Two-Path Test: Understanding Variable Function Limits

  • Context: Undergrad 
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Discussion Overview

The discussion revolves around the evaluation of limits for variable functions, specifically addressing the conditions under which a limit exists or does not exist based on the chosen path. Participants explore the implications of path dependence in limit calculations and the application of the path test in this context.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant notes that if the limit depends on the path chosen, it typically indicates that the limit does not exist, yet presents a specific limit that appears to exist despite path dependence.
  • The same participant questions whether they overlooked any important hypothesis or condition in applying the path test correctly.
  • Another participant points out that the chosen path becomes undefined as the variables approach the limit point, suggesting that the path must be valid for the limit evaluation.
  • A later reply acknowledges the oversight and emphasizes that such mistakes are common among learners at various levels.

Areas of Agreement / Disagreement

Participants generally agree on the importance of choosing valid paths for limit evaluation, but the initial confusion regarding the existence of the limit based on path dependence remains unresolved.

Contextual Notes

The discussion highlights the necessity of ensuring that paths chosen for limit evaluation do not lead to undefined expressions as variables approach the limit point.

Who May Find This Useful

This discussion may be useful for students and educators in mathematics, particularly those studying calculus and limit evaluation techniques.

assed
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Well, I've been calculating some limits of several variable functions and got confused with something: everybody knows that if taking a path the limit depends on the path chosen then the limit doesn't exist. But, if you consider the limit

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

that can be easily calculated( function that converges to 0 multiplied by limited function ). ALthough, if you choose the path

y=\sqrt{\frac{c}{x-c}}x

you'll end up with a limit equals to c, in other words, it depends oof the path you choose what would be enough for us to say the limit doesn't exist. The problem is we know it exists and my question is: did i forget to consider any important hypothesis or condition? How can the path test be applied correctly then?
I appreciate your attention.
 
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assed said:
Well, I've been calculating some limits of several variable functions and got confused with something: everybody knows that if taking a path the limit depends on the path chosen then the limit doesn't exist. But, if you consider the limit

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

that can be easily calculated( function that converges to 0 multiplied by limited function ). ALthough, if you choose the path

y=\sqrt{\frac{c}{x-c}}x

you'll end up with a limit equals to c, in other words, it depends oof the path you choose what would be enough for us to say the limit doesn't exist. The problem is we know it exists and my question is: did i forget to consider any important hypothesis or condition? How can the path test be applied correctly then?
I appreciate your attention.



The path must be chosen as to make sense when the variable(s) approach the wanted point. When \,x\to 0\, , the expression
\sqrt{\frac{c}{x-c}}\,
becomes undefined (over the reals) as the fraction into the square root becomes negative, no matter what number \,c\neq 0\, is.

DonAntonio
 
Thanks DonAntonio.
It's so obvious yet i didn't noticed it.
 
assed said:
Thanks DonAntonio.
It's so obvious yet i didn't noticed it.


Don't worry about it (it occurs to us all, no matter whether you're a beginning undergraduate or a graduate one.

Just learn this lesson and try to remeber it in the future...where new mistakes await for us all, too. :)

DonAntonio
 

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