Lt x→1 (x^n - n) / (x - 1 - log x)

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) $.

Steps to Solve

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.

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} )

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.
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} )

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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