The limit you want to evaluate is:

$\lim_{x \to 2} \frac{\sqrt{3 + x + x^2} - \sqrt{9 - 2x + x^2}}{x^2 - 3x + 2}$

Step 1: Check Direct Substitution

Substitute $x = 2$ directly into the expression to check if the limit can be evaluated easily.

1. Numerator:

$\sqrt{3 + 2 + 2^2} - \sqrt{9 - 2(2) + 2^2} = \sqrt{3 + 2 + 4} - \sqrt{9 - 4 + 4} = \sqrt{9} - \sqrt{9} = 3 - 3 = 0$

2. Denominator:

$x^2 - 3x + 2 = 2^2 - 3(2) + 2 = 4 - 6 + 2 = 0$

So, the limit is in the indeterminate form $\frac{0}{0}$, meaning we need to simplify further.

Step 2: Simplify Using Conjugates (for the numerator)

To simplify the numerator, we multiply and divide by the conjugate of the numerator:

$\frac{\sqrt{3 + x + x^2} - \sqrt{9 - 2x + x^2}}{x^2 - 3x + 2} \cdot \frac{\sqrt{3 + x + x^2} + \sqrt{9 - 2x + x^2}}{\sqrt{3 + x + x^2} + \sqrt{9 - 2x + x^2}}$

This results in:

$\frac{\left(\sqrt{3 + x + x^2} - \sqrt{9 - 2x + x^2}\right)\left(\sqrt{3 + x + x^2} + \sqrt{9 - 2x + x^2}\right)}{(x^2 - 3x + 2)\left(\sqrt{3 + x + x^2} + \sqrt{9 - 2x + x^2}\right)}$

Using the difference of squares in the numerator:

$\text{Numerator:} \quad (3 + x + x^2) - (9 - 2x + x^2) = 3 + x + x^2 - 9 + 2x - x^2 = 3x - 6$

So, the expression becomes:

$\frac{3(x - 2)}{(x^2 - 3x + 2)\left(\sqrt{3 + x + x^2} + \sqrt{9 - 2x + x^2}\right)}$

Step 3: Factor the Denominator

Notice that $x^2 - 3x + 2$ factors as:

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

Thus, the expression becomes:

$\frac{3(x - 2)}{(x - 2)(x - 1)\left(\sqrt{3 + x + x^2} + \sqrt{9 - 2x + x^2}\right)}$

Step 4: Cancel Common Factors

You can cancel the $(x - 2)$ terms in the numerator and denominator, leaving:

$\frac{3}{(x - 1)\left(\sqrt{3 + x + x^2} + \sqrt{9 - 2x + x^2}\right)}$

Step 5: Evaluate the Limit

Now substitute $x = 2$ into the simplified expression:

$\frac{3}{(2 - 1)\left(\sqrt{3 + 2 + 2^2} + \sqrt{9 - 2(2) + 2^2}\right)} = \frac{3}{(1)\left(\sqrt{9} + \sqrt{9}\right)} = \frac{3}{1 \times (3 + 3)} = \frac{3}{6} = \frac{1}{2}$

Final Answer:

$\lim_{x \to 2} \frac{\sqrt{3 + x + x^2} - \sqrt{9 - 2x + x^2}}{x^2 - 3x + 2} = \frac{1}{2}$

Would you like more details or have any questions about this solution?

Here are some related questions you might find interesting:

1. How do you handle indeterminate forms like $\frac{0}{0}$ in limits?
2. What is the role of the conjugate when simplifying square roots in limits?
3. How does factoring the denominator help in evaluating limits?
4. Why is the difference of squares formula so useful in calculus?
5. How do you approach limits involving square roots or other complex expressions?

Tip: When dealing with square roots in limits, multiplying by the conjugate is often a useful method for simplifying the expression.