Lt x→1 (x^n - n) / (x - 1 - log x)

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Understand the Problem

The question is asking to find the limit of the given expression as x approaches 1. It involves analyzing the limits of the functions in the numerator and denominator.

Answer

The limit is \( n(n-1) \).
Answer for screen readers

The limit is ( n(n-1) ).

Steps to Solve

  1. Identify the Limit Form As ( x ) approaches 1, evaluate the numerator and denominator:
  • Numerator: ( x^n - n ) approaches ( 1^n - n = 1 - n )
  • Denominator: ( x - 1 - \log x ) approaches ( 1 - 1 - \log(1) = 0 )

This indicates a ( \frac{0}{0} ) form, which suggests we can apply L'Hôpital's rule.

  1. Apply L'Hôpital's Rule Differentiate the numerator and denominator:
  • Derivative of numerator: ( \frac{d}{dx}(x^n - n) = n x^{n-1} )
  • Derivative of denominator: ( \frac{d}{dx}(x - 1 - \log x) = 1 - \frac{1}{x} )
  1. Re-evaluate the Limit Now substitute and find the limit again: $$ \lim_{x \to 1} \frac{n x^{n-1}}{1 - \frac{1}{x}} = \frac{n \cdot 1^{n-1}}{1 - 1} = \frac{n}{0} $$ This still leads to an indeterminate form. So we apply L'Hôpital's rule again.

  2. Differentiate Again Differentiate the new numerator and denominator:

  • New derivative of numerator: ( \frac{d}{dx}(n x^{n-1}) = n(n-1)x^{n-2} )
  • New derivative of denominator: ( \frac{d}{dx}(1 - \frac{1}{x}) = \frac{1}{x^2} )
  1. Final Limit Calculation Now we calculate the limit: $$ \lim_{x \to 1} \frac{n(n-1)x^{n-2}}{\frac{1}{x^2}} = n(n-1) \cdot x^2 \text{ at } x = 1 = n(n-1) \cdot 1^2 = n(n-1) $$

The limit is ( n(n-1) ).

More Information

This limit demonstrates the effectiveness of L'Hôpital's rule for resolving indeterminate forms, especially when dealing with polynomials and logarithmic functions.

Tips

  • Forgetting to check if the limit form is ( \frac{0}{0} ) or ( \frac{\infty}{\infty} ) before applying L'Hôpital's Rule.
  • Not differentiating both the numerator and denominator appropriately when applying L'Hôpital's Rule.

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