Having trouvle with a quick limit provlem.

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In summary, the conversation revolves around the function g(x,y) = (x^2+y^2)/(x^2+y^2-2) and determining if it has a limit as (x,y) approaches (1,1). Strategies for discovering the limit are discussed, including trying different curves and using Wolfram Alpha to plot the function. The concept of one-sided limits is also brought up, with the conclusion that the limit does not exist for this function.
  • #1
ozone
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[itex]F(x,y)-> (1,1)[/itex]
[itex]
(x^2 + y^2) / (x^2+y^2-2)

[/itex]

I am just having an issue as I don't know how to deal with the -2 in the denominator.

Thank you
 
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  • #2
What is the problem? Are we to assume that the second line is the definition of F? Is the problem to determine a value of (x, y) such that F(x,y)= (1, 1)? x]

But that can't be right because [/itex]F(x,y)= (x^2+ y^2)/(x^2+ y^2+1)[/itex] maps pairs of numbers, (x, y) to a single number, not a pair of numbers.

So please got back, read the problem again, and tell us what it really says.
 
  • #3
One part of the problem is discovering if the limit exists or not, then you have to prove it.

The discovery itself can be difficult. Here are some strategies for that. Try some curves like straight lines with slope m and parabolas that go through the point, so that it is a one dimensional limit along those curves. You may find that the limit doesn't exist this way. But If you find no issue here, this does not prove that the limit exists, for the limit relies on any curve, which is hard to consider all, so we let x,y go to x0,y0 in an arbitrary way (compare proofs of 2-d limits that do exist).

The algebra was not telling me anything quickly, so I cheated and looked at the plot on wolfram alpha (on my phone yo!). That told me that the limit didn't exist, and to try the curve y=x, and that for x<1, the denominator is negative, while for x>1, the denominator is positive. Oh, and sometimes when the limit goes to infinity, we say the limit exists, but not here even since different paths give positive and negative infinity.
 
  • #4
algebrat said:
One part of the problem is discovering if the limit exists or not, then you have to prove it.

The discovery itself can be difficult. Here are some strategies for that. Try some curves like straight lines with slope m and parabolas that go through the point, so that it is a one dimensional limit along those curves. You may find that the limit doesn't exist this way. But If you find no issue here, this does not prove that the limit exists, for the limit relies on any curve, which is hard to consider all, so we let x,y go to x0,y0 in an arbitrary way (compare proofs of 2-d limits that do exist).

The algebra was not telling me anything quickly, so I cheated and looked at the plot on wolfram alpha (on my phone yo!). That told me that the limit didn't exist, and to try the curve y=x, and that for x<1, the denominator is negative, while for x>1, the denominator is positive. Oh, and sometimes when the limit goes to infinity, we say the limit exists, but not here even since different paths give positive and negative infinity.

You can simplify the problem a little bit by noting the symmetry that reduces it to the limit as [itex]\xi[/itex] goes to 2, where [itex]\xi[/itex] = x[itex]^{2} [/itex] + y[itex]^{2} [/itex]. It may be instructive to plot the function (taking care to limit the results to constrain the blow-up as [itex]\xi[/itex]-2 becomes < 1 (as in the attached Mathcad-derived image of a cross-section through the curve)).

As you note, the the limits are different depending upon whether [itex]\xi[/itex] [itex]\stackrel{+}{\rightarrow}[/itex] 2 or [itex]\xi[/itex] [itex]\stackrel{-}{\rightarrow}[/itex] 2.
 

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  • #5
Thanks for the responses guys.

The actualy question reads
[itex] g(x,y)=(x^2+y^2)/(x^2+y^2-2) [/itex]
The question asks if this function has a limit as (x,y) approaches (1,1)

To me it doesn't mean anything if you look at the value slightly left of the function and the value slightly right of it. As this means you could be approaching the point from above and from below respectively.

However I remember reading about that in my book, but I thought it only made a difference if it was at the Specific point. I'm not sure if I am completely getting it yet.
 
  • #6
ozone said:
Thanks for the responses guys.

The actualy question reads
[itex] g(x,y)=(x^2+y^2)/(x^2+y^2-2) [/itex]
The question asks if this function has a limit as (x,y) approaches (1,1)

To me it doesn't mean anything if you look at the value slightly left of the function and the value slightly right of it. As this means you could be approaching the point from above and from below respectively.

However I remember reading about that in my book, but I thought it only made a difference if it was at the Specific point. I'm not sure if I am completely getting it yet.
OK, if the question is after terminological exactitude, then there is no limit as the one-sided limits have to be equal for The Limit to exist. (If you look at the Mathcad image you should find that Mathcad states The Limit is undefined, even though the one-sided limits are.)

NR
 
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  • #7
ozone said:
To me it doesn't mean anything if you look at the value slightly left of the function and the value slightly right of it. As this means you could be approaching the point from above and from below respectively.

However I remember reading about that in my book, but I thought it only made a difference if it was at the Specific point. I'm not sure if I am completely getting it yet.

Ozone, I am not sure I understand exactly where you are having problems. Please explain further your thinking.
 
  • #8
I was just having some trouble dealing with this in 3 space, but if the limit has to be identical and either side of the point, then this is a good enough way for me to prove that it doesn't exist. I just wanted to make sure I knew how to sufficiently prove this on my upcoming test. Thank you everyone
 

FAQ: Having trouvle with a quick limit provlem.

What is a quick limit problem?

A quick limit problem is a mathematical exercise that involves finding the limit of a function as the input approaches a specific value. It is a common topic in calculus and is used to study the behavior of functions at certain points.

How do I solve a quick limit problem?

To solve a quick limit problem, you need to first identify the value that the input is approaching. Then, you can use various techniques such as substitution, factoring, or L'Hopital's rule to evaluate the limit. Practice and familiarity with different types of functions will also help in solving these problems.

What are the common challenges in solving quick limit problems?

One of the main challenges in solving quick limit problems is identifying the correct approach to use. Some problems may require algebraic manipulation while others may require the use of special rules or theorems. Additionally, keeping track of all the steps and avoiding mistakes can also be a challenge.

How can I improve my skills in solving quick limit problems?

The best way to improve your skills in solving quick limit problems is through practice. Start with simpler problems and gradually work your way up to more complex ones. It can also be helpful to study and understand the different techniques and rules used in solving these problems.

What are the real-world applications of quick limit problems?

Quick limit problems have many real-world applications, especially in fields such as physics, engineering, and economics. For example, they can be used to model the behavior of a moving object, the rate of change of a system, or the maximum profit of a business. Understanding quick limit problems can also help in making predictions and informed decisions based on mathematical data.

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