Determining Limits for Multivariable Functions

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SUMMARY

The limit of the multivariable function as (x, y) approaches (2, 2) is analyzed through various approaches, including substitution and polar coordinates. The function is expressed as f(x,y) = (2x^2 + 2xy + 2x - xy^2 - y^3 - y^2) / (2x^3 - 2x^2y + 2x - x^2y^2 + xy^3 - y^2). Factoring reveals that both the numerator and denominator share a common factor of (2x - y^2), leading to the conclusion that the limit does not exist due to the degrees of the numerator and denominator. The discussion emphasizes the importance of recognizing indeterminate forms and factoring techniques in multivariable calculus.

PREREQUISITES
  • Understanding of multivariable limits
  • Familiarity with factoring polynomials
  • Knowledge of indeterminate forms in calculus
  • Experience with polar coordinates in limit evaluation
NEXT STEPS
  • Study the concept of indeterminate forms in depth
  • Learn advanced factoring techniques for polynomials
  • Explore the application of polar coordinates in multivariable calculus
  • Investigate the behavior of limits in cases of differing degrees of numerator and denominator
USEFUL FOR

Students and educators in calculus, particularly those focusing on multivariable functions and limit evaluation techniques.

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Homework Statement




[tex]\lim_{\substack{x\rightarrow 2\\y\rightarrow 2}} f(x,y)=\frac{2x^2+2xy+2x-xy^2-y^3-y^2}{2x^3-2x^2y+2x-x^2y^2+xy^3-y^2}[/tex]


Homework Equations



N/A

The Attempt at a Solution



Well, I tried using the line y = 2, and let x approach 2 (as well as making x=2 and let y approach 2), and nothing really seems to cancel when I do that.

I also tried converting everything to polar form and let r approach 2[itex]\sqrt{2}[/itex], but again, everything just looks really messy.

This is a previous year's test question, and it's not worth much marks, so I think I'm missing some trick/procedure to simplify this question.

Also, I think this does not have a limit because the degree of the top (3) is less than the degree of the bottom (4); however, I'm not sure of how to prove this.
 
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What do you mean by "nothing really seems to cancel". You should get the indeterminate form, 0/0 .

Factoring the numerator gives [itex](x+y+1) (2 x-y^2)\,.[/itex]

Factoring the denominator gives [itex](x^2-x y+1) (2 x-y^2)\,.[/itex]
 
When I said "nothing seems to cancel", I just meant that the variables don't cancel out. For example, we can sometimes sub in y=mx^2, into a function to balance out the powers and allow the x to divide out, leaving a limit that changes depending on m.

Anyway, thank you very much for helping me regarding this problem.
 

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