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The given limit problem is:

\[
\lim_{x \to 1} \frac{1 - x^2}{\sqrt{x} - 1}
\]

To solve this, we can start by simplifying the expression.

### Step 1: Factorize the numerator
The numerator \( 1 - x^2 \) can be factorized using the difference of squares:

\[
1 - x^2 = (1 - x)(1 + x)
\]

Now the expression becomes:

\[
\lim_{x \to 1} \frac{(1 - x)(1 + x)}{\sqrt{x} - 1}
\]

### Step 2: Substitute \( x = 1 \) and evaluate directly
Direct substitution of \( x = 1 \) results in the indeterminate form \( \frac{0}{0} \). Thus, we need to simplify further, possibly using algebraic manipulation or L'Hôpital's rule.

### Step 3: Apply L'Hôpital's Rule
Since the limit is in the form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which involves differentiating the numerator and denominator with respect to \( x \).

#### Derivatives:
- Derivative of the numerator \( 1 - x^2 \): \( -2x \)
- Derivative of the denominator \( \sqrt{x} - 1 \): \( \frac{1}{2\sqrt{x}} \)

Thus, the limit becomes:

\[
\lim_{x \to 1} \frac{-2x}{\frac{1}{2\sqrt{x}}}
\]

### Step 4: Simplify and substitute
Simplifying the expression:

\[
\lim_{x \to 1} \frac{-2x}{\frac{1}{2\sqrt{x}}} = \lim_{x \to 1} \frac{-2x \cdot 2\sqrt{x}}{1} = \lim_{x \to 1} -4x\sqrt{x}
\]

Now, substitute \( x = 1 \):

\[
-4(1)(\sqrt{1}) = -4
\]

Thus, the value of the limit is:

\[
\boxed{-4}
\]

Would you like more details on any step or have questions about this process?

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Here are five related questions to explore:
1. What is L'Hôpital's Rule, and when is it applicable?
2. How can algebraic manipulation help in solving limits without L'Hôpital's Rule?
3. What are the common forms of indeterminate limits?
4. How do you differentiate composite functions like \( \sqrt{x} \)?
5. How would this limit problem change if the denominator were \( \sqrt{x+1} - 1 \)?

**Tip:** Always check for factorization or algebraic simplifications before applying L'Hôpital's Rule—it can save time and effort!

Learn how to solve the limit lim (x -> 1) (1 - x^2) / (sqrt(x) - 1) using algebraic manipulation and L'Hôpital's Rule. This calculus problem involves factoring and applying derivatives for indeterminate forms. Ideal for students in Grades 11-12 or early college-level calculus courses.

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