Problem 73 - Calculus Early Transcendentals

Question

Calculate the following limits using the factorization formula \(x^{n}-a^{n}=(x-a)\left(x^{n-1}+x^{n-2} a+x^{n-3} a^{2}+\cdots+x a^{n-2}+a^{n-1}\right)\) where \(n\) is a positive integer and a is a real number. \(\lim _{x \rightarrow a} \frac{x^{5}-a^{5}}{x-a}\)

Step-by-Step Solution

Answer

Question: Determine the limit as x approaches a for the function f(x) = (x^5 - a^5) / (x - a), using the factorization formula x^n - a^n = (x - a)(x^(n-1) + x^(n-2)a + ... + xa^(n-2) + a^(n-1)). Answer: The limit as x approaches a for the function f(x) = (x^5 - a^5) / (x - a) is equal to 5a^4.

Step 1: Analyzing the problem

We have to calculate the limit \(\lim _{x \rightarrow a} \frac{x^{5}-a^{5}}{x-a}\). We can see that this expression is similar to the factorization formula given in the problem, as it involves the subtraction of two terms with the same exponent. So, we can use the factorization formula to simplify this expression.

Step 2: Applying the factorization formula

Using the factorization formula for \(n = 5\), we can write \(x^5 - a^5\) as: \(x^5 - a^5 = (x-a)\left(x^4 + x^3a + x^2a^2 + xa^3 + a^4\right)\) Now, we can substitute this expression for \(x^5 - a^5\) in the limit: \(\lim _{x \rightarrow a} \frac{x^{5}-a^{5}}{x-a} = \lim _{x \rightarrow a} \frac{(x-a)\left(x^4 + x^3a + x^2a^2 + xa^3 + a^4\right)}{x-a}\)

Step 3: Simplifying the expression

We can now simplify the expression by canceling out the (x-a) terms: \(\lim _{x \rightarrow a} \frac{(x-a)\left(x^4 + x^3a + x^2a^2 + xa^3 + a^4\right)}{x-a} = \lim _{x \rightarrow a} \left(x^4 + x^3a + x^2a^2 + xa^3 + a^4\right)\)

Step 4: Finding the limit

Now, we can find the limit as \(x \rightarrow a\): \(\lim _{x \rightarrow a} \left(x^4 + x^3a + x^2a^2 + xa^3 + a^4\right) = a^4 + a^3a + a^2a^2 + aa^3 + a^4\)

Step 5: Simplifying the final result

Lastly, we can simplify the expression: \(a^4 + a^3a + a^2a^2 + aa^3 + a^4 = a^4 + a^4 + a^4 + a^4 + a^4 = 5a^4\) Therefore, the result is: \(\lim _{x \rightarrow a} \frac{x^{5}-a^{5}}{x-a} = 5a^4\)

Key Concepts

Factorization Formula

Understanding the factorization formula is crucial for solving many problems in calculus, particularly when dealing with limits involving indeterminate forms such as \(0/0\). The formula \(x^n - a^n = (x - a)(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \cdots + xa^{n-2} + a^{n-1})\) elegantly simplifies expressions where a term is raised to a power and then subtracted by another term raised to the same power.

In the given exercise, the expression \(x^5 - a^5\) is factored using the formula for \(n = 5\), resulting in a product of \(x - a\) and a sum of terms where the powers of \(x\) decrement while the powers of \(a\) increment. This factorization leads to the cancellation of common terms, thus resolving the indeterminate nature of the original limit by revealing the continuous behavior of the function as \(x\) approaches \(a\).

Indeterminate Forms

In calculus, indeterminate forms are expressions that do not have a well-defined limit as they approach a certain point. The most common indeterminate form is \(0/0\), but there are others, such as \(\infty/\infty\), \(0 \cdot \infty\), and \(\infty - \infty\).

When we confront an indeterminate form while calculating limits, it usually indicates that additional work is needed to find a determinate form that can be evaluated. Techniques to resolve these forms include factoring, conjugation, and L'Hôpital's Rule. In our exercise, the indeterminate form \(0/0\) arises when \(x = a\), and it's resolved by factoring the numerator and canceling out the common \(x - a\) term, clearing the path to find a finite limit.

Algebraic Simplification

Once an expression is factored, we engage in algebraic simplification to reduce it to its simplest form. Simplifying expressions is not only crucial for finding limits but also for making complicated problems more manageable.

After canceling common factors, as seen in the step-by-step solution for our exercise, we end up with an expression that is less complex and can be directly evaluated by substitution. In our exercise, simplifying the expression after canceling the \(x - a\) term involves substituting \(x\) with \(a\) and adding up the like terms to finally arrive at the limit. The algebraic simplification uncovers the value \(5a^4\), which is the limit of the original expression as \(x\) approaches \(a\).