Problem 20 - Calculus and its applications

Question

The initial substitution of \(x=a\) yields the form \(0 / 0 .\) Look for ways to simplify the function algebraically, or use a table or graph to determine the limit. When necessary, state that the limit does not exist. $$ \lim _{x \rightarrow 5} \frac{x^{2}-25}{x-5} $$

Step-by-Step Solution

Answer

The limit is 10.

Step 1: Substitution Check

First, substitute \( x = 5 \) into the function \( \frac{x^2 - 25}{x - 5} \) to check if direct substitution gives a determinate value. Calculate: \( 5^2 - 25 = 0 \) and \( 5 - 5 = 0 \), resulting in \( \frac{0}{0} \). This is an indeterminate form, indicating that we need further analysis.

Step 2: Simplify the Expression

Recognize that the numerator \( x^2 - 25 \) is a difference of squares. Factor it as \( (x - 5)(x + 5) \). The expression then becomes: \( \frac{(x - 5)(x + 5)}{x - 5} \).

Step 3: Cancel Common Terms

Notice that \( x - 5 \) is a common term in the numerator and the denominator. Cancel out \( x - 5 \) from both the numerator and denominator for all \( x eq 5 \). The simplified expression is \( x + 5 \).

Step 4: Evaluate the Limit

With the simplified expression \( x + 5 \), substitute \( x = 5 \) to determine the limit. This results in \( 5 + 5 = 10 \). Thus, the limit is \( 10 \).

Key Concepts

Indeterminate Forms

Indeterminate forms are expressions where substitution into a limit results in an undefined or ambiguous result, such as \( \frac{0}{0} \). These often arise in calculus when both the numerator and the denominator of a rational function become zero as they approach a specific value. This can indicate that the function's behavior needs further exploration to determine the limit.

The expression \( \frac{x^2 - 25}{x - 5} \) becomes \( \frac{0}{0} \) when \( x = 5 \), which is a classic indeterminate form. Instead of concluding that the limit doesn't exist, we apply algebraic methods to investigate further. Simplifying the expression often reveals the function's limit as it approaches the specific point.

Factoring in Algebra

Factoring in algebra is a technique used to simplify expressions by expressing them as a product of their factors. This is especially handy in calculus when dealing with limits, as it can help eliminate indeterminate forms and simplify computations.

Consider the expression \( x^2 - 25 \). This can be factored using the difference of squares method, which states that any expression of the form \( a^2 - b^2 \) can be written as \( (a - b)(a + b) \). For \( x^2 - 25 \), think about 25 as \( 5^2 \), so the expression can be refactored as \( (x - 5)(x + 5) \).

With this factorization, the expression \( \frac{x^2 - 25}{x - 5} \) transforms into \( \frac{(x - 5)(x + 5)}{x - 5} \), allowing us to cancel common terms. This leaves us with \( x + 5 \) for \( x eq 5 \), simplifying the process of finding the limit.

Difference of Squares

The difference of squares is a special factoring technique used to simplify algebraic expressions. It's characterized by a binomial with the form \( a^2 - b^2 \), and it can be factored into \( (a - b)(a + b) \). Understanding how to apply this method is crucial for factoring certain types of polynomials effectively.

In our problem, \( x^2 - 25 \) fits the difference of squares pattern because \( x^2 \) and \( 25 \) can be viewed as perfect squares. By recognizing this, we can restructure the expression into \( (x - 5)(x + 5) \).

The insight gained from identifying the difference of squares is essential for solving limits involving polynomials. By simplifying \( \frac{(x - 5)(x + 5)}{x - 5} \), we can eliminate the common factor \( x - 5 \), making it possible to compute the limit of the remaining expression, \( x + 5 \), as \( x \) approaches 5.