https://file.aiquickdraw.com/mathbot/file/1728598847223nup0om3f.jpg

Let's analyze the limits given in the image:

### Problem 6:
\[
\lim_{x \to 1} \frac{\sqrt{2x} - \sqrt{x+1}}{x - 1}
\]

### Solution for Problem 6:
1. Direct substitution of \( x = 1 \) results in an indeterminate form \( \frac{0}{0} \).
2. To resolve this, use the rationalization technique by multiplying the numerator and denominator by the conjugate of the numerator:
   \[
   \text{Multiply by } \frac{\sqrt{2x} + \sqrt{x+1}}{\sqrt{2x} + \sqrt{x+1}}.
   \]
3. The expression becomes:
   \[
   \lim_{x \to 1} \frac{(\sqrt{2x} - \sqrt{x+1})(\sqrt{2x} + \sqrt{x+1})}{(x - 1)(\sqrt{2x} + \sqrt{x+1})}
   \]
   Which simplifies the numerator:
   \[
   (\sqrt{2x})^2 - (\sqrt{x+1})^2 = 2x - (x+1) = x - 1
   \]
4. The \( x - 1 \) terms in the numerator and denominator cancel out, leading to:
   \[
   \lim_{x \to 1} \frac{x - 1}{(x - 1)(\sqrt{2x} + \sqrt{x+1})} = \lim_{x \to 1} \frac{1}{\sqrt{2x} + \sqrt{x+1}}
   \]
5. Substituting \( x = 1 \):
   \[
   \frac{1}{\sqrt{2 \cdot 1} + \sqrt{1+1}} = \frac{1}{\sqrt{2} + \sqrt{2}} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}
   \]
   Thus, the limit is \( \frac{\sqrt{2}}{4} \).

---

### Problem 7:
\[
\lim_{x \to 4} \frac{x^2 - 3x - 4}{x - 4}
\]

### Solution for Problem 7:
1. Direct substitution of \( x = 4 \) results in an indeterminate form \( \frac{0}{0} \).
2. Factor the numerator:
   \[
   x^2 - 3x - 4 = (x - 4)(x + 1)
   \]
3. The expression becomes:
   \[
   \lim_{x \to 4} \frac{(x - 4)(x + 1)}{x - 4}
   \]
4. Cancel the \( x - 4 \) terms:
   \[
   \lim_{x \to 4} (x + 1) = 4 + 1 = 5
   \]
   Thus, the limit is \( 5 \).

---

### Final Answers:
1. Problem 6: \(\lim_{x \to 1} \frac{\sqrt{2x} - \sqrt{x+1}}{x - 1} = \frac{\sqrt{2}}{4}\)
2. Problem 7: \(\lim_{x \to 4} \frac{x^2 - 3x - 4}{x - 4} = 5\)

Would you like a more detailed explanation for any of these steps or have any other questions?

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**Related Questions:**
1. How do you rationalize expressions involving square roots?
2. What are common indeterminate forms in limits?
3. How do you factor quadratic expressions effectively?
4. What is the conjugate of a binomial expression?
5. How does L'Hôpital's rule apply to indeterminate forms like \( \frac{0}{0} \)?

**Tip:** When solving limits that result in \( \frac{0}{0} \), consider using algebraic techniques like factoring or rationalization before applying L'Hôpital's rule.

Learn how to solve limit problems with square roots and quadratics. Problem 6 uses rationalization to simplify indeterminate forms, while Problem 7 involves factoring a quadratic expression. Suitable for students in Grades 11-12.

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