To evaluate the limit $\lim_{x \to 3} \frac{\sqrt{x^3 - 7}}{\sqrt{x} - 3}$, we'll proceed with the following steps:

Step 1: Direct Substitution

Substituting $x = 3$ directly into the expression:

$\frac{\sqrt{3^3 - 7}}{\sqrt{3} - 3} = \frac{\sqrt{27 - 7}}{\sqrt{3} - 3} = \frac{\sqrt{20}}{\sqrt{3} - 3}$

This expression is not of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, which are indeterminate forms. So, we can evaluate it directly.

Step 2: Simplify the Expression

The expression can be simplified by multiplying the numerator and denominator by the conjugate of the denominator:

$\frac{\sqrt{20}}{\sqrt{3} - 3} \times \frac{\sqrt{3} + 3}{\sqrt{3} + 3} = \frac{\sqrt{20}(\sqrt{3} + 3)}{(\sqrt{3})^2 - 3^2}$

This simplifies to:

$\frac{\sqrt{20}(\sqrt{3} + 3)}{3 - 9} = \frac{\sqrt{20}(\sqrt{3} + 3)}{-6}$

Further simplifying:

$\frac{\sqrt{20}(\sqrt{3} + 3)}{-6} = \frac{-\sqrt{20}(\sqrt{3} + 3)}{6}$

Final Answer

The limit is:

$\lim_{x \to 3} \frac{\sqrt{x^3 - 7}}{\sqrt{x} - 3} = \frac{-\sqrt{20}(\sqrt{3} + 3)}{6}$

Would you like any more details or have any questions?

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Here are 5 related questions you might find interesting:

1. How would you handle the limit if the expression was indeterminate?
2. Can you simplify the expression further by rationalizing the numerator?
3. What if $x$ approached a different value, say $x \to 2$? How would that affect the limit?
4. How can L'Hôpital's rule be applied to more complex limits involving square roots?
5. How does the presence of square roots affect the convergence of limits?

Tip: Always check if the expression simplifies directly before applying more advanced techniques like L'Hôpital's rule.