1. ## Multivariable Limit Problem

I need to find:
lim as (x,y) => (0,0) of ((x - y)^2) / (abs(x) + abs(y))

I know that the limit should be 0, and I am searching for a bounding function B(x) to use the squeeze theorem with.

I have so far:

0 <= abs[f(x,y) - L] = abs[((x - y)^2) / (abs(x) + abs(y))]
<= ((x - y)^2) / abs(x + y)

dropping the absolute value bars since all terms are positive, and combining abs(x) and abs(y) using the triangle inequality

Where can I go from here to remove the denominator and find a bounding function?

2. Here you can convert to polars...

I agree that you can bound this function with

$0 \leq \frac{(x-y)^2}{|x| + |y|} \leq \frac{(x-y)^2}{|x+y|}$.

Now to squeeze the function, you need to evaluate $\lim_{(x,y) \to (0,0)}\frac{(x-y)^2}{|x+y|}$ which you can do by converting to polars...

$\lim_{(x,y) \to (0,0)}\frac{(x-y)^2}{|x+y|} = \lim_{r \to 0}\frac{(r\cos{\theta} - r\sin{\theta})^2}{|r\cos{\theta} + r\sin{\theta}|}$

$= \lim_{r \to 0}\frac{r^2(\cos{\theta} - \sin{\theta})}{|r||\cos{\theta} + \sin{\theta}|}$

$= \lim_{r \to 0}\frac{|r|^2(\cos{\theta} - \sin{\theta})}{|r||\cos{\theta} + \sin{\theta}|}$

$= \lim_{r \to 0}\frac{|r|(\cos{\theta} - \sin{\theta})}{|\cos{\theta} + \sin{\theta}|}$

$= 0$.