Problem 4 - University Calculus: Early Transcendentals

Question

Use I'Hópital'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 Indeterminate Form

First, we should check if the limit leads to an indeterminate form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). Substitute \( x = 1 \) into the expression \( \frac{x^{3}-1}{4 x^{3}-x-3} \).\[ \frac{1^3-1}{4(1)^3-1-3} = \frac{0}{0}. \] Hence, the given limit is in an indeterminate form, allowing the use of L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule

Since the limit is in the form \( \frac{0}{0} \), we apply L'Hôpital's Rule, which involves differentiating the numerator and the denominator separately. Calculate the derivatives: * Numerator: \( \frac{d}{dx}(x^3 - 1) = 3x^2 \) * Denominator: \( \frac{d}{dx}(4x^3 - x - 3) = 12x^2 - 1 \). Now evaluate the limit using these derivatives: \[ \lim_{x \to 1} \frac{3x^2}{12x^2 - 1}. \]

Step 3: Re-evaluate the Limit after Differentiation

Substitute \( x = 1 \) into the differentiated expression: \[ \frac{3(1)^2}{12(1)^2 - 1} = \frac{3}{12 - 1} = \frac{3}{11}. \] Therefore, the limit using L'Hôpital's Rule is \( \frac{3}{11} \).

Step 4: Confirm Using Factoring Method

Now, confirm the limit using the polynomial factoring method studied in Chapter 2. The numerator \( x^3 - 1 \) can be factored as \((x - 1)(x^2 + x + 1)\), and the denominator \( 4x^3 - x - 3 \) can be factored (or simplified if possible) to \( (x-1)(4x^2 + 4x + 3)\). The limit becomes:\[ \lim_{x \to 1} \frac{(x - 1)(x^2 + x + 1)}{(x - 1)(4x^2 + 4x + 3)}. \] Cancel \( x - 1 \) from the numerator and denominator:\[ \lim_{x \to 1} \frac{x^2 + x + 1}{4x^2 + 4x + 3}. \] Substituting \( x = 1 \):\[ \frac{1^2 + 1 + 1}{4(1)^2 + 4(1) + 3} = \frac{3}{11}. \] Thus, the limit is confirmed as \( \frac{3}{11}. \)

Key Concepts

Indeterminate Forms

When we encounter a limit problem like \( \lim _{x \rightarrow 1} \frac{x^{3}-1}{4 x^{3}-x-3} \), the first step is to check whether substituting the limit point, here \( x = 1 \), leads to an indeterminate form. Indeterminate forms occur when a mathematical expression doesn't have a clear-cut value, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms indicate that straightforward substitution won't work, and special techniques are needed to resolve them.

• An expression turns into \( \frac{0}{0} \) if both the numerator and denominator equal zero at the point of substitution.
• Similarly, expressions like \( \frac{\infty}{\infty} \) arise from both parts tending towards infinity.

Recognizing these situations allows the use of methods such as L'Hôpital's Rule, which is specifically designed to evaluate the limits of these sorts of indeterminate forms, providing a way forward when numbers seem impossible to handle.

Polynomial Factoring

Polynomial factoring can simplify complex expressions, making limit evaluation more straightforward. In the given limit problem, \( x^3 - 1 \) and \( 4x^3 - x - 3 \) are polynomials that lead us to an indeterminate form initially. Factoring helps break these polynomials into simpler components.

• The numerator \( x^3 - 1 \) is a difference of cubes, which factors as \((x - 1)(x^2 + x + 1)\).
• The denominator \( 4x^3 - x - 3 \) involves more steps and can be simplified using known equations or a guessing method to \((x-1)(4x^2 + 4x + 3)\).

By factoring out a common term \((x - 1)\), the polynomials reduce, allowing a straightforward substitution of \( x = 1 \).

In limit problems, this method not only simplifies calculations but also confirms results derived from other techniques like L'Hôpital's Rule.

Limit Evaluation

The main goal in evaluating limits is to find the value that a function approaches as the input approaches a given number, like \( x \to 1 \) in our example. Approaching this evaluation can be done through various techniques, such as substitution, factoring, or using calculus-based methods like L'Hôpital's Rule.

• In simple cases, direct substitution might work, providing an immediate answer if no indeterminate form arises.
• When encountering an indeterminate form, methods like factoring and applying L'Hôpital's Rule are often necessary to simplify the expression and evaluate the limit.

L'Hôpital's Rule, specifically, is applied when the limit results in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). By differentiating the numerator and denominator separately, a new expression emerges, often resolving the indeterminate form. In our practice problem, both methods, factoring and L'Hôpital's Rule, agree and confirm the result to be \( \frac{3}{11} \).

Understanding the versatility and applicability of these strategies ensures accurate limit evaluation, providing clear insights into how functions behave at specific points.