Math Problem Statement

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

Solution

The problem is:

limx1(xx11lnx)\lim_{x \to 1} \left( \frac{x}{x - 1} - \frac{1}{\ln x} \right)

Step-by-step solution:

1. Identify the indeterminate form:

  • As x1x \to 1, the first term xx1\frac{x}{x - 1} approaches 10\frac{1}{0} (infinity).
  • The second term 1lnx\frac{1}{\ln x} also approaches infinity since lnx0\ln x \to 0.

Thus, the overall expression is in the indeterminate form \infty - \infty. We must simplify further.


2. Combine the terms into a single fraction:

Write the expression with a common denominator: xx11lnx=xlnx(x1)(x1)lnx.\frac{x}{x - 1} - \frac{1}{\ln x} = \frac{x \ln x - (x - 1)}{(x - 1) \ln x}.

Now the limit becomes: limx1xlnx(x1)(x1)lnx.\lim_{x \to 1} \frac{x \ln x - (x - 1)}{(x - 1) \ln x}.


3. Expand and analyze the numerator:

  • As x1x \to 1, expand lnx\ln x using a Taylor series: lnx(x1)(x1)22+O((x1)3).\ln x \approx (x - 1) - \frac{(x - 1)^2}{2} + \mathcal{O}((x - 1)^3).

Thus: xlnx=x((x1)(x1)22+).x \ln x = x \big((x - 1) - \frac{(x - 1)^2}{2} + \cdots \big).

For small x1x - 1, x1+(x1)x \approx 1 + (x - 1), so: xlnx(x1)(x1)22+.x \ln x \approx (x - 1) - \frac{(x - 1)^2}{2} + \cdots.

The numerator becomes: xlnx(x1)(x1)22+.x \ln x - (x - 1) \approx -\frac{(x - 1)^2}{2} + \cdots.


4. Simplify the denominator:

The denominator is: (x1)lnx(x1)((x1)(x1)22+).(x - 1) \ln x \approx (x - 1)((x - 1) - \frac{(x - 1)^2}{2} + \cdots).

This simplifies to: (x1)2(x1)32+.(x - 1)^2 - \frac{(x - 1)^3}{2} + \cdots.


5. Simplify the fraction:

Divide the numerator by the denominator: xlnx(x1)(x1)lnx(x1)22(x1)2.\frac{x \ln x - (x - 1)}{(x - 1) \ln x} \approx \frac{-\frac{(x - 1)^2}{2}}{(x - 1)^2}.

Cancel (x1)2(x - 1)^2: 12.\approx -\frac{1}{2}.


Final Answer:

limx1(xx11lnx)=12.\lim_{x \to 1} \left( \frac{x}{x - 1} - \frac{1}{\ln x} \right) = -\frac{1}{2}.

Would you like further clarification or details? Let me know!


Related Questions:

  1. What is the general approach to solving \infty - \infty limits?
  2. How do Taylor series expansions simplify indeterminate forms?
  3. Can L'Hôpital's Rule be applied to this problem? If yes, how?
  4. How does lnx\ln x behave as x1x \to 1?
  5. What are some other examples of combining fractions to resolve limits?

Tip:

When faced with \infty - \infty forms, always look for ways to combine terms into a single fraction for simplification.

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Math Problem Analysis

Mathematical Concepts

Limits
Indeterminate Forms
Taylor Series Expansion

Formulas

Taylor Series of ln(x): ln(x) ≈ (x - 1) - (x - 1)^2 / 2 + O((x - 1)^3)
Simplification of indeterminate forms: ∞ - ∞
Combination of terms into a single fraction

Theorems

Taylor Series Expansion
L'Hôpital's Rule (optional)

Suitable Grade Level

Undergraduate Calculus / Advanced High School