The image contains several mathematical problems that involve evaluating the limits of functions as the variable $x$ approaches a certain value. These are common exercises in calculus. Below is the detailed step-by-step solution for each problem:

a) $\lim_{x \to 2} \frac{x - 2}{x^2 - 4}$

To solve this limit, observe that the denominator can be factored: $x^2 - 4 = (x - 2)(x + 2)$ Thus, the expression becomes: $\lim_{x \to 2} \frac{x - 2}{(x - 2)(x + 2)}$ We can cancel the $x - 2$ term in the numerator and denominator: $\lim_{x \to 2} \frac{1}{x + 2} = \frac{1}{2 + 2} = \frac{1}{4}$ So the answer is $\frac{1}{4}$.

b) $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$

Notice that the numerator can be factored: $x^2 - 1 = (x - 1)(x + 1)$ Thus, the expression becomes: $\lim_{x \to 1} \frac{(x - 1)(x + 1)}{x - 1}$ Cancel the $x - 1$ term: $\lim_{x \to 1} (x + 1) = 1 + 1 = 2$ So the answer is $2$.

c) $\lim_{x \to 0} \frac{x^2 + x}{x}$

Simplify the expression by factoring the numerator: $\frac{x^2 + x}{x} = x + 1$ Now, compute the limit: $\lim_{x \to 0} (x + 1) = 0 + 1 = 1$ So the answer is $1$.

d) $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$

Factor the numerator: $x^2 - 9 = (x - 3)(x + 3)$ Now the expression becomes: $\lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3}$ Cancel the $x - 3$ term: $\lim_{x \to 3} (x + 3) = 3 + 3 = 6$ So the answer is $6$.

e) $\lim_{x \to 1} \frac{x^3 - 1}{x - 1}$

Factor the numerator using the difference of cubes: $x^3 - 1 = (x - 1)(x^2 + x + 1)$ Now the expression becomes: $\lim_{x \to 1} \frac{(x - 1)(x^2 + x + 1)}{x - 1}$ Cancel the $x - 1$ term: $\lim_{x \to 1} (x^2 + x + 1) = 1^2 + 1 + 1 = 3$ So the answer is $3$.

f) $\lim_{x \to -1} \frac{x^2 - 1}{x + 1}$

Factor the numerator: $x^2 - 1 = (x - 1)(x + 1)$ Now the expression becomes: $\lim_{x \to -1} \frac{(x - 1)(x + 1)}{x + 1}$ Cancel the $x + 1$ term: $\lim_{x \to -1} (x - 1) = -1 - 1 = -2$ So the answer is $-2$.

g) $\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x}$

This is a tricky limit that requires multiplying by the conjugate: $\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x} \cdot \frac{\sqrt{x + 4} + 2}{\sqrt{x + 4} + 2} = \lim_{x \to 0} \frac{(x + 4) - 4}{x(\sqrt{x + 4} + 2)} = \lim_{x \to 0} \frac{x}{x(\sqrt{x + 4} + 2)} = \lim_{x \to 0} \frac{1}{\sqrt{x + 4} + 2}$ Now, evaluate the limit: $\lim_{x \to 0} \frac{1}{\sqrt{0 + 4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}$ So the answer is $\frac{1}{4}$.

Would you like more details on any of these steps or have any further questions?

Here are 5 related questions for practice:

1. Evaluate $\lim_{x \to -2} \frac{x^2 + 4x + 4}{x + 2}$.
2. Solve $\lim_{x \to 2} \frac{x^2 - 5x + 6}{x - 2}$.
3. Compute $\lim_{x \to 1} \frac{2x^2 + 3x - 5}{x - 1}$.
4. Find $\lim_{x \to 0} \frac{\sqrt{x + 9} - 3}{x}$.
5. Determine $\lim_{x \to 4} \frac{x^2 - 16}{x - 4}$.

Tip: Always check if the function can be factored or simplified before substituting the limit value directly. This often resolves indeterminate forms like $\frac{0}{0}$.