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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.