$$ \lim_{x\rarr1} $$$$ \frac{\sqrt{x}-1}{x-1} $$

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?

---

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.

Explore the limit lim_{x→1} (sqrt{x} - 1) / (x - 1) and learn how to handle indeterminate forms using rationalization techniques. A detailed step-by-step solution with formulas and theorems is provided. This problem is suitable for students in Grades 10-12.

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