We are asked 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

We are asked 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

Let $u = x^2 + y^2$. Then as $(x, y) \to (0, 0)$, we have $u \to 0$. Thus, we can rewrite the limit as:

$$$$

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

$$$$

We know that $\lim_{x \to 0} \frac{\tan x}{x} = 1$. Thus, we have

$$$$

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

$$$$

To show this, we can use L'Hopital's Rule:

$$$$

\lim_{u \to 0} \frac{\tan(u)}{u} = \lim_{u \to 0} \frac{\sec^2(u)}{1} = \sec^2(0) = \frac{1}{\cos^2(0)} = \frac{1}{1^2} = 1

$$$$

Alternatively, we can use the small angle approximation for $\tan(u)$ where $\tan(u) \approx u$ as $u \to 0$.

$$$$

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

$$$$

3. Final Answer

The limit is 1.