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.