Problem 70 - 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. \(\lim _{x \rightarrow 2} \frac{x^{5}-32}{x-2}\)

Step-by-Step Solution

Answer

Question: Determine the limit of the following expression: \(\lim _{x \rightarrow 2} \frac{x^{5}-32}{x-2}\) Answer: 80

Step 1: Identify given values

The given formula is \(x^{n}-a^{n}=(x-a)\left(x^{n-1}+x^{n-2} a+x^{n-3} a^{2}+\cdots+x a^{n-2}+a^{n-1}\right)\) with \(n\) as a positive integer and \(a\) as a real number. The limit to be calculated is \(\lim _{x \rightarrow 2} \frac{x^{5}-32}{x-2}\). In this case, \(n = 5\) and \(a = 2\).

Step 2: Apply the factorization formula

Use the given factorization formula to subtract the given numerator expression: $$x^{5} - 2^{5} = (x-2)\left(x^{5-1}+x^{5-2}\cdot2+x^{5-3}\cdot2^{2}+\cdots+x\cdot2^{5-2}+2^{5-1}\right)$$ Now simplify the expression: $$x^5 - 32 = (x-2)(x^4 + 2x^3 + 4x^2 + 8x + 16)$$

Step 3: Calculate the limit

With the expression simplified, we can now substitute it into the limit: $$\lim _{x \rightarrow 2} \frac{x^5 - 32}{x-2} = \lim _{x \rightarrow 2} \frac{(x-2)(x^4 + 2x^3 + 4x^2 + 8x + 16)}{x-2}$$ We can cancel out the \((x-2)\) terms: $$\lim _{x \rightarrow 2} \frac{(x-2)(x^4 + 2x^3 + 4x^2 + 8x + 16)}{x-2} = \lim _{x \rightarrow 2} (x^4 + 2x^3 + 4x^2 + 8x + 16)$$ Now, substitute \(x=2\) into the remaining expression to find the limit: $$\lim _{x \rightarrow 2} (x^4 + 2x^3 + 4x^2 + 8x + 16) = 2^4 + 2(2^3) + 4(2^2) + 8(2) + 16$$ Simplify the expression to get the value of the limit: $$\lim _{x \rightarrow 2} (x^4 + 2x^3 + 4x^2 + 8x + 16) = 16 + 16 + 16 + 16 + 16 = \boxed{80}$$

Key Concepts

Factorization Formula

The factorization formula is an essential mathematical tool that simplifies the process of dealing with polynomial expressions, particularly when calculating limits in calculus. A standard form is \[x^{n} - a^{n} = (x - a)(x^{n-1} + x^{n-2}a + \ldots + xa^{n-2} + a^{n-1})\], where, \(n\) is a positive integer and \(a\) is a real number. This formula is crucial when a polynomial function has a common term that can be factored out. In such cases, factorization provides a straightforward pathway to simplifying the expression and facilitating easier computation of limits.

Importance of Factorization

By breaking down complex polynomials into more manageable pieces, factorization allows you to cancel out common factors in the numerator and denominator, which is often a key step in finding the limit of a function. In the context of our given exercise, factorization enables us to cancel out the \((x-2)\) term, simplifying the process of finding \(\lim _{x \rightarrow 2} \frac{x^{5} - 32}{x - 2}\).

Benefits in Limit Calculation

Utilizing the factorization formula is advantageous because it can often transform an indefinite form, such as \(\frac{0}{0}\), into a definite form and hence, allows the calculation of the limit without the need for complex rules such as L'Hôpital's rule.

Limits of Polynomial Functions

When calculating limits for polynomial functions, one often arrives at results directly by substituting the value to which \(x\) approaches into the polynomial, assuming the function is continuous and well-behaved at that point. For instance, with a simple polynomial like \(x^2\), calculating \(\lim _{x \rightarrow 3} x^2\) merely requires substituting 3 for \(x\), resulting in 9.

Evaluating Complex Polynomials

However, when evaluating more complex polynomials, especially those that produce indeterminate forms like \(\frac{0}{0}\) when directly substituted, factorization becomes vital. As seen in the textbook solution, the given polynomial \(x^5 - 32\) initially generates an indeterminate form. By employing the factorization formula, the limit of the polynomial function can be reformulated into one that no longer results in the troublesome \(0/0\) scenario.

Continuity and Polynomial Limits

It's essential to note that polynomial functions are continuous for all real numbers. This means that their limits at any point are equal to their values at that point if the function is defined there. Using this principle, we can evaluate polynomial function limits quite confidently, knowing the limit exists as long as we avoid indeterminate forms.

Limit Calculation

Calculating limits is a foundational component of calculus, as it deals with understanding the behavior of functions as they approach specific points. The calculation of limits often involves determining the value that a function approaches as the input approaches some point. In practice, limit calculation can range from straightforward direct substitution to more complex strategies involving factorization, algebraic manipulation, and the application of limit laws and theorems.

Strategies for Indeterminate Forms

If direct substitution leads to an indeterminate form, additional techniques, such as factorization or conjugate multiplication, may be required. In the textbook solution provided, we use factorization to simplify an indeterminate form and facilitate the computation of the limit. Once the factorizable part is removed, the limit calculation becomes straightforward, highlighting the power of algebraic manipulation in calculus.

Applying Limits to Real-world Problems

Understanding how to calculate limits is crucial not only for solving textbook exercises but also for applications in science, engineering, and economics where modeling change and understanding behavior near specific points is essential. Mastering the various methods of limit calculation prepares students to handle the wide array of problems they will encounter in more advanced studies and professional applications.