# Evaluating the limit of a multivariable function with paths?

In summary, evaluating the limit of a multivariable function with paths is important for understanding the behavior of the function as it approaches a certain point in the domain. This can be determined by approaching the point along different paths and observing the function's behavior. Some common techniques for evaluating the limit include using polar coordinates, parametric equations, and substitution. The limit may not always exist if the function approaches different values or is discontinuous at the point of interest. This concept is useful in real-world applications such as predicting the behavior of a system or optimizing designs.

## Homework Statement

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

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

## 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?

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.

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.

## Homework Statement

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

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

## 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?##

Mark44 said:
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!

Ray Vickson said:
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..

Ray Vickson said:
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.

Mark44 said:
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?

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.

## 1. What is the purpose of evaluating the limit of a multivariable function with paths?

The purpose of evaluating the limit of a multivariable function with paths is to determine the behavior of the function as it approaches a certain point in the domain. This can help in understanding the overall behavior of the function and its relationship to other variables.

## 2. How do you determine the limit of a multivariable function with paths?

The limit of a multivariable function with paths can be determined by approaching the point of interest along different paths and observing the behavior of the function. If the function approaches the same value regardless of the path, then the limit exists. However, if the function approaches different values along different paths, then the limit does not exist.

## 3. What are some common techniques for evaluating the limit of a multivariable function with paths?

Some common techniques for evaluating the limit of a multivariable function with paths include using polar coordinates, using parametric equations, and using substitution to simplify the function. It is also helpful to graph the function in order to visualize its behavior along different paths.

## 4. Can the limit of a multivariable function with paths always be determined?

No, the limit of a multivariable function with paths may not always be determined. If the function approaches different values along different paths, then the limit does not exist. Additionally, if the function is discontinuous at the point of interest, the limit may not exist.

## 5. How is evaluating the limit of a multivariable function with paths useful in real-world applications?

Evaluating the limit of a multivariable function with paths is useful in real-world applications as it can help in understanding the behavior of a system or process. For example, in physics, it can help in predicting the behavior of a moving object as it approaches a certain point. In economics, it can help in analyzing the relationship between variables in a complex system. In engineering, it can aid in optimizing designs for maximum efficiency.

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