To solve the limit

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

we'll examine the behavior of the expression as $(x, y) \to (0, 0)$. This limit has a fraction with square roots, so we'll apply standard limit techniques, such as series expansion and rationalization.

Step 1: Rewrite the expression in a simpler form

Let's break down the numerator:

$\sqrt{x^{y^2} + 2} - \sqrt{2}.$

For very small $x$ and $y$, we can use a first-order approximation of the square root function. Using the expansion for small $h$, $\sqrt{a + h} \approx \sqrt{a} + \frac{h}{2\sqrt{a}}$, we apply this to the expression with $a = 2$ and $h = x^{y^2}$:

$\sqrt{x^{y^2} + 2} \approx \sqrt{2} + \frac{x^{y^2}}{2\sqrt{2}}.$

Thus, the numerator becomes:

$\left(\sqrt{2} + \frac{x^{y^2}}{2\sqrt{2}}\right) - \sqrt{2} = \frac{x^{y^2}}{2\sqrt{2}}.$

Step 2: Simplify the limit expression

Now the limit becomes:

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

Step 3: Analyze the behavior as $(x, y) \to (0, 0)$

We observe the term $x^{y^2}$. As $y^2 \to 0$, this term approaches $x^0 = 1$, so $x^{y^2} \approx 1$ near the origin. Therefore, the limit simplifies to:

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

Step 4: Check if the limit exists

The denominator $x^2 + y^2$ approaches 0 as $(x, y) \to (0, 0)$. Hence, the whole expression tends to infinity because the numerator is a constant, and the denominator goes to 0.

Thus, the limit does not exist, as the expression tends to infinity.

Conclusion

The limit diverges to infinity:

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

---

Would you like to see any more details or ask further questions?

Here are 5 related questions to deepen understanding:

1. How do we derive series expansions for square roots around small values?
2. What alternative methods can be used to approach limits involving two variables?
3. How does rationalization help in simplifying square root limits?
4. What are the conditions for a limit in two variables to exist?
5. How do polar coordinates help in analyzing limits at the origin?

Tip: When evaluating limits in two variables, converting to polar coordinates can simplify the process, especially near the origin.