The problem asks for the value of the limit:

$\lim_{x \to 2} \frac{x^2 - 3x + 2}{\sqrt{x} - \sqrt{2}}.$

Let's solve it step by step.

Step 1: Factor the numerator

We need to factor the quadratic expression in the numerator, $x^2 - 3x + 2$.

$x^2 - 3x + 2 = (x - 1)(x - 2).$

So, the original expression becomes:

$\lim_{x \to 2} \frac{(x - 1)(x - 2)}{\sqrt{x} - \sqrt{2}}.$

Step 2: Apply the limit

When we directly substitute $x = 2$, we get an indeterminate form $\frac{0}{0}$, so we need to simplify further.

Step 3: Rationalize the denominator

To eliminate the square root in the denominator, we multiply both the numerator and denominator by the conjugate of the denominator, which is $\sqrt{x} + \sqrt{2}$:

$\frac{(x - 1)(x - 2)}{\sqrt{x} - \sqrt{2}} \times \frac{\sqrt{x} + \sqrt{2}}{\sqrt{x} + \sqrt{2}}.$

This simplifies the expression to:

$\lim_{x \to 2} \frac{(x - 1)(x - 2)(\sqrt{x} + \sqrt{2})}{x - 2}.$

Step 4: Cancel common factors

The factor $x - 2$ appears in both the numerator and denominator, so we can cancel it:

$\lim_{x \to 2} (x - 1)(\sqrt{x} + \sqrt{2}).$

Step 5: Evaluate the limit

Now, substitute $x = 2$:

$(2 - 1)(\sqrt{2} + \sqrt{2}) = 1 \times 2\sqrt{2} = 2\sqrt{2}.$

Final Answer:

The value of the limit is $2\sqrt{2}$.

Would you like more details on any step, or have any further questions?

Here are 5 questions related to this:

1. What other methods can be used to evaluate indeterminate limits like $\frac{0}{0}$?
2. How do you factor a quadratic expression like $x^2 - 3x + 2$?
3. Why do we multiply by the conjugate when simplifying square roots in limits?
4. What are the common forms of indeterminate limits, and how are they handled?
5. Can L'Hopital's Rule be applied to this limit? If so, how?

Tip: Always check for indeterminate forms before applying algebraic simplifications when solving limits.