Problem 4 - Thomas Calculus

Question

In Exercises \(1-6,\) use 1'Hopital's Rule to evaluate the limit. Then evaluate the limit using a method studied in Chapter 2 . $$ \lim _{x \rightarrow 1} \frac{x^{3}-1}{4 x^{3}-x-3} $$

Step-by-Step Solution

Answer

The limit is \( \frac{3}{11} \).

Step 1: Check the Indeterminate Form

First, verify if the function is in an indeterminate form by plugging the limit point into the given expression. For \( f(x) = x^3 - 1 \), substituting \( x = 1 \) gives \( f(1) = 1^3 - 1 = 0 \). For \( g(x) = 4x^3 - x - 3 \), substituting \( x = 1 \) gives \( 4(1)^3 - 1 - 3 = 0 \). Therefore, the given expression is in the indeterminate form \( \frac{0}{0} \).

Step 2: Apply L'Hôpital's Rule

L'Hôpital's Rule states that if \( \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \). For \( f(x) = x^3 - 1 \), the derivative \( f'(x) = 3x^2 \). For \( g(x) = 4x^3 - x - 3 \), the derivative \( g'(x) = 12x^2 - 1 \). Now, evaluate the limit using the derivatives:\[\lim_{x \to 1} \frac{3x^2}{12x^2 - 1}\]

Step 3: Evaluate the Limit of Derivatives

Substitute \( x = 1 \) into the derivative expressions:\[\lim_{x \to 1} \frac{3(1)^2}{12(1)^2 - 1} = \frac{3}{12 - 1} = \frac{3}{11}\]So, the limit using L'Hôpital's Rule is \( \frac{3}{11} \).

Step 4: Alternative Method - Factorization

Using an alternative method (studied in Chapter 2), factor the numerator and denominator. The numerator \( x^3 - 1 = (x - 1)(x^2 + x + 1) \) and the denominator \( 4x^3 - x - 3 = (x - 1)(4x^2 + 4x + 3) \). Simplifying the expression by canceling \( x - 1 \):\[\lim_{x \to 1} \frac{x^2 + x + 1}{4x^2 + 4x + 3}\]

Step 5: Evaluate Simplified Expression

Substitute \( x = 1 \) into the simplified expression:\[\lim_{x \to 1} \frac{1^2 + 1 + 1}{4(1)^2 + 4(1) + 3} = \frac{1 + 1 + 1}{4 + 4 + 3} = \frac{3}{11}\]Thus, using factorization also gives the limit \( \frac{3}{11} \).

Key Concepts

Indeterminate Forms

When dealing with limits in calculus, not all expressions yield straightforward results. Sometimes, substituting a value directly into a function results in expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These are known as indeterminate forms. Understanding these forms is crucial because they indicate that the limit can't be directly evaluated simply by substitution.

For example, in the provided exercise, when you plug \( x = 1 \) into the expression \( \frac{x^3-1}{4x^3-x-3} \), both the numerator and the denominator become zero, resulting in the \( \frac{0}{0} \) indeterminate form. This shows that we need to apply more advanced techniques, such as L'Hôpital's Rule or algebraic manipulation, to find the limit.

Limit Evaluation

Evaluating limits involves finding the value that a function approaches as the input approaches some point. Often, limits are integral to understanding the behavior of functions and are a foundational concept in calculus.

In the context of indeterminate forms, techniques such as L'Hôpital's Rule become invaluable. According to L'Hôpital's Rule, if an expression is in the \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) form, the limit of the ratio of the functions can be found by taking the limit of the ratio of their derivatives. This provides a way to solve limits that would otherwise seem unsolvable through direct substitution.

This method simplifies the evaluation process, allowing for a clearer path to find the limit. However, it's important to verify that you're working with an indeterminate form before applying the rule to ensure accuracy.

Factorization

Factorization is an algebraic method that can simplify complex expressions, making it easier to evaluate limits. By breaking down expressions into products of simpler factors, it's often possible to cancel common terms and reduce the complexity of the function.

In our exercise, factorizing both the numerator \( x^3 - 1 = (x - 1)(x^2 + x + 1) \) and the denominator \( 4x^3 - x - 3 = (x - 1)(4x^2 + 4x + 3) \) reveals a common factor \( x - 1 \). By canceling this factor, the expression simplifies to \( \frac{x^2 + x + 1}{4x^2 + 4x + 3} \). This simplification allows for an easy direct substitution of \( x = 1 \) to find the limit. Factorization is an alternative to L'Hôpital's Rule and can sometimes be more straightforward depending on the complexity of the derivatives involved.

Calculus Techniques

Calculus offers various techniques for solving limits, each with its own advantages. Beyond L'Hôpital's Rule and factorization, other methods like rationalization, conjugates, and trigonometric identities can also be used.

These techniques emphasize the creativity and flexibility required in calculus. Choosing the right method depends on the problem's specific circumstances and the form presented. For limits involving polynomials, factorization might be simpler. Meanwhile, rational expressions might benefit from L'Hôpital's Rule.

Developing a good intuition for selecting the most efficient method is part of mastering calculus. Practice exploring these techniques can enhance problem-solving skills, ensuring you're well-equipped to tackle any limit-related challenge. The goal is to arrive at an understanding of how a function behaves as it approaches a particular point, which is a powerful tool in mathematical analysis and application.