Solving Limits Using L'Hopital's Rule and Polynomial Factorization: A Practical Guide

Introduction

Understanding and solving limits is a fundamental aspect of calculus, often appearing in various real-world applications such as physics, engineering, and economics. This article focuses on how to evaluate a specific limit using both algebraic manipulation and L'Hopital's rule. We provide detailed steps and an in-depth explanation to help clarify the process for students and professionals alike.

The Problem

Consider the problem:

[ lim_{{x to 1}} left(frac{{x^{100} - 2x - 1}}{{x^{50} - 1}}right) ]

This limit is easy to solve if you are familiar with L'Hopital's rule. However, for a deeper understanding, let's explore both methods: algebraic factorization and the L'Hopital's rule.

Method 1: Factorization

First, let's try to factorize the expression:

Let L [ lim_{{x to 1}} left(frac{{x^{100} - 2x - 1}}{{x^{50} - 1}}right) ]

Note that both the numerator and the denominator have a factor of (x - 1). This simplifies the problem significantly.

To factorize, start with the denominator:

(x^{50} - 1 (x - 1)(x^{49} x^{48} ldots x 1))

For the numerator, notice that:

(x^{100} - 2x - 1 x^{100} - x^{99} - x^{99} x^{98} ldots x^2 - x - x - 1)

Add and subtract the terms in a strategic way to factor out (x - 1):

(x^{100} - 2x - 1 (x^{100} - x^{99}) - (x^{99} - x^{98}) - (x^2 - x) - (x - 1))

This results in:

(x^{100} - 2x - 1 (x - 1)(x^{99} x^{98} ldots x 1) - (x - 1)(x^{98} ldots x 1) - (x - 1)(x 1) - (x - 1))

Combining the factors, we get:

(x^{100} - 2x - 1 (x - 1)(x^{98} 2x^{97} ldots x 1))

Thus, the limit becomes:

(L lim_{{x to 1}} left(frac{{(x - 1)(x^{98} 2x^{97} ldots x 1)}}{{(x - 1)(x^{49} x^{48} ldots x 1)}}right))

Cancel out the common factor (x - 1):

(L lim_{{x to 1}} left(frac{{x^{98} 2x^{97} ldots x 1}}{{x^{49} x^{48} ldots x 1}}right))

Evaluating the limit:

(L frac{{1^{98} 2(1^{97}) ldots 1 1}}{{1^{49} 1^{48} ldots 1 1}} frac{{1 2 1}}{{1 1}} frac{{4}}{{2}} frac{{98}}{{50}} frac{{49}}{{25}})

Method 2: L'Hopital's Rule

Another method to solve the same limit is by applying L'Hopital's rule. According to this rule, if the limit of the ratio of two functions results in an indeterminate form (frac{0}{0}) or (frac{infty}{infty}), we can differentiate the numerator and the denominator separately and then take the limit.

(L lim_{{x to 1}} frac{{f'(x)}}{{g'(x)}})

Where (f(x) x^{100} - 2x - 1) and (g(x) x^{50} - 1).

Differentiating (f(x)) and (g(x)) with respect to (x):

(f'(x) 10^{99} - 2)

(g'(x) 5^{49})

Applying the rule:

(L lim_{{x to 1}} frac{{10^{99} - 2}}{{5^{49}}})

Evaluating the limit:

(L frac{{100(1^{99}) - 2}}{{50(1^{49})}} frac{{100 - 2}}{{50}} frac{{98}}{{50}} frac{{49}}{{25}})

Conclusion

Both methods lead to the same result: (frac{{49}}{{25}}). Understanding the algebraic manipulation and L'Hopital's rule not only helps in solving the problem but also enhances your problem-solving skills.

This guide provides a comprehensive approach to solving limits using algebraic techniques and calculus rules. Whether you prefer factorization or L'Hopital's rule, both methods are powerful tools in your mathematical toolkit.