To determine whether the limit

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

exists, we can analyze the limit along different paths approaching the origin $(0, 0)$. If the limit is different for two different paths, the limit does not exist.

Path 1: $y = x$

Substitute $y = x$ into the limit:

$\lim_{x \to 0} \frac{\sin(x \cdot x)}{x + x} = \lim_{x \to 0} \frac{\sin(x^2)}{2x}$

As $x \to 0$, $x^2$ also approaches 0. We can use the approximation $\sin(u) \approx u$ for small $u$:

$\lim_{x \to 0} \frac{x^2}{2x} = \lim_{x \to 0} \frac{x}{2} = 0$

So along the path $y = x$, the limit is 0.

Path 2: $y = -x$

Substitute $y = -x$ into the limit:

$\lim_{x \to 0} \frac{\sin(x \cdot (-x))}{x - x} = \lim_{x \to 0} \frac{\sin(-x^2)}{0}$

Here, the expression is undefined because we have division by zero in the denominator.

Path 3: $y = 0$

Substitute $y = 0$ into the limit:

$\lim_{x \to 0} \frac{\sin(x \cdot 0)}{x + 0} = \lim_{x \to 0} \frac{0}{x} = 0$

So along the path $y = 0$, the limit is 0.

Conclusion

The limit yields different behavior depending on the path taken. For $y = -x$, the limit is undefined, whereas for other paths, such as $y = x$ or $y = 0$, the limit appears to be 0.

Since the limit depends on the path, the limit does not exist.

Would you like any more details or have any further questions?

Here are 5 related questions:

1. How do you formally prove the limit does not exist using the epsilon-delta definition?
2. What happens if you change the function to $\frac{\sin(xy)}{x^2 + y^2}$?
3. How can polar coordinates help analyze limits like this?
4. What is an example of a limit that exists even though different paths yield the same value?
5. How does this problem relate to the concept of continuity at a point?

Tip: When dealing with limits in multiple variables, checking different paths is a common strategy to see if the limit exists uniformly.