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The given problem is:

\[
\lim_{x \to 1} \left( 1 - x^2 \right) \left[ \frac{1}{\ln(1 - x)} \right].
\]

Let's proceed step by step to solve the limit.

### Step 1: Simplify the expression
First, express \( 1 - x^2 \) as:
\[
1 - x^2 = (1 - x)(1 + x).
\]
Now, the limit becomes:
\[
\lim_{x \to 1} \frac{(1 - x)(1 + x)}{\ln(1 - x)}.
\]

### Step 2: Analyze the behavior as \( x \to 1 \)
As \( x \to 1 \):
- \( 1 - x \to 0 \),
- \( 1 + x \to 2 \),
- \( \ln(1 - x) \to \ln(0) = -\infty \).

This suggests the limit might be an indeterminate form \( \frac{0}{0} \), so we can apply **L'Hopital's Rule**.

### Step 3: Apply L'Hopital's Rule
We apply L'Hopital's Rule, which states that if a limit is in the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can take the derivatives of the numerator and denominator and then compute the limit again.

The expression is:
\[
\lim_{x \to 1} \frac{(1 - x)(1 + x)}{\ln(1 - x)}.
\]
We differentiate the numerator and denominator with respect to \( x \).

- Derivative of the numerator \( (1 - x)(1 + x) \) is:
\[
\frac{d}{dx} \left( (1 - x)(1 + x) \right) = \frac{d}{dx} \left( 1 - x^2 \right) = -2x.
\]
At \( x = 1 \), this becomes \( -2 \).

- Derivative of the denominator \( \ln(1 - x) \) is:
\[
\frac{d}{dx} \ln(1 - x) = \frac{-1}{1 - x}.
\]
At \( x = 1 \), this becomes \( -\infty \).

### Step 4: Compute the limit
Now, applying L'Hopital's Rule:
\[
\lim_{x \to 1} \frac{-2x}{\frac{-1}{1 - x}} = \lim_{x \to 1} \frac{-2x(1 - x)}{-1}.
\]
Simplifying:
\[
\lim_{x \to 1} 2x(1 - x) = 0.
\]

Thus, the value of the limit is **0**.

Would you like more details on any step, or do you have any questions?

### Related Questions:
1. How does L'Hopital's Rule work for different indeterminate forms like \( \frac{\infty}{\infty} \)?
2. What are the common situations where limits lead to indeterminate forms?
3. Can you explain how to differentiate logarithmic functions in general?
4. Why does \( \ln(0) \) lead to a limit of negative infinity?
5. How would this limit change if we replaced \( \ln(1 - x) \) with another logarithmic expression?

### Tip:
When dealing with limits that approach 0 or infinity, checking for indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) is key, as it signals that L'Hopital's Rule might be applicable!

Evaluate lim_{x \to 1} \left(1 - x^2\right) \left[ \frac{1}{\ln(1 - x)} \right].

This problem explores how to evaluate the limit lim_{x \to 1} (1 - x^2)[1 / ln(1 - x)]. It involves algebraic simplification and applying L'Hopital's Rule for a 0/0 indeterminate form. A step-by-step solution is provided, suitable for advanced calculus students.

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