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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 \( \frac{1}{4} \).

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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 \(\frac{0}{0}\).

Evaluate the following limits: 
(a) lim(x→2) (x - 2)/(x² - 4) 
(b) lim(x→1) (x² - 1)/(x - 1) 
(c) lim(x→0) (x² + x)/x 
(d) lim(x→3) (x² - 9)/(x - 3) 
(e) lim(x→1) (x³ - 1)/(x - 1) 
(f) lim(x→-1) (x² - 1)/(x + 1) 
(g) lim(x→0) (sqrt(x + 4) - 2)/x

Learn how to solve various limit problems using algebraic factoring and the conjugate method. This detailed solution covers the evaluation of limits as x approaches a specific value, suitable for students in Grades 11-12 or College-level Calculus.

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