has to be the same whether (x,y) = (0,0) or (x,y) = (x,0) or (x,y) = (x,x^3).

No matter where you approach the limit from, if not all the answers are the same, then your limit does not exist and hence f is not a continuous function.

 

[Mdhiggenz]

So approaching this problem, should I always do 3 just to be safe?

For instance what if both (x,y) = (0,0) or (x,y) = (x,0) contain the same value but (x,y) = (x,x^3) Was the odd ball out. Seems like a good way to trick students on an examination. What do you think?

 

[SammyS]

So approaching this problem, should I always do 3 just to be safe?

For instance what if both (x,y) = (0,0) or (x,y) = (x,0) contain the same value but (x,y) = (x,x^3) Was the odd ball out. Seems like a good way to trick students on an examination. What do you think?

(Regarding the title of thread: It's clear that you mean, more than one variable, rather than more than one function.)

Using three paths is not safe, nor any particular number.

If you do get a different limit along two different paths, then, you are correct to say, the limit doesn't exist.

When approaching (0,0), a good general method is to use polar coordinates, (r, θ), and consider the limit as r → 0 .

In the end you may need to consider an epsilon-delta argument in however many coordinates are involved.

 

[Mdhiggenz]

As r approaches 0 from the right you mean?

 

[SammyS]

As r approaches 0 from the right you mean?

As r approaches 0 from the positive.

 

[HallsofIvy]

In polar coordinates, r is always positive.

 

[Mdhiggenz]

So let me get this straight. If I'm given a limit of more than one variable and I'm told to solve the limit. It is best to change to polar coordinates, since I already know that it exist.

 

[SammyS]

So let me get this straight. If I'm given a limit of more than one variable and I'm told to solve the limit. It is best to change to polar coordinates, since I already know that it exist.

If you already know that the limit exists, then using any path will work.

If you want to prove that such a limit exists, changing to polar coordinates is generally a good choice, but circumstances may dictate some other method.

 

[jbunniii]

Homework Statement

For some reason I'm not grasping the concept of limits when another function is included.

For example,

lim (x²-y²)/(x²+y²)
(x,y)---->(0,0)So pretty much what I did was took the Limit as (x,y=x)------>0 and got 0. Then

I took the limit as (x,y=0)------>0 and got 1.

They are not equal so the limit does not exist.

However the person who solved the question instead of taking the Limit as (x,y=x) they took the limit as (x=0,y) and got -1 and then from there stated that the limit did not exist.

Was my method wrong?

Your method is correct, and so is the other person's method. You found that approaching (0,0) by two different paths gives you two different results. Therefore the limit does not exist. The other person did the same thing as you, but chose a different pair of paths. That's fine. As long as you get two DIFFERENT answers using two different paths, it doesn't matter what those paths were. You can conclude that the limit does NOT exist.

As others have pointed out, however, if you get the SAME answer using two different paths, that's not enough to conclude that the limit DOES exist.

 

[Mdhiggenz]

Gotcha so in order to prove that the limit exist I would have to use the definition. F(x,y)-L... ext ext...