Math Problem Statement

Evaluate the limit: lim(x → 1) ((3x)/(x - 1) - (3)/(ln(x)))

Solution

The given problem involves evaluating the limit:

limx1(3xx13ln(x)).\lim_{x \to 1} \left( \frac{3x}{x - 1} - \frac{3}{\ln(x)} \right).

Let’s analyze this step-by-step.

Step 1: Check for indeterminate forms

When x1x \to 1, both terms 3xx1\frac{3x}{x - 1} and 3ln(x)\frac{3}{\ln(x)} can lead to indeterminate expressions. Specifically:

  • For 3xx1\frac{3x}{x - 1}: When x1x \to 1, this becomes 3111=30\frac{3 \cdot 1}{1 - 1} = \frac{3}{0}, which suggests potential divergence or a more subtle behavior.
  • For 3ln(x)\frac{3}{\ln(x)}: As x1x \to 1, ln(1)=0\ln(1) = 0, so this also takes the form 30\frac{3}{0}.

Both terms suggest that they individually blow up, so we need to analyze their combined behavior carefully.


Step 2: Simplify using series expansions

To handle limits where x1x \to 1, we can apply Taylor expansions around x=1x = 1.

  • For x1x - 1 near 0, we approximate:
    x1+(x1)x \approx 1 + (x - 1).

  • For ln(x)\ln(x) near 1, the Taylor expansion gives:
    ln(x)(x1)(x1)22+\ln(x) \approx (x - 1) - \frac{(x - 1)^2}{2} + \cdots.
    So, ln(x)(x1)\ln(x) \approx (x - 1) when higher-order terms are neglected.


Step 3: Substitute the expansions into the limit

  1. First term:
    3xx1=3(1+(x1))x1=3+3(x1)x1=3+3.\frac{3x}{x - 1} = \frac{3(1 + (x - 1))}{x - 1} = \frac{3 + 3(x - 1)}{x - 1} = 3 + 3.

  2. Second term:
    3ln(x)3x1.\frac{3}{\ln(x)} \approx \frac{3}{x - 1}.


Step 4: Combine the terms

The expression inside the limit becomes:

3xx13ln(x)(3+3)3=3.\frac{3x}{x - 1} - \frac{3}{\ln(x)} \approx (3 + 3) - 3 = 3.


Step 5: Evaluate the limit

Since the expression simplifies to a constant 3, the limit is:

3.\boxed{3}.


Would you like further clarifications or additional details?


Here are 5 related questions:

  1. How do Taylor expansions help in limit calculations?
  2. What are some other indeterminate forms besides 00\frac{0}{0}?
  3. How can L'Hôpital's rule be applied to similar limits?
  4. What are the conditions for applying Taylor series approximations?
  5. What other techniques can handle limits with logarithms?

Tip: When limits result in indeterminate forms involving 0/00/0 or \infty - \infty, Taylor expansions or L'Hôpital's rule are often effective strategies.

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Math Problem Analysis

Mathematical Concepts

Limits
Indeterminate Forms
Series Expansions

Formulas

Taylor Expansion for ln(x) around 1
Basic limit properties

Theorems

L'Hôpital's Rule
Taylor Expansion

Suitable Grade Level

Grades 11-12 or College level Calculus