Evaluating the limit of a multivariable function with paths?

[Rijad Hadzic]

Homework Statement

From here, question C.
http://tutorial.math.lamar.edu/Classes/CalcIII/Limits.aspx

lim (x,y) -> (0,0) \frac {x^2y^2}{x^4 + 3y^4}

Homework Equations

The Attempt at a Solution

So if we approach along the x axis, we know y will be 0, so we get

lim (x,0) -> (0,0) \frac {x^2(0)^2}{x^4 + 3(0)^4}lim (x,0) -> (0,0) \frac {x^2(0)^2}{x^4 + 3(0)^4}

After applying direct substitution after this, the reason they get the answer 0, and not undefined, is because the limit is APPROACHING zero, meaning its not actually getting evaluated there, and since they got 0 on the numerator, they get zero? Because if we apply direct substitution it will be undefined, but since its the limit its not actually zero is an infinitely close value to zero?

 

[andrewkirk]

After applying direct substitution after this, the reason they get the answer 0, and not undefined, is because the limit is APPROACHING zero, meaning its not actually getting evaluated there, and since they got 0 on the numerator, they get zero? Because if we apply direct substitution it will be undefined, but since its the limit its not actually zero is an infinitely close value to zero?

That's right, the value at the actual point is irrelevant. In evaluating the limit, all we are interested in is the value at points near (0,0), not at (0,0) itself.

The value at the point itself becomes relevant if we ask whether the function is continuous at the point. But that is not what is being asked here.

 

[Mark44]

After applying direct substitution after this, the reason they get the answer 0, and not undefined

If you read the example more carefully, you'll see that they DON'T get 0 as the limit. In fact, along a different path, they get a value of 1/4. Since the limit along different paths yields different results, the limit as (x, y) approaches (0, 0) doesn't exist.

 

[Ray Vickson]

Homework Statement

From here, question C.
http://tutorial.math.lamar.edu/Classes/CalcIII/Limits.aspx

lim (x,y) -> (0,0) \frac {x^2y^2}{x^4 + 3y^4}

Homework Equations

The Attempt at a Solution

So if we approach along the x axis, we know y will be 0, so we get

lim (x,0) -> (0,0) \frac {x^2(0)^2}{x^4 + 3(0)^4}lim (x,0) -> (0,0) \frac {x^2(0)^2}{x^4 + 3(0)^4}

After applying direct substitution after this, the reason they get the answer 0, and not undefined, is because the limit is APPROACHING zero, meaning its not actually getting evaluated there, and since they got 0 on the numerator, they get zero? Because if we apply direct substitution it will be undefined, but since its the limit its not actually zero is an infinitely close value to zero?

Sometimes questions like this one are most easily answered by going to polar coordinates: $x = r \cos(\theta), y = r \sin(\theta)$, and you want to take $r \to 0.$ Try it and see: does the limit depend on $\theta?$

 

[Rijad Hadzic]

If you read the example more carefully, you'll see that they DON'T get 0 as the limit. In fact, along a different path, they get a value of 1/4. Since the limit along different paths yields different results, the limit as (x, y) approaches (0, 0) doesn't exist.

Sorry I meant to say the value of that path gives you 0.

But I think I'm satisfied with the answers given, thank you guys!

 

[Rijad Hadzic]

Sometimes questions like this one are most easily answered by going to polar coordinates: $x = r \cos(\theta), y = r \sin(\theta)$, and you want to take $r \to 0.$ Try it and see: does the limit depend on $\theta?$

No, the limit depends entirely on r, right? So essentially what we're doing is making a multivariable function, just one variable? Which is similar to using paths, and going along a certain axis?

And because when using polar coordinates, you find that there is no r after simplification, which means that the limit does not exist? Let me know if I'm right please, because if so, using polar coordinates sound like a very important technique to know..

 

[Mark44]

Sometimes questions like this one are most easily answered by going to polar coordinates: $x = r \cos(\theta), y = r \sin(\theta)$, and you want to take $r \to 0.$ Try it and see: does the limit depend on $\theta?$

No, the limit depends entirely on r, right?

In this case, no.

So essentially what we're doing is making a multivariable function, just one variable? Which is similar to using paths, and going along a certain axis?

Not really. In changing a limit from $\lim_{(x, y) \to (0, 0)} \dots$ to one of the form $\lim_{r \to 0}\dots$, you aren't specifying any path. You're just saying that r is getting smaller, regardless of the path taken.

And because when using polar coordinates, you find that there is no r after simplification, which means that the limit does not exist?

That's somewhat oversimplified. If you find that the limit ends up in an expression that involves only $\theta$ (which can take on arbitrary values), then you can conclude that the limit doesn't exist.

Let me know if I'm right please, because if so, using polar coordinates sound like a very important technique to know..

Yes, it is.

 

[Rijad Hadzic]

In this case, no.

What do you mean in this case no? In this case no, that the limit does not depend entirely on r, or that it does?

 

[Mark44]

What do you mean in this case no? In this case no, that the limit does not depend entirely on r, or that it does?

In this case (i.e., changing to r --> 0) the limit doesn't depend on r at all. That's what I meant.