Understand Multivariable Limit "As (x,y) \rightarrow (0,0), r \rightarrow

In summary, a multivariable limit is a mathematical concept that describes the behavior of a function as two or more variables approach a certain point simultaneously. To identify the limit of a multivariable function, the point towards which the variables are approaching must be determined and the function can then be evaluated at that point. Understanding multivariable limits is important in analyzing and predicting the behavior of functions in complex systems. Common techniques for evaluating multivariable limits include substitution, approaching along specific paths, using polar coordinates, and using the Squeeze Theorem.
  • #1
Bashyboy
1,421
5
"As [itex](x,y) \rightarrow (0,0)[/itex], [itex]r \rightarrow[/itex]," is a fact that I am given, in order to solve a problem. I simply want to know if I properly understand why this fact is true.

I know that [itex]x = rcos \theta [/itex] and [itex]y=rsin \theta[/itex]. If I were to look at the individual limits, as [itex]x \rightarrow 0[/itex] and [itex]y \rightarrow 0[/itex], then I see each relation becomes [itex] 0 = rcos \theta [/itex] and [itex]0=rsin \theta[/itex]. Now, as x and y approach 0, there is no angle that [itex]\theta[/itex] approaches that make [itex]rcos \theta [/itex] and [itex]rsin \theta[/itex] simultaneously zero; hence, it must be that r approaches 0 as to make each expression zero simultaneously.

In summary, [itex]\lim_{(x,y) \rightarrow (0,0)}~(x+y) = lim_{r \rightarrow 0} ~ (rcos \theta + rsin \theta)[/itex].

Are my arguments correct?
 
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  • #2
Bashyboy said:
"As [itex](x,y) \rightarrow (0,0)[/itex], [itex]r \rightarrow[/itex],"

You mean ##r\rightarrow 0##, right?

is a fact that I am given, in order to solve a problem. I simply want to know if I properly understand why this fact is true.

I know that [itex]x = rcos \theta [/itex] and [itex]y=rsin \theta[/itex]. If I were to look at the individual limits, as [itex]x \rightarrow 0[/itex] and [itex]y \rightarrow 0[/itex], then I see each relation becomes [itex] 0 = rcos \theta [/itex] and [itex]0=rsin \theta[/itex]. Now, as x and y approach 0, there is no angle that [itex]\theta[/itex] approaches that make [itex]rcos \theta [/itex] and [itex]rsin \theta[/itex] simultaneously zero; hence, it must be that r approaches 0 as to make each expression zero simultaneously.

In summary, [itex]\lim_{(x,y) \rightarrow (0,0)}~(x+y) = lim_{r \rightarrow 0} ~ (rcos \theta + rsin \theta)[/itex].

Are my arguments correct?

What does ##x+y## have to do with anything? You are making something very simple seem very complicated. When you say ##(x,y)\rightarrow (0,0)## that means the distance from ##(x,y)## to ##(0,0)## goes to zero. That is ##r##. Of course, maybe what you posted and what you were actually given aren't the same.
 
  • #3
From my understanding, [itex](x,y) \rightarrow (0,0)[/itex] means that the point (x,y) assumes the geometric location (0,0), not that the distance between two points becomes zero.

Also, x + y is just the addition of the two equations [itex]x = r\cos \theta[/itex] and [itex]y=rsin \theta[/itex]
 
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  • #4
Bashyboy said:
From my understanding, [itex](x,y) \rightarrow (0,0)[/itex] means that the point (x,y) assumes the geometric location (0,0), not that the distance between two points becomes zero.

Also, x + y is just the addition of the two equations [itex]x = r\cos \theta[/itex] and [itex]y=rsin \theta[/itex]
If we want to consider it as a limit here, we use distance. It can't just "assume" the location. We can make it arbitrarily close, though. :tongue:

Once you recognize that the limit is implying that the distance is going to zero, it becomes clear why your equation is correct, because r is the distance.
 
  • #5
So, the distance becomes zero because the point (x,y) moves, as you said, arbitrarily close to (0,0)?
 
  • #6
Bashyboy said:
So, the distance becomes zero because the point (x,y) moves, as you said, arbitrarily close to (0,0)?
No. This is a limit! The distance is allowed to become arbitrarily close to zero, thus making (x,y) arbitrarily close to (0,0).
 
  • #7
I do not see any distinguishable qualities between what you just said and what I said in post #5. In short, you said precisely the same thing I did.
 
  • #8
Bashyboy said:
I do not see any distinguishable qualities between what you just said and what I said in post #5. In short, you said precisely the same thing I did.
The difference is your statement implies that ##(x,y)=(0,0)## and r=0, where r is the distance of the point (x,y) from (0,0). My statement implies that ##\displaystyle\lim_{r\rightarrow 0}(x,y)=(0,0)##.
 
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  • #9
Yes, but isn't it sometimes the case that the limit of a function at a particular value is equal to what the function is evaluated at that approaching value, thereby implying that the coordinate (x,y) has not just approached the limit, but is directly on said limit? This is decidedly true of polynomial functions.
 
  • #10
Bashyboy said:
So, the distance becomes zero because the point (x,y) moves, as you said, arbitrarily close to (0,0)?

"The distance" never actually becomes zero. Are you familiar with the epsilon-delta definition of the limit? To truly understand the concept of a limit, you need to understand that definition.
 
  • #11
Yes, I am acquainted with the epsilon-delta definition of a limit.
 
  • #12
The point is that there is a difference between being "arbitrarily close to (0, 0)" and "being at (0, 0)".
 
  • #13
Bashyboy said:
"As [itex](x,y) \rightarrow (0,0)[/itex], [itex]r \rightarrow 0[/itex]," is a fact that I am given, in order to solve a problem.

I am still wondering if you posted what you meant to post originally. Were you given ##(x,y) \rightarrow (0,0)## as you have written, or were you given that ##x\rightarrow 0## and ##y\rightarrow 0##? They aren't the same statement and your question has more content if you were actually given the latter.
 
  • #14
I was given that [itex](x,y) \rightarrow (0,0)[/itex] implies that [itex]r \rightarrow 0[/itex]
 

What is a multivariable limit?

A multivariable limit is a mathematical concept that describes the behavior of a function as two or more variables approach a certain point simultaneously. It is used to determine the value that a function approaches as the variables get closer and closer to the specified point.

How do I identify the limit of a multivariable function?

To identify the limit of a multivariable function, you must first determine the point towards which the variables are approaching. Then, you can evaluate the function at that point to find the limit. However, if the function is undefined at that point, other methods such as approaching along a specific path or using polar coordinates may be necessary.

What is the difference between a multivariable limit and a single variable limit?

The main difference between a multivariable limit and a single variable limit is the number of variables involved. In a single variable limit, only one variable is approaching a specific point, while in a multivariable limit, two or more variables are approaching a specific point simultaneously.

Why is it important to understand multivariable limits?

Understanding multivariable limits is important because it allows us to analyze and predict the behavior of a function in complex systems where multiple variables are involved. This is especially useful in fields such as physics, engineering, and economics.

What are some common techniques for evaluating multivariable limits?

Some common techniques for evaluating multivariable limits include substitution, approaching along specific paths, using polar coordinates, and using the Squeeze Theorem. These techniques can help simplify the evaluation of the limit and provide a more accurate result.

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