Limit of function of two variables

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Homework Help Overview

The discussion revolves around finding conditions on real numbers a and b such that the limit of a function of two variables exists and is finite as (x,y) approaches (0,0). The function in question is given by \(\frac{\left\vert x \right\vert ^a \left\vert y \right\vert ^b}{x^2+y^2}\).

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore various paths to evaluate the limit, including substituting specific values for x and y, and using polar coordinates. There are discussions about the conditions under which the limit exists, particularly focusing on inequalities involving a and b.

Discussion Status

Some participants have provided insights into the implications of negative exponents and the need to justify boundary points. Others are questioning whether additional paths need to be examined and how to interpret the results from different approaches.

Contextual Notes

One participant notes that this is an assignment question from a specific course, which raises concerns about adherence to forum guidelines regarding assistance with assignments.

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Homework Statement


Find all real a and b such that the limit [tex]\lim_{(x,y) \to (0,0)} \frac{\left\vert x \right\vert ^a \left\vert y \right\vert ^b}{x^2+y^2}[/tex] exists and is finite. (Hint: it is not enough to find the limit or prove it exists in the region found. You must also prove that the limit is infinite or nonexistent outside the region, and also justify your decision about which boundary points to include.)


Homework Equations





The Attempt at a Solution



Since this is a limit in 2D, I tried the common paths first.

Substituting x = 0 gives 0 on the numerator (if a =/= 0), and y^2 on the denominator. I'm not sure how I could evaluate this limit further.

Similarly for y = 0

Substituting y = kx gives [tex]\lim_{x \to 0} \left\vert x \right\vert ^{a+b-2} \frac{\left\vert k\right\vert ^b}{k^2+1}[/tex] thus a + b - 2 > 0 to ensure the existence of the limit

Substituting y = mx^2 gives [tex]\lim_{x \to 0} \left\vert x\right\vert ^{a+2b-2} \frac{\left\vert m \right\vert ^b}{1+m^2x^2}[/tex] thus a + 2b - 2 > 0

then use the polar coordinates [tex]x=rcos\theta , y=rsin\theta[/tex] the limit becomes [tex]\lim_{r \to 0} r^{a+b-2} \left\vert cos\theta \right\vert ^a \left\vert sin\theta \right\vert ^b[/tex]. hence a + b - 2 > 0 as above

I can't think of any other approaches to evaluate this limit though, and I have no idea how to show limit is nonexistent outside (a+b>2 and a+2b>2) and justify the boundary points chosen. It would be great if anyone could share their thoughts on this. Thanks in advance.
 
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You must also prove that the limit is infinite or nonexistent outside the region, and also justify your decision about which boundary points to include.)

Outside the region you have found that the exponent of r is negative. Investigate the different cases where the negative exponent is even and when it is odd.
 
It suffices to show the exponent of base r is negative when a + b < 2? Do I need to examine other paths as approaches to evaluate the limit? Since the polar coordinate substitution seems to encompass more general cases, would I still need the a + 2b > 0 from the parabola path though?

Also could you elaborate on the second part - about even and odd negative exponents. I can't really see the difference - wouldn't they all make the limit of r^a (with a < 0) as r approaches 0 become nonexistent? Thanks.
 
This is an ASSIGNMENT question from the MATH1901 course at the University of Sydney, and therefore a violation of PF guidelines to not at least inform us. Please refrain from helping this user, this assignment counts towards the final mark.
 
Please delete this post if this constitutes a violation of the forum rules. I did not realize that. Sorry.
 

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