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hallar los limites de la funciones siguientes

The image contains a set of exercises asking to find the limits of various functions. I'll guide you through the process of solving a few of them, but first, let's focus on a selection of the problems from the image. If you want a specific number explained, feel free to ask, and I’ll provide a step-by-step solution.

I'll start with solving a few problems:

### Problem 25: \(\lim_{x \to 3} \frac{x^2 - 9}{x - 3}\)
To solve this, we first notice that directly substituting \(x = 3\) gives a \(0/0\) indeterminate form. This suggests that we should try to simplify the expression.

\[ \frac{x^2 - 9}{x - 3} \]
Notice that the numerator can be factored as:
\[ x^2 - 9 = (x - 3)(x + 3) \]

So, the expression becomes:
\[ \frac{(x - 3)(x + 3)}{x - 3} \]

For \(x \neq 3\), the \(x - 3\) terms cancel out, leaving:
\[ \lim_{x \to 3} (x + 3) \]

Now, substituting \(x = 3\):
\[ \lim_{x \to 3} (x + 3) = 6 \]

### Problem 31: \(\lim_{x \to 0} \frac{\sin(2x)}{x}\)
To solve this, we can use the standard limit result:
\[ \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \]

Here, the argument of the sine function is \(2x\), so we rewrite the limit as:
\[ \lim_{x \to 0} \frac{\sin(2x)}{x} = \lim_{x \to 0} \frac{2\sin(2x)}{2x} = 2 \lim_{x \to 0} \frac{\sin(2x)}{2x} = 2 \times 1 = 2 \]

### Problem 34: \(\lim_{x \to 1} \frac{x^3 - 1}{x - 1}\)
We can simplify this limit by noticing that the numerator is a difference of cubes:
\[ x^3 - 1 = (x - 1)(x^2 + x + 1) \]

So the limit becomes:
\[ \lim_{x \to 1} \frac{(x - 1)(x^2 + x + 1)}{x - 1} \]

For \(x \neq 1\), the \(x - 1\) terms cancel, leaving:
\[ \lim_{x \to 1} (x^2 + x + 1) \]

Substituting \(x = 1\):
\[ 1^2 + 1 + 1 = 3 \]

### Problem 36: \(\lim_{x \to 2} \frac{x^2 + x - 6}{x - 2}\)
This limit gives an indeterminate form \(0/0\) when you substitute \(x = 2\), so let's factor the numerator:
\[ x^2 + x - 6 = (x - 2)(x + 3) \]

Thus, the expression simplifies to:
\[ \lim_{x \to 2} \frac{(x - 2)(x + 3)}{x - 2} \]

Cancel out the \(x - 2\) terms (for \(x \neq 2\)):
\[ \lim_{x \to 2} (x + 3) = 5 \]

### Problem 39: \(\lim_{x \to \infty} \frac{5x^2 - 3x + 2}{2x^2 + x - 1}\)
When \(x\) approaches infinity, the highest degree terms in the numerator and denominator dominate the behavior of the function.

\[ \lim_{x \to \infty} \frac{5x^2 - 3x + 2}{2x^2 + x - 1} \approx \lim_{x \to \infty} \frac{5x^2}{2x^2} = \frac{5}{2} \]

If you need further explanations or solutions for specific problems from this set, let me know!

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Do you want details on any particular problem? Here are some questions that can further deepen your understanding:

1. How can L'Hôpital's Rule be used for limits that yield indeterminate forms?
2. What are the special limits involving trigonometric functions?
3. Why do limits involving infinity often focus on the highest degree terms?
4. How do factorization techniques help in solving limits involving polynomials?
5. What are the key differences between evaluating limits at finite points versus infinity?

**Tip:** Always check if an expression can be simplified or factored before directly substituting the limit value to avoid indeterminate forms.

Explore how to find limits of various functions, including polynomial and trigonometric limits, with step-by-step solutions. Learn about L'Hôpital's Rule, factorization techniques for polynomials, and special limits involving trigonometric functions. Suitable for advanced high school students (Grades 11-12).

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