\lim _{x\to \:2}\left(\frac{\sqrt{x-2}+x^2-3x+2}{\sqrt{x^2-4}}\right)

We are tasked with evaluating the limit:

\[
\lim_{x \to 2} \left( \frac{\sqrt{x - 2} + x^2 - 3x + 2}{\sqrt{x^2 - 4}} \right)
\]

### Step 1: Simplify the expression
First, let's analyze the denominator and numerator separately.

#### Denominator:
\[
\sqrt{x^2 - 4}
\]
This can be factored as:
\[
\sqrt{x^2 - 4} = \sqrt{(x - 2)(x + 2)}
\]

#### Numerator:
\[
\sqrt{x - 2} + x^2 - 3x + 2
\]
Now, simplify the quadratic part \( x^2 - 3x + 2 \):
\[
x^2 - 3x + 2 = (x - 1)(x - 2)
\]
So the numerator becomes:
\[
\sqrt{x - 2} + (x - 1)(x - 2)
\]

Thus, the entire expression now is:
\[
\lim_{x \to 2} \frac{\sqrt{x - 2} + (x - 1)(x - 2)}{\sqrt{(x - 2)(x + 2)}}
\]

### Step 2: Apply the limit as \( x \to 2 \)

When \( x \to 2 \), both the numerator and denominator contain the term \( \sqrt{x - 2} \), which tends to zero. Hence, we have an indeterminate form of type \( \frac{0}{0} \), so we need to simplify further or apply L'Hôpital's Rule.

### Step 3: Factor and Cancel

Factor \( (x - 2) \) from the numerator and denominator:
\[
\lim_{x \to 2} \frac{(x - 2)(\frac{1}{\sqrt{x - 2}} + (x - 1))}{\sqrt{(x - 2)(x + 2)}}
\]

We can cancel \( \sqrt{x - 2} \) from both the numerator and denominator:
\[
\lim_{x \to 2} \frac{\frac{1}{\sqrt{x - 2}} + (x - 1)}{\sqrt{x + 2}}
\]

### Step 4: Apply the limit

Now, substitute \( x = 2 \) directly:
\[
\frac{1}{\sqrt{2 - 2}} + (2 - 1) = \text{undefined because the first term involves division by zero}.
\]

Thus, the limit does not exist because the expression blows up as \( x \to 2 \).

The final result is:

\[
\boxed{\text{The limit does not exist.}}
\]

Would you like further explanation or details?

---

Here are 5 related questions:
1. How does L'Hôpital's Rule help resolve indeterminate forms?
2. What are common methods for factoring expressions involving square roots?
3. What is the difference between limits that exist and those that do not?
4. How do you approach limits where square roots appear in both the numerator and denominator?
5. Can a limit approaching infinity ever result in a defined value?

**Tip:** Always try substituting directly before applying advanced techniques like L'Hôpital's Rule or factoring.

\\lim _{x\to 2}\\left(\\frac{\\sqrt{x-2}+x^2-3x+2}{\\sqrt{x^2-4}}\\right)

Explore how to evaluate the limit \(\lim_{x \to 2} \left( \frac{\sqrt{x - 2} + x^2 - 3x + 2}{\sqrt{x^2 - 4}} \right)\) using factoring techniques and L'Hôpital's Rule. This problem is suitable for students studying undergraduate calculus.

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