https://r2.erweima.ai/mathbot/image/1726447164669fn5ppivg.jpg

Let's solve the two limits step by step from the image you uploaded.

### First limit:
\[
\lim_{x \to 2} \frac{\sqrt{x + 2} - 2}{x - 2}
\]

#### Step 1: Direct substitution
If we substitute \(x = 2\), we get:
\[
\frac{\sqrt{2 + 2} - 2}{2 - 2} = \frac{\sqrt{4} - 2}{0} = \frac{2 - 2}{0} = \frac{0}{0}
\]
This is an indeterminate form, so we need to simplify the expression.

#### Step 2: Multiply by the conjugate
To eliminate the square root, we multiply both the numerator and denominator by the conjugate of the numerator:
\[
\frac{\sqrt{x + 2} - 2}{x - 2} \times \frac{\sqrt{x + 2} + 2}{\sqrt{x + 2} + 2} = \frac{(\sqrt{x + 2})^2 - 2^2}{(x - 2)(\sqrt{x + 2} + 2)}
\]
Simplifying the numerator:
\[
(\sqrt{x + 2})^2 - 2^2 = (x + 2) - 4 = x - 2
\]
Now the expression becomes:
\[
\frac{x - 2}{(x - 2)(\sqrt{x + 2} + 2)}
\]
We can cancel out \(x - 2\) from the numerator and denominator:
\[
\frac{1}{\sqrt{x + 2} + 2}
\]

#### Step 3: Substitute \(x = 2\)
Now, substituting \(x = 2\) gives:
\[
\frac{1}{\sqrt{2 + 2} + 2} = \frac{1}{2 + 2} = \frac{1}{4}
\]
Thus, the value of the limit is:
\[
\boxed{\frac{1}{4}}
\]

---

### Second limit:
\[
\lim_{x \to 2^+} \frac{1}{\sqrt{x} - 2 + \sqrt{x^2 - 4}}
\]

#### Step 1: Direct substitution
Substitute \(x = 2\):
\[
\frac{1}{\sqrt{2} - 2 + \sqrt{2^2 - 4}} = \frac{1}{\sqrt{2} - 2 + \sqrt{4 - 4}} = \frac{1}{\sqrt{2} - 2 + 0}
\]
This results in:
\[
\frac{1}{\sqrt{2} - 2}
\]
Since \(\sqrt{2} \approx 1.414\), we get:
\[
\frac{1}{1.414 - 2} = \frac{1}{-0.586}
\]
Thus, the limit is:
\[
\boxed{-1.705}
\]

---

Let me know if you'd like any additional details or clarification!

### Related questions:
1. How do you find limits involving square roots?
2. What does the conjugate method do to simplify indeterminate limits?
3. How do you handle limits that result in indeterminate forms like \(\frac{0}{0}\)?
4. What is the significance of approaching from the right (\(x \to 2^+\)) in a limit?
5. Can limits be evaluated using L'Hopital's Rule, and when is it applicable?

### Tip:
Always check if substituting directly into a limit gives an indeterminate form before applying any algebraic manipulations like factoring or multiplying by the conjugate.

Calculate the following limits: 
 1. lim (x -> 2) (sqrt(x + 2) - 2) / (x - 2) 
 2. lim (x -> 2+) 1 / (sqrt(x) - 2 + sqrt(x^2 - 4))

Learn how to calculate limits involving square roots, like lim (x -> 2) (sqrt(x + 2) - 2) / (x - 2) and lim (x -> 2+) 1 / (sqrt(x) - 2 + sqrt(x^2 - 4)), using techniques such as the conjugate method. Perfect for high school students tackling advanced limit problems.

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