The given limit is:

$\lim_{x \to 1} \frac{\sqrt{x} - 1}{x - 1}$

Step-by-Step Solution:

1. Direct Substitution: Plugging $x = 1$ directly into the expression, we get:

   $\frac{\sqrt{1} - 1}{1 - 1} = \frac{1 - 1}{1 - 1} = \frac{0}{0}$

   This is an indeterminate form, so we must simplify the expression further.

2. Rationalizing the Numerator: To simplify, we can multiply both the numerator and denominator by the conjugate of the numerator, $\sqrt{x} + 1$:

   $\frac{\sqrt{x} - 1}{x - 1} \times \frac{\sqrt{x} + 1}{\sqrt{x} + 1} = \frac{(\sqrt{x} - 1)(\sqrt{x} + 1)}{(x - 1)(\sqrt{x} + 1)}$

   The numerator simplifies using the difference of squares:

   $(\sqrt{x} - 1)(\sqrt{x} + 1) = x - 1$

   So the expression becomes:

   $\frac{x - 1}{(x - 1)(\sqrt{x} + 1)}$

3. Canceling Terms: Now, we can cancel the $x - 1$ terms from both the numerator and denominator (as long as $x \neq 1$):

   $\frac{1}{\sqrt{x} + 1}$

4. Substitute $x = 1$: Now that the indeterminate form is resolved, substitute $x = 1$:

   $\frac{1}{\sqrt{1} + 1} = \frac{1}{1 + 1} = \frac{1}{2}$

Final Answer:

$\lim_{x \to 1} \frac{\sqrt{x} - 1}{x - 1} = \frac{1}{2}$

Would you like further clarification or additional details?

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Here are 5 related questions to explore further:

1. How would you handle limits involving cube roots instead of square roots?
2. What happens when the indeterminate form $\frac{0}{0}$ arises in other contexts?
3. Can you apply L'Hôpital's Rule to solve this limit? How would it work?
4. How does the process change if the limit is $\lim_{x \to 0} \frac{\sqrt{x} - 1}{x - 1}$?
5. What are the differences between rationalizing the numerator and the denominator in limits?

Tip: Always check for indeterminate forms first to decide the best method for evaluating limits.