Math Problem Statement

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

Solution

The given problem is:

limx1(1x2)[1ln(1x)].\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 1x21 - x^2 as: 1x2=(1x)(1+x).1 - x^2 = (1 - x)(1 + x). Now, the limit becomes: limx1(1x)(1+x)ln(1x).\lim_{x \to 1} \frac{(1 - x)(1 + x)}{\ln(1 - x)}.

Step 2: Analyze the behavior as x1x \to 1

As x1x \to 1:

  • 1x01 - x \to 0,
  • 1+x21 + x \to 2,
  • ln(1x)ln(0)=\ln(1 - x) \to \ln(0) = -\infty.

This suggests the limit might be an indeterminate form 00\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 00\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: limx1(1x)(1+x)ln(1x).\lim_{x \to 1} \frac{(1 - x)(1 + x)}{\ln(1 - x)}. We differentiate the numerator and denominator with respect to xx.

  • Derivative of the numerator (1x)(1+x)(1 - x)(1 + x) is: ddx((1x)(1+x))=ddx(1x2)=2x.\frac{d}{dx} \left( (1 - x)(1 + x) \right) = \frac{d}{dx} \left( 1 - x^2 \right) = -2x. At x=1x = 1, this becomes 2-2.

  • Derivative of the denominator ln(1x)\ln(1 - x) is: ddxln(1x)=11x.\frac{d}{dx} \ln(1 - x) = \frac{-1}{1 - x}. At x=1x = 1, this becomes -\infty.

Step 4: Compute the limit

Now, applying L'Hopital's Rule: limx12x11x=limx12x(1x)1.\lim_{x \to 1} \frac{-2x}{\frac{-1}{1 - x}} = \lim_{x \to 1} \frac{-2x(1 - x)}{-1}. Simplifying: limx12x(1x)=0.\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)\ln(0) lead to a limit of negative infinity?
  5. How would this limit change if we replaced ln(1x)\ln(1 - x) with another logarithmic expression?

Tip:

When dealing with limits that approach 0 or infinity, checking for indeterminate forms like 00\frac{0}{0} or \frac{\infty}{\infty} is key, as it signals that L'Hopital's Rule might be applicable!

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Math Problem Analysis

Mathematical Concepts

Limits
L'Hopital's Rule
Logarithmic Functions
Algebraic Expressions

Formulas

L'Hopital's Rule: \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}
Algebraic Identity: 1 - x^2 = (1 - x)(1 + x)

Theorems

L'Hopital's Rule

Suitable Grade Level

College Calculus or Advanced High School Calculus