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?

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.