Problem 71 - 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. \(\lim _{x \rightarrow 1} \frac{x^{6}-1}{x-1}\)

Step-by-Step Solution

Answer

Question: Determine the limit of the function \(\frac{x^6 - 1}{x - 1}\) as \(x\) approaches \(1\), using the given factorization formula. Answer: The limit of the function \(\frac{x^6 - 1}{x - 1}\) as \(x\) approaches \(1\) is 6.

Step 1: Identify the given formula and expressions in the problem

In this exercise, we are given the factorization formula: \(x^{n}-a^{n}=(x-a)\left(x^{n-1}+x^{n-2} a+x^{n-3} a^{2}+\cdots+x a^{n-2}+a^{n-1}\right)\) We have the expression \(\frac{x^{6}-1}{x-1}\) and we need to find the limit as \(x\) approaches \(1\). We can notice that the expression in the numerator resembles the left side of the given factorization formula with \(n = 6\) and \(a = 1\).

Step 2: Apply the factorization formula to the given expression

Using the factorization formula provided, when \(n = 6\) and \(a = 1\), we have: \(x^6 - 1 = (x - 1)\left(x^5 + x^4 \cdot1 + x^3 \cdot1^2+ \cdots + x\cdot1^4+1^5\right)\) So, \(x^6 - 1 = (x - 1)(x^5 + x^4 + x^3 + x^2 + x + 1)\)

Step 3: Simplify the given expression using the factorization

Substitute the factorized form for \(x^6 - 1\) into the given expression: \(\frac{x^6 - 1}{x - 1} = \frac{(x - 1)(x^5 + x^4 + x^3 + x^2 + x + 1)}{x - 1}\) Now we can cancel the common factors \((x - 1)\) in the numerator and denominator: \(\frac{(x - 1)(x^5 + x^4 + x^3 + x^2 + x + 1)}{x - 1} = x^5 + x^4 + x^3 + x^2 + x + 1\)

Step 4: Compute the limit

Now that our expression is simplified, we'll calculate the limit as \(x\) approaches \(1\): \(\lim_{x \rightarrow 1}(x^5 + x^4 + x^3 + x^2 + x + 1)\) Substitute \(x = 1\) into the simplified expression: \(1^5 + 1^4 + 1^3 + 1^2 + 1 + 1 = 1 + 1 + 1 + 1 + 1 + 1 = 6\) So, the limit is: \(\lim_{x \rightarrow 1} \frac{x^{6}-1}{x-1} = 6\)

Key Concepts

Factorization Formula

The factorization formula is a powerful algebraic tool that helps break down complex polynomial expressions into simpler, more manageable components. This formula takes the form \(x^{n}-a^{n}=(x-a)(x^{n-1}+x^{n-2} a+x^{n-3} a^{2}+\cdots+x a^{n-2}+a^{n-1})\) where \(n\) represents a positive integer, and \(a\) is a real number.

In the context of calculating limits, this formula can be instrumental in reducing expressions to simpler forms that are easier to handle. For instance, in the example provided, \(x^{6}-1\) can be seen as a difference of sixth powers with \(a\) being 1. By applying the factorization formula, we are able to convert this expression into a product of factors, thoroughly simplifying the initial polynomial.

Algebraic Simplification

Algebraic simplification involves manipulating and reorganizing algebraic expressions to make them shorter, clearer, and often to expose other properties that can be valuable in solving mathematical problems. Simplification often involves expanding products, combining like terms, and canceling common factors.

In our exercise, after applying the factorization formula, we obtained \(\frac{(x - 1)(x^5 + x^4 + x^3 + x^2 + x + 1)}{x - 1}\). The next step is simplifying by canceling the common \(x - 1\) term in the numerator and denominator, which is a crucial step to avoid indeterminate forms when we compute the limit.

Limit Computation

Limit computation is a fundamental concept in calculus involving finding the value that a function approaches as the input approaches some value. Limits are often used to determine the behavior of functions at points where they are not explicitly defined.

To compute \(\lim_{x \rightarrow a} f(x)\), we need to consider how function \(f(x)\) behaves as \(x\) gets arbitrarily close to \(a\). Algebraic simplification, as shown in our example, makes it easier to substitute the value directly into the function to compute the limit, as we do not deal with complex expressions that can make direct substitution impossible or arduous.

Indeterminate Form Resolution

In calculus, certain expressions cannot be evaluated in their initial form as they produce indeterminate forms such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These require special techniques to resolve.

Our example initially presents us with an expression that can lead to an indeterminate form as \(x\) approaches 1. By algebraically simplifying the expression after applying the factorization formula, we are able to resolve the indeterminate form. This enables us to substitute \(x = 1\) directly into the simplified expression and find that the limit is indeed 6 without the troublesome division by zero obstacle.