# Limits in multivariable calculus

• Chris18
In summary, the conversation discusses finding the limit of a function using polar coordinates and small angle approximation. One approach involves using the factorization technique and the other involves simplifying the expression. The limit is shown to exist and be equal to 0, but there is a discussion about whether the function is valid at certain points.
Chris18
1. Find if the limit exist: sin (x^3 + y^3) / (x + y)
(x,y)-> (0,0)
So I am starting solving this by using polar coordinates form and I get to lim= sin r^3 ( cos^3θ + sin^3θ) / r ( cosθ + sinθ) = lim r^2 ( cos^2Θ + sin^2Θ) My question is ok so far and how should I procceed from now on?

An alternative (and, I suspect, easier) approach is to use the factorisation ##x^3+y^3=(x+y)(x^2-xy+y^2)## together with the small angle approximation ##\lim_{\theta\to 0}\frac{\sin\theta}\theta=1##.

That's a good solution indeed thanks a lot but still I am interested to see how my way will go on? If I have r^2( cos^2θ + sin^2θ) can I say that the () is zero or should I do something more? Thanks in advance

Chris18 said:
1. Find if the limit exist: sin (x^3 + y^3) / (x + y)
(x,y)-> (0,0)
So I am starting solving this by using polar coordinates form and I get to lim= sin r^3 ( cos^3θ + sin^3θ) / r ( cosθ + sinθ) = lim r^2 ( cos^2Θ + sin^2Θ) My question is ok so far and how should I procceed from now on?
No: ##(\cos^3 \theta + \sin^3 \theta) / (\cos \theta + \sin \theta) \neq \cos^2 \theta + \sin^2 \theta##.

Chris18 said:
That's a good solution indeed thanks a lot but still I am interested to see how my way will go on? If I have r^2( cos^2θ + sin^2θ) can I say that the () is zero or should I do something more? Thanks in advance
As @Ray Vickson pointed out, your expression is not correct.
$$\frac{\cos^3 \theta + \sin^3 \theta}{\cos \theta + \sin \theta} = \cos^2 \theta - \cos\theta\sin\theta + \sin^2 \theta = 1 - \frac 12 \sin\theta$$
which is effectively what @andrewkirk suggested.

It is slightly different from yours, but the following argument is the same.
We can see that whatever ##\theta## is, the absolute value of the expression is less than or equal to 3/2.
When we multiply it by ##r^2## the result will go to zero when r does, as does the sine that is taken subsequently.
Therefore the limit exists and is 0.

Last edited:
I like Serena said:
As @Ray Vickson pointed out, your expression is not correct.
$$\frac{\cos^3 \theta + \sin^3 \theta}{\cos \theta + \sin \theta} = \cos^2 \theta - \cos\theta\sin\theta + \sin^2 \theta = 1 - \frac 12 \sin\theta$$
which is effectively what @andrewkirk suggested.

It is slightly different from yours, but the following argument is the same.
We can see that whatever ##\theta## is, the absolute value of the expression is less than or equal to 3/2.
When we multiply it by ##r^2## the result will go to zero when r does.
Therefore the limit exists and is 0.

Yes, indeed----almost. The function has the form 0/0 along the line ##y = -x##. I don't know whether that invalidates the whole concept of a limit, since when we take a limit in 2 dimensions we are supposed to allow ##(x,y) \to (0,0)## and any direction or along any curve. The limit-point (0,0) itself is not a problem, even though it, too, gives 0/0. I am genuinely puzzled by this question.

Ray Vickson said:
Yes, indeed----almost. The function has the form 0/0 along the line ##y = -x##. I don't know whether that invalidates the whole concept of a limit, since when we take a limit in 2 dimensions we are supposed to allow ##(x,y) \to (0,0)## and any direction or along any curve. The limit-point (0,0) itself is not a problem, even though it, too, gives 0/0. I am genuinely puzzled by this question.
Since those points are not part of the domain of the function, they do not participate in the evaluation of the limit do they?
We would only pick directions or curves that are inside the domain.

## 1. What is a limit in multivariable calculus?

A limit in multivariable calculus is a fundamental concept that describes the behavior of a function as the input values approach a specific point or value. It is used to determine the value of a function at a given point by examining its behavior near that point.

## 2. How do you evaluate limits in multivariable calculus?

Limits in multivariable calculus can be evaluated by taking the limit of the function along different paths or directions approaching the point in question. This involves taking the limit along the x-axis, y-axis, and other relevant directions to see if the function approaches the same value or not. If it does, then the limit exists; if not, then the limit does not exist.

## 3. Can the limit of a multivariable function be undefined?

Yes, the limit of a multivariable function can be undefined if the function behaves differently along different paths approaching the point in question. In this case, the limit does not exist, and the function is said to be discontinuous at that point.

## 4. What is the relationship between continuity and limits in multivariable calculus?

In multivariable calculus, a function is said to be continuous at a point if its limit at that point exists and is equal to the value of the function at that point. In other words, continuity is closely related to limits, as it ensures that a function is well-behaved and has no sudden jumps or breaks at a particular point.

## 5. How are limits used in real-world applications?

Limits in multivariable calculus are used in various real-world applications, such as optimization problems, where the goal is to find the maximum or minimum value of a function. They are also used in physics and engineering to describe the behavior of physical systems and to solve problems involving rates of change and motion.

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