Problem 4 - Thomas Calculus in SI Units

Question

Use \(I^{\prime}\) Hôpital'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: Identify Indeterminate Form

First, evaluate the expression at the limit point to check if it is an indeterminate form. Substitute \(x = 1\) into \(\frac{x^{3}-1}{4x^{3}-x-3}\) to see if it results in \(\frac{0}{0}\):\[\frac{1^3 - 1}{4(1)^3 - 1 - 3} = \frac{0}{0}\]Since it results in \(\frac{0}{0}\), it is an indeterminate form.

Step 2: Apply L'Hôpital's Rule

L'Hôpital's Rule states that if a limit \(\frac{f(x)}{g(x)}\) results in an indeterminate form \(\frac{0}{0}\), the limit can be evaluated as:\[\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\]Differentiate the numerator and denominator with respect to \(x\):- Derivative of the numerator \(f(x) = x^3 - 1\) is \(f'(x) = 3x^2\).- Derivative of the denominator \(g(x) = 4x^3 - x - 3\) is \(g'(x) = 12x^2 - 1\).

Step 3: Evaluate Limit After Differentiation

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

Step 4: Factor the Original Expression

To evaluate the limit using Chapter 2 methods, factor the numerator and denominator if possible. Rewrite the original expression:The numerator \(x^3 - 1\) can be factored as \((x-1)(x^2 + x + 1)\).The denominator \(4x^3 - x - 3\) can be rewritten and factored: \(4x^3 - x - 3 = (x-1)(4x^2 + 4x + 3)\).Thus:\[\frac{x^3 - 1}{4x^3 - x - 3} = \frac{(x-1)(x^2 + x + 1)}{(x-1)(4x^2 + 4x + 3)}\]

Step 5: Cancel Common Factors and Evaluate

Cancel the common factor \(x-1\) from the numerator and denominator:\[\frac{x^2 + x + 1}{4x^2 + 4x + 3}\]Now evaluate the limit as \(x \to 1\):\[\frac{1^2 + 1 + 1}{4(1)^2 + 4(1) + 3} = \frac{3}{11}\]Thus, the limit using factorization is \(\frac{3}{11}\).

Key Concepts

Indeterminate Forms

When tackling calculus problems, you often encounter expressions that don't initially make sense, like dividing zero by zero or infinity by infinity. These are called "indeterminate forms." They arise when substituting a limit value results in a form like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), among others. These forms do not directly provide an answer, and typically, special techniques are used to evaluate them.

In the exercise provided, when substituting \(x = 1\) into the expression \(\frac{x^3-1}{4x^3-x-3}\), the result is \(\frac{0}{0}\). This indicates an indeterminate form, pointing to the need for methods such as L'Hôpital's Rule or factorization to find the actual limit.

Limit Evaluation

Limit evaluation is at the core of many calculus problems and involves finding the value that a function approaches as the input approaches some point. When direct substitution gives an indeterminate form, it's a signal to use additional strategies to find the limit.

L'Hôpital's Rule is one of the go-to strategies for dealing with limits that result in \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). To use it, differentiate the numerator and the denominator separately and then take the limit of these new expressions. In our example, L'Hôpital's Rule was applied by differentiating the numerator \(f(x) = x^3 - 1\) to get \(f'(x) = 3x^2\) and the denominator \(g(x) = 4x^3 - x - 3\) to get \(g'(x) = 12x^2 - 1\). The limit of \(\frac{3x^2}{12x^2 - 1}\) as \(x\) approaches 1 turned out to be \(\frac{3}{11}\).

Factorization Method

The factorization method is another valuable technique for evaluating limits when faced with indeterminate forms. This strategy involves simplifying the expression by breaking it down into factors so that common terms can be canceled, thus transforming the indeterminate form into one that can be evaluated directly.

In the specific exercise, both the numerator \(x^3 - 1\) and the denominator \(4x^3 - x - 3\) were factored to reveal a common term involving \(x-1\). This enabled the cancellation of \(x-1\), simplifying the expression to \(\frac{x^2 + x + 1}{4x^2 + 4x + 3}\). After simplification, substituting \(x = 1\) yields \(\frac{3}{11}\), confirming the result obtained through L'Hôpital's Rule.