Problem 52 - Calculus Early Transcendentals

Question

Evaluate the following limits, where a and \(b\) are fixed real numbers. \(\lim _{x \rightarrow a} \frac{x^{3}-a^{3}}{x-a}\)

Step-by-Step Solution

Answer

Question: Evaluate the limit \(\lim_{x\rightarrow a} \frac{x^3 - a^3}{x-a}\). Answer: The limit is equal to \(3a^2\).

Step 1: Recognize the formula for the difference of cubes

The expression in the limit has a denominator of x-a and a numerator with the form of difference of cubes. The formula for the difference of cubes is given as: \[A^3 - B^3 = (A - B)(A^2 + AB + B^2)\] where A and B are any expressions, in our case, A=x and B=a.

Step 2: Apply the difference of cubes formula to the numerator

Using the difference of cubes formula, we can rewrite the numerator as: \[x^3 - a^3 = (x - a)(x^2 + ax + a^2)\]

Step 3: Simplify the expression in the limit

Now that the numerator is factored, we can cancel out the common factor (x-a) from the numerator and denominator. This simplifies the expression as follows: \[\frac{x^3 - a^3}{x-a} = \frac{(x-a)(x^2 + ax + a^2)}{x-a} = x^2 + ax + a^2\]

Step 4: Evaluate the limit

Now that the expression is simplified, we can evaluate the limit as x approaches a: \[\lim_{x\rightarrow a} (x^2 + ax + a^2) = a^2 + a(a) + a^2 = a^2 + a^2 + a^2 = 3a^2\] The limit is equal to \(3a^2\).

Key Concepts

Difference of Cubes

The concept of "difference of cubes" is essential in understanding how to factor certain polynomial expressions. Specifically, it pertains to expressions of the form \(A^3 - B^3\). This can be decomposed using the formula:

• \(A^3 - B^3 = (A - B)(A^2 + AB + B^2)\)

This formula provides a simple, structured way to break down a complex cubic polynomial into more manageable parts. In the case of our exercise, with \(x = A\) and \(a = B\), it helps to transform the cubic expression into a form that can be simplified more easily later.
Applying this formula converts the difference of two cubes into a product of a binomial and a trinomial, facilitating further operations, such as canceling terms in a quotient. Deconstructing a cubic expression like this is powerful in calculus, especially when evaluating limits, as it often allows for the simplification necessary to resolve indeterminate forms.

Factoring

Factoring is a critical algebraic tool in mathematics that involves rewriting expressions as products of simpler components. When handling limits featuring polynomials, factoring is often used to simplify complex expressions that otherwise would be challenging to evaluate directly.

In our problem, the difference of cubes formula allowed us to factor the numerator \(x^3 - a^3\) into \((x - a)(x^2 + ax + a^2)\). This factorization is crucial because it exposes common factors in the numerator and denominator that can be canceled.

• By canceling the \((x-a)\) terms, the expression becomes \(x^2 + ax + a^2\).

Factoring simplifies the complex fraction, making it possible to evaluate the limit without encountering division by zero issues. In calculus, this technique often enables smoother evaluation of limits.

Evaluating Limits

Evaluating limits involves determining the value that a function approaches as the input approaches a particular point. In calculus, limits are vital for understanding behavior around points of discontinuity or indeterminate forms. In our example, the initial direct substitution into \(\lim_{x \rightarrow a} \frac{x^3-a^3}{x-a}\) could lead to division by zero, indicative of an indeterminate form.

The goal is to simplify the expression so that this doesn't occur, allowing for straightforward evaluation. Through factoring and cancelation, the limit became:

• \(\lim_{x\rightarrow a} (x^2 + ax + a^2)\)

With this expression, direct substitution is possible: replacing \(x\) with \(a\) results in \(3a^2\). This calculation avoids the original indeterminate form and provides the correct limit. Evaluating limits accurately involves both algebraic manipulation and understanding how functions behave as inputs change.