https://r2.erweima.ai/mathbot/image/17265473559134j9kv6ts.jpg

Let's solve the limits presented in the image without using L'Hopital's rule:

### a. \(\lim_{x \to 4} \frac{x^3 + 64}{x^2 - 16}\)

1. First, factor both the numerator and denominator:
   - Numerator: \(x^3 + 64 = (x + 4)(x^2 - 4x + 16)\) (sum of cubes)
   - Denominator: \(x^2 - 16 = (x - 4)(x + 4)\) (difference of squares)

2. The expression becomes:
   \[
   \lim_{x \to 4} \frac{(x + 4)(x^2 - 4x + 16)}{(x - 4)(x + 4)}
   \]

3. Simplify by canceling out \((x + 4)\) from both the numerator and denominator:
   \[
   \lim_{x \to 4} \frac{x^2 - 4x + 16}{x - 4}
   \]

4. Substituting \(x = 4\) directly leads to:
   \[
   \frac{16 - 16 + 16}{4 - 4} = \frac{16}{0}
   \]
   This expression leads to an infinite value (positive or negative depending on the sign).

So, the limit does not exist because it tends toward infinity.

---

### b. \(\lim_{x \to -8} \frac{x + 8}{\sqrt{x + 8}}\)

1. Let \(u = x + 8\), so as \(x \to -8\), \(u \to 0\).

2. The expression becomes:
   \[
   \lim_{u \to 0} \frac{u}{\sqrt{u}}
   \]

3. Simplify the fraction:
   \[
   \frac{u}{\sqrt{u}} = \sqrt{u}
   \]

4. Substituting \(u = 0\):
   \[
   \sqrt{0} = 0
   \]

So, the limit is \(0\).

---

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

This limit can be solved by multiplying both the numerator and denominator by the conjugate of the numerator:

1. Multiply by \(\frac{\sqrt{3 - x} + \sqrt{3 + x}}{\sqrt{3 - x} + \sqrt{3 + x}}\):
   \[
   \lim_{x \to 0} \frac{(\sqrt{3 - x} - \sqrt{3 + x})(\sqrt{3 - x} + \sqrt{3 + x})}{x (\sqrt{3 - x} + \sqrt{3 + x})}
   \]

2. The numerator simplifies using the difference of squares:
   \[
   (\sqrt{3 - x})^2 - (\sqrt{3 + x})^2 = (3 - x) - (3 + x) = -2x
   \]

3. Now the expression becomes:
   \[
   \lim_{x \to 0} \frac{-2x}{x (\sqrt{3 - x} + \sqrt{3 + x})}
   \]

4. Simplify by canceling \(x\):
   \[
   \lim_{x \to 0} \frac{-2}{\sqrt{3 - x} + \sqrt{3 + x}}
   \]

5. Substituting \(x = 0\):
   \[
   \frac{-2}{\sqrt{3} + \sqrt{3}} = \frac{-2}{2\sqrt{3}} = \frac{-1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}
   \]

So, the limit is \(-\frac{\sqrt{3}}{3}\).

---

### d. \(\lim_{x \to 81} \frac{\sqrt{x} - 9}{\sqrt[4]{x} - 3}\)

1. Substitute \(u = \sqrt{x}\), so when \(x \to 81\), \(u \to 9\). This changes the expression:
   \[
   \lim_{u \to 9} \frac{u - 9}{\sqrt{u} - 3}
   \]

2. Multiply by the conjugate of the denominator:
   \[
   \lim_{u \to 9} \frac{(u - 9)(\sqrt{u} + 3)}{(\sqrt{u} - 3)(\sqrt{u} + 3)}
   \]

3. The denominator becomes:
   \[
   (\sqrt{u})^2 - 9 = u - 9
   \]

4. The expression simplifies to:
   \[
   \lim_{u \to 9} \sqrt{u} + 3
   \]

5. Substituting \(u = 9\):
   \[
   \sqrt{9} + 3 = 3 + 3 = 6
   \]

So, the limit is \(6\).

---

If you have any questions or need further details, feel free to ask!

### Follow-up Questions:
1. How do conjugates help simplify limits involving square roots?
2. Can you explain the difference between indeterminate forms and infinite limits?
3. How can you detect when to factor a limit problem?
4. What strategies work best for limits that approach infinity?
5. Why does the limit in part (c) involve multiplying by a conjugate?

**Tip:** Always check if direct substitution results in an indeterminate form before applying further techniques like factoring or conjugates.

Sin utilizar la regla de L'Hopital, calcular los siguientes límites: a. lim(x -> 4) (x^3 + 64)/(x^2 - 16), b. lim(x -> -8) (x + 8)/(sqrt(x + 8)), c. lim(x -> 0) (sqrt(3 - x) - sqrt(3 + x))/x, d. lim(x -> 81) (sqrt(x) - 9)/(sqrt[4]{x} - 3)

Learn how to solve limit problems without L'Hopital's Rule. This guide covers algebraic factorization, conjugate multiplication, and dealing with indeterminate forms. Includes detailed solutions for limits approaching zero, infinity, and specific points.

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