The given function is:

$f(x) = \frac{x^2 - 4}{\sqrt{x} - \sqrt{2}}$

We are asked to find the limit as $x \to 2$:

$\lim_{x \to 2} f(x)$

Let's solve this step by step. The function appears to have the indeterminate form $\frac{0}{0}$ at $x = 2$, so we can apply algebraic manipulation to resolve this indeterminate form.

The expression $x^2 - 4$ can be factored as:

$x^2 - 4 = (x - 2)(x + 2)$

Thus, the function becomes:

$f(x) = \frac{(x - 2)(x + 2)}{\sqrt{x} - \sqrt{2}}$

We now need to simplify this further to resolve the limit at $x = 2$. To do this, we'll rationalize the denominator $\sqrt{x} - \sqrt{2}$ by multiplying both the numerator and denominator by $\sqrt{x} + \sqrt{2}$:

$\frac{(x - 2)(x + 2)}{\sqrt{x} - \sqrt{2}} \times \frac{\sqrt{x} + \sqrt{2}}{\sqrt{x} + \sqrt{2}} = \frac{(x - 2)(x + 2)(\sqrt{x} + \sqrt{2})}{x - 2}$

The $x - 2$ terms cancel out, leaving:

$f(x) = (x + 2)(\sqrt{x} + \sqrt{2})$

Now, we can substitute $x = 2$ directly into the simplified expression:

$f(2) = (2 + 2)(\sqrt{2} + \sqrt{2}) = 4(2\sqrt{2}) = 8\sqrt{2}$

Thus, the limit is:

$\lim_{x \to 2} f(x) = 8\sqrt{2}$

If you have any questions or need further clarification, let me know!

Here are 5 related questions for further practice:

1. How would you evaluate $\lim_{x \to 2} \frac{x^2 - 9}{\sqrt{x} - 3}$?
2. How do you rationalize the denominator in $\frac{1}{\sqrt{5} - 2}$?
3. What are the indeterminate forms in limits, and how do you recognize them?
4. How can L'Hopital's Rule be applied to resolve indeterminate limits?
5. What is the general approach to factoring differences of squares?

Tip: When limits yield indeterminate forms like $\frac{0}{0}$, always check for algebraic simplifications such as factoring or rationalizing.