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!