The problem asks us to prove that the limit of the sum of two functions $f(x, y)$ and $g(x, y)$ as $(x, y)$ approaches $(a, b)$ is equal to the sum of their individual limits, provided that these individual limits exist. That is, we want to prove: $\lim_{(x, y) \to (a, b)} [f(x, y) + g(x, y)] = \lim_{(x, y) \to (a, b)} f(x, y) + \lim_{(x, y) \to (a, b)} g(x, y)$

1. Problem Description

The problem asks us to prove that the limit of the sum of two functions $f(x, y)$ and $g(x, y)$ as $(x, y)$ approaches $(a, b)$ is equal to the sum of their individual limits, provided that these individual limits exist. That is, we want to prove:

$\lim_{(x, y) \to (a, b)} [f(x, y) + g(x, y)] = \lim_{(x, y) \to (a, b)} f(x, y) + \lim_{(x, y) \to (a, b)} g(x, y)$

2. Solution Steps

Let $L_1 = \lim_{(x, y) \to (a, b)} f(x, y)$ and $L_2 = \lim_{(x, y) \to (a, b)} g(x, y)$.

Since both limits exist, for any $\epsilon > 0$, we can find $\delta_1 > 0$ and $\delta_2 > 0$ such that:

If $0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta_1$, then $|f(x, y) - L_1| < \frac{\epsilon}{2}$.

If $0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta_2$, then $|g(x, y) - L_2| < \frac{\epsilon}{2}$.

Now, let $\delta = \min(\delta_1, \delta_2)$. Then, if $0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta$, both of the above inequalities hold. We can write:

$|f(x, y) + g(x, y) - (L_1 + L_2)| = |(f(x, y) - L_1) + (g(x, y) - L_2)|$

Using the triangle inequality $|a + b| \le |a| + |b|$, we have:

$|f(x, y) + g(x, y) - (L_1 + L_2)| \le |f(x, y) - L_1| + |g(x, y) - L_2|$

Since $0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta \le \delta_1$ and $\delta \le \delta_2$, we can substitute the bounds for $|f(x, y) - L_1|$ and $|g(x, y) - L_2|$:

$|f(x, y) + g(x, y) - (L_1 + L_2)| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

Therefore, for any $\epsilon > 0$, there exists a $\delta > 0$ such that if $0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta$, then $|f(x, y) + g(x, y) - (L_1 + L_2)| < \epsilon$. This means:

$\lim_{(x, y) \to (a, b)} [f(x, y) + g(x, y)] = L_1 + L_2 = \lim_{(x, y) \to (a, b)} f(x, y) + \lim_{(x, y) \to (a, b)} g(x, y)$

3. Final Answer

$\lim_{(x, y) \to (a, b)} [f(x, y) + g(x, y)] = \lim_{(x, y) \to (a, b)} f(x, y) + \lim_{(x, y) \to (a, b)} g(x, y)$