The problem asks to find the limit of the function $\frac{\tan(x^2 + y^2)}{x^2 + y^2}$ as $(x, y)$ approaches $(0, 0)$.

1. Problem Description

The problem asks to find the limit of the function $\frac{\tan(x^2 + y^2)}{x^2 + y^2}$ as $(x, y)$ approaches $(0, 0)$.

2. Solution Steps

We want to evaluate the limit

$$\lim_{(x, y) \to (0, 0)} \frac{\tan(x^2 + y^2)}{x^2 + y^2}$$

Let $u = x^2 + y^2$. As $(x, y) \to (0, 0)$, $u \to 0$. Therefore, we can rewrite the limit as

$$\lim_{u \to 0} \frac{\tan(u)}{u}$$

We know that

$$\lim_{u \to 0} \frac{\tan(u)}{u} = \lim_{u \to 0} \frac{\sin(u)}{u \cos(u)} = \lim_{u \to 0} \frac{\sin(u)}{u} \cdot \lim_{u \to 0} \frac{1}{\cos(u)}$$

Since $\lim_{u \to 0} \frac{\sin(u)}{u} = 1$ and $\lim_{u \to 0} \cos(u) = 1$, we have

$$\lim_{u \to 0} \frac{\tan(u)}{u} = 1 \cdot \frac{1}{1} = 1$$

3. Final Answer

The limit is 1.