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The image contains several limit problems. Let's transcribe and solve them one by one.

### 1. \( \lim_{x \to 2} \frac{\sqrt{5x + 6}}{x - 4} \)

This is an indeterminate form at \( x = 2 \), so we can use L'Hôpital's Rule:
\[
\text{Differentiate the numerator: } \frac{d}{dx} \sqrt{5x + 6} = \frac{5}{2\sqrt{5x + 6}}
\]
\[
\text{Differentiate the denominator: } \frac{d}{dx}(x - 4) = 1
\]
Using L'Hôpital's Rule:
\[
\lim_{x \to 2} \frac{\frac{5}{2\sqrt{5x + 6}}}{1} = \frac{5}{2\sqrt{5(2) + 6}} = \frac{5}{2\sqrt{10 + 6}} = \frac{5}{2\sqrt{16}} = \frac{5}{2 \times 4} = \frac{5}{8}
\]
So, the solution is:
\[
\boxed{\frac{5}{8}}
\]

### 2. \( \lim_{x \to 1} \frac{(2x + 3x^2) - 2}{x - 1} \)

At \( x = 1 \), we have a \( \frac{0}{0} \) indeterminate form. Let's simplify by factoring or applying L'Hôpital's Rule. First, rewrite and differentiate:

Differentiate the numerator:
\[
\frac{d}{dx} (2x + 3x^2 - 2) = 2 + 6x
\]
Denominator derivative:
\[
\frac{d}{dx}(x - 1) = 1
\]
Now apply L'Hôpital's Rule:
\[
\lim_{x \to 1} \frac{2 + 6x}{1} = 2 + 6(1) = 2 + 6 = 8
\]
Thus, the limit is:
\[
\boxed{8}
\]

### 3. \( \lim_{x \to 0} \frac{x}{1 - \sqrt{1 - x}} \)

At \( x = 0 \), we have the indeterminate form \( \frac{0}{0} \), so we'll apply L'Hôpital's Rule again.

Differentiate the numerator:
\[
\frac{d}{dx} (x) = 1
\]
Differentiate the denominator:
\[
\frac{d}{dx} (1 - \sqrt{1 - x}) = \frac{1}{2\sqrt{1 - x}}
\]
Now apply L'Hôpital's Rule:
\[
\lim_{x \to 0} \frac{1}{\frac{1}{2\sqrt{1 - x}}} = 2\sqrt{1 - 0} = 2
\]
Thus, the limit is:
\[
\boxed{2}
\]

### 4. \( \lim_{x \to 0} \frac{\sqrt{x^2 + 1} - \sqrt{2}}{x} \)

At \( x = 0 \), we have the indeterminate form \( \frac{0}{0} \), so again we apply L'Hôpital's Rule.

Differentiate the numerator:
\[
\frac{d}{dx} \left( \sqrt{x^2 + 1} - \sqrt{2} \right) = \frac{x}{\sqrt{x^2 + 1}}
\]
Denominator derivative:
\[
\frac{d}{dx}(x) = 1
\]
Now apply L'Hôpital's Rule:
\[
\lim_{x \to 0} \frac{\frac{x}{\sqrt{x^2 + 1}}}{1} = \frac{0}{\sqrt{0^2 + 1}} = 0
\]
Thus, the limit is:
\[
\boxed{0}
\]

### 5. \( \lim_{x \to 2} \frac{x^2 - 4}{\sqrt{3x + 2} - 2} \)

At \( x = 2 \), we have a \( \frac{0}{0} \) indeterminate form. Let's multiply by the conjugate to simplify.

Multiply numerator and denominator by \( \sqrt{3x + 2} + 2 \):
\[
\frac{x^2 - 4}{\sqrt{3x + 2} - 2} \cdot \frac{\sqrt{3x + 2} + 2}{\sqrt{3x + 2} + 2} = \frac{(x - 2)(x + 2)}{(\sqrt{3x + 2})^2 - 2^2}
\]
Simplifying the denominator:
\[
(\sqrt{3x + 2})^2 - 2^2 = (3x + 2) - 4 = 3x - 2
\]
Thus, the expression becomes:
\[
\lim_{x \to 2} \frac{(x - 2)(x + 2)}{3x - 2}
\]
At \( x = 2 \), the denominator does not go to 0, so we can directly substitute:
\[
\frac{(2 - 2)(2 + 2)}{3(2) - 2} = \frac{0}{6 - 2} = 0
\]
Thus, the limit is:
\[
\boxed{0}
\]

---

Would you like a detailed explanation for any of the steps? Here are five related problems to further explore these concepts:

1. Find \( \lim_{x \to 3} \frac{x^2 - 9}{x - 3} \).
2. Compute \( \lim_{x \to 1} \frac{\sqrt{x} - 1}{x - 1} \).
3. Evaluate \( \lim_{x \to 0} \frac{x}{e^x - 1} \).
4. Solve \( \lim_{x \to 0} \frac{1 - \cos x}{x^2} \).
5. Determine \( \lim_{x \to 1} \frac{x^3 - 1}{x - 1} \).

**Tip:** When you encounter indeterminate forms like \( \frac{0}{0} \), try applying L'Hôpital's Rule or algebraic simplifications such as factoring or multiplying by conjugates.

Evaluate the following limits: 1. lim (x -> 2) sqrt(5x + 6) / (x - 4) 2. lim (x -> 1) (2x + 3x^2 - 2) / (x - 1) 3. lim (x -> 0) x / (1 - sqrt(1 - x)) 4. lim (x -> 0) (sqrt(x^2 + 1) - sqrt(2)) / x 5. lim (x -> 2) (x^2 - 4) / (sqrt(3x + 2) - 2)

Solve limit problems involving indeterminate forms with the application of L'Hôpital's Rule and algebraic simplifications. The problems include finding the limits of functions with square roots and factorizations, suitable for students in advanced calculus courses.

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