The expression provided is:

$\lim_{(x, y) \to (0, 0)} \frac{xy}{x + y}$

To solve this using polar coordinates, we convert $x$ and $y$ to polar form:

• $x = r \cos \theta$
• $y = r \sin \theta$

Then, rewrite the expression in terms of $r$ and $\theta$:

$\frac{xy}{x + y} = \frac{(r \cos \theta)(r \sin \theta)}{r \cos \theta + r \sin \theta} = \frac{r^2 \cos \theta \sin \theta}{r (\cos \theta + \sin \theta)}$

We can simplify by canceling $r$ from the numerator and the denominator (assuming $r \neq 0$):

$= \frac{r \cos \theta \sin \theta}{\cos \theta + \sin \theta}$

Now, consider the limit as $r \to 0$. The expression depends on $r$ in the numerator:

$\lim_{r \to 0} \frac{r \cos \theta \sin \theta}{\cos \theta + \sin \theta} = 0$

This is because as $r \to 0$, the entire expression approaches zero regardless of the value of $\theta$.

Final Answer:

$\lim_{(x, y) \to (0, 0)} \frac{xy}{x + y} = 0$

Would you like a detailed explanation on any step?

Here are five related questions for further exploration:

1. How would this limit change if we used a different coordinate system, like cylindrical or spherical?
2. What happens if we approach $(0, 0)$ along paths where $x = y$ or $x = -y$?
3. Can we use L'Hôpital's Rule to solve this limit in Cartesian coordinates?
4. What other limits can be simplified by converting to polar coordinates?
5. How would this problem change if the expression were $\frac{x^2 + y^2}{x + y}$?

Tip: When dealing with limits approaching the origin, switching to polar coordinates can often simplify expressions, especially if both $x$ and $y$ appear in multiplicative terms.