Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(\lim _{p \rightarrow x} \frac{p^{3}-x^{3}}{p-x}\)

Step-by-Step Solution

Answer

The limit is \( 3x^{2} \).

Step 1: Identify the Limitation

We need to find the limit of the expression as \( p \) approaches \( x \). Specifically, we have the form \( \lim _{p \rightarrow x} \frac{p^{3}-x^{3}}{p-x} \). This indicates that we will need to simplify the expression, as direct substitution leads to an indeterminate form \( \frac{0}{0} \).

Step 2: Apply the Difference of Cubes Formula

Recall the difference of cubes formula: \( a^{3} - b^{3} = (a-b)(a^{2} + ab + b^{2}) \). Applying this, set \( a = p \) and \( b = x \) which gives us \( p^{3} - x^{3} = (p-x)(p^{2} + px + x^{2}) \).

Step 3: Simplify the Expression

Substitute the expression from the difference of cubes formula into the limit: \( \lim _{p \rightarrow x} \frac{(p-x)(p^{2} + px + x^{2})}{p-x} \). We can now cancel \((p-x)\) from the numerator and the denominator, simplifying the limit to \( \lim _{p \rightarrow x} (p^{2} + px + x^{2}) \).

Step 4: Evaluate the Limit

With the simplified expression \( p^{2} + px + x^{2} \), substitute \( p = x \) to determine the limit: \( x^{2} + x \times x + x^{2} = x^{2} + x^{2} + x^{2} = 3x^{2} \).

Step 5: State the Result

Finally, the limit of the original expression as \( p \) approaches \( x \) is found to be \( 3x^{2} \).

Key Concepts

Difference of Cubes Formula

The difference of cubes formula is a handy algebraic tool to simplify expressions like

• \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \)

It's particularly useful when you're dealing with expressions that seem complicated at first. Here, the cube of one term is subtracted from the cube of another. In our original exercise, the challenge was to simplify \( p^3 - x^3 \). By recognizing this as a difference of cubes, we could immediately apply the formula:

• \( p^3 - x^3 = (p-x)(p^2 + px + x^2) \)

This step transforms the expression, simplifying the analysis and evaluation of limits. If you remember this formula, you can switch intimidating algebraic terms into something far more manageable.

Indeterminate Forms

In calculus, you frequently encounter expressions that result in

• indeterminate forms like \( \frac{0}{0} \).

These forms are undefined in standard arithmetic, requiring additional steps for clarification. In our exercise, substituting \( p = x \) directly into the limit expression gives us \( \frac{0}{0} \). Recognizing this indicates a need for expression simplification before evaluation.
Understanding indeterminate forms helps direct your approach to simplification and potentially the application of other calculus techniques like L'Hôpital's Rule, when appropriate. However, in our case, algebraic simplification was sufficient.

Limit Simplification

Simplifying limits involves rewriting expressions to eliminate terms that cause indeterminate forms or complexity. By applying the difference of cubes formula, we rewrote the expression in the

• numerator: \((p-x)(p^2 + px + x^2)\)

This allows us to cancel \((p-x)\) from both the numerator and the denominator in the original limit expression.
This simplification results in a much clearer expression:

• \( \lim_{p \to x} (p^2 + px + x^2) \)

This step transforms the problem into something much easier to resolve, making it straightforward to substitute and evaluate the limit.

Polynomial Division

Polynomial division can be an essential technique when dealing with complex algebraic expressions. In our original problem, dividing within the limit context by canceling common factors may not involve traditional polynomial long division, but it operates on a similar principle.
Simply, you're looking to "factor and reduce" complexity by

• eliminating shared polynomial factors between the numerator and the denominator.

In this exercise, the polynomial division was more of a conceptual step. However, the ultimate goal is the same: to simplify and reach a point where you no longer have indeterminate forms, enabling you to evaluate the limit directly. Understanding this concept helps us deal with both simple cancellations and more complex algebraic manipulations.