Solving the Limit of an Indeterminate Form: A Comprehensive Guide

In the field of calculus, solving limits is an essential skill that often requires recognizing and working through indeterminate forms such as (frac{0}{0}). This guide provides a detailed explanation on how to solve the limit:

Introduction and Problem Statement

We are given the limit:

[lim_{x to 1} frac{x^n - nx^{n-1}}{(x-1)^2}]

Here, we explore various methods to solve this limit, including the use of L'H?pital's Rule, Maclaurin series expansion, and direct substitution. Let's break down the steps and methods to understand how to approach this problem.

Step-by-Step Solution Using L'H?pital's Rule

Step 1: Direct Substitution

To start, let's substitute (x 1) directly:

The numerator becomes: [1^n - n1^{n-1}) This simplifies to: [1 - n 0) The denominator becomes: [1^2 - 1 0) Thus, we have an indeterminate form (frac{0}{0}).

Step 2: Applying L'H?pital's Rule

Since we have an indeterminate form, we apply L'H?pital's Rule by differentiating both the numerator and the denominator:

Numerator

[f(x) x^n - nx^{n-1}) The derivative is: [f'(x) nx^{n-1} - n)

Denominator

[g(x) (x-1)^2) The derivative is: [g'(x) 2x - 2)

Applying L'H?pital's Rule:

[lim_{x to 1} frac{f'(x)}{g'(x)} lim_{x to 1} frac{nx^{n-1} - n}{2x - 2}]

Substituting (x 1):

The numerator becomes: [n1^{n-1} - n) This simplifies to: [n - n 0) The denominator becomes: [21 - 2 0) Thus, we again have an indeterminate form (frac{0}{0}).

Step 3: Applying L'H?pital's Rule a Second Time

Applying L'H?pital's Rule a second time involves differentiating the numerator and denominator once more:

Numerator

[f''(x) n(n-1)x^{n-2})

Denominator

[g''(x) 2)

Applying L'H?pital's Rule again:

[lim_{x to 1} frac{f''(x)}{g''(x)} lim_{x to 1} frac{n(n-1)x^{n-2}}{2}]

Substituting (x 1):

[frac{n(n-1)1^{n-2}}{2} frac{n(n-1)}{2})

Thus, the limit is:

[lim_{x to 1} frac{x^n - nx^{n-1}}{(x-1)^2} frac{n(n-1)}{2})

Alternative Methods

Maclaurin Series Expansion

Another approach to solve this limit problem is through the Maclaurin series expansion. For (n geq 2):

Using the binomial theorem and the Maclaurin series expansion:

[lim_{x to 1} frac{x^n - nx^{n-1}}{(x-1)^2} lim_{y to 0} frac{y^n - ny y^{n-1}}{y^2} lim_{y to 0} frac{n(n-1)y^{n-2}}{2} frac{n(n-1)}{2})

Conclusion

Through both L'H?pital's Rule and Maclaurin series expansion, we have shown that the limit:

[lim_{x to 1} frac{x^n - nx^{n-1}}{(x-1)^2} frac{n(n-1)}{2})

The limit does not exist for (n 0) since the numerator does not tend to zero. For (n in mathbb{N}) and (n geq 2), the limit exists and is:

[frac{n(n-1)}{2})

This comprehensive guide should help you understand and solve similar limit problems involving indeterminate forms.