We are tasked with finding the limit of the following expression as $(x, y)$ approaches $(0, 0)$:

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

Step 1: Polar coordinates

Since the expression involves $x^2 + y^2$, it's natural to switch to polar coordinates:

• $x = r \cos(\theta)$,
• $y = r \sin(\theta)$, where $r = \sqrt{x^2 + y^2}$.

In polar coordinates, the given expression becomes:

$\frac{\tan(\sqrt{x^2 + y^2})}{\sqrt{x^2 + y^2}} = \frac{\tan(r)}{r}.$

Thus, we need to find the limit as $r \to 0$ of:

$\lim_{r \to 0} \frac{\tan(r)}{r}.$

Step 2: Apply the known limit

We know that:

$\lim_{r \to 0} \frac{\tan(r)}{r} = 1.$

This is a standard limit from calculus. Therefore, the limit of the original expression is:

$\lim_{(x, y) \to (0, 0)} \frac{\tan(\sqrt{x^2 + y^2})}{\sqrt{x^2 + y^2}} = 1.$

Final Answer:

$\boxed{1}.$

Would you like further details or explanations on any part of this solution?

Here are 5 related questions to expand your understanding:

1. What are polar coordinates, and how are they useful in solving limits?
2. Why is $\lim_{r \to 0} \frac{\tan(r)}{r} = 1$ considered a standard limit?
3. How do we solve limits involving two variables like $(x, y)$ using polar coordinates?
4. Can you apply a similar method for $\lim_{(x, y) \to (0, 0)} \frac{\sin(\sqrt{x^2 + y^2})}{\sqrt{x^2 + y^2}}$?
5. What are the conditions for applying L'Hopital's Rule in finding limits?

Tip: Polar coordinates are extremely useful when dealing with limits involving expressions like $\sqrt{x^2 + y^2}$, as they simplify the analysis by reducing two variables to one (the radius).