Problem 13 - Calculus for Biology and Medicine

Question

In Problems 1-32, use a table or a graph to investigate each limit. $$ \lim _{x \rightarrow 3} \frac{x^{2}-16}{x-4} $$

Step-by-Step Solution

Answer

The limit is 7.

Step 1: Understand the Problem

We are given the expression \( \lim _{x \rightarrow 3} \frac{x^{2}-16}{x-4} \). Our task is to investigate this limit using a table or a graph. Since the expression has a denominator that becomes zero when \( x \) approaches 4, we suspect there might be a removable discontinuity.

Step 2: Simplify the Expression

Rewrite the numerator to see if the limit can be simplified. The numerator \( x^2 - 16 \) can be factored using the difference of squares: \( x^2 - 16 = (x - 4)(x + 4) \). So the expression becomes: \( \frac{(x - 4)(x + 4)}{x - 4} \).

Step 3: Cancel Common Factors

Notice that \( x - 4 \) is present in both the numerator and the denominator. We can cancel the \( x - 4 \) terms, as these are common factors, simplifying to \( x + 4 \).

Step 4: Evaluate the Simplified Expression

After simplification, the expression becomes \( x + 4 \). Now, evaluate the limit as \( x \) approaches 3: \( \lim _{x \rightarrow 3} (x + 4) = 3 + 4 = 7 \). The limit does not depend on the graph or table due to the algebraic simplification.

Step 5: Confirm with Graph or Table

Although we found the limit algebraically, we can confirm it by observing a graph of \( y = \frac{x^2 - 16}{x-4} \) or a table of values as \( x \) approaches 3. By plugging values close to 3 (e.g., 2.9, 3.1) into the unsimplified expression, and noticing that they approach 7, it confirms our limit.

Key Concepts

Removable Discontinuity

A removable discontinuity occurs when a function is not defined at a certain point, yet could be defined at that point by filling in a single value. When we calculate limits, we often encounter expressions that become undefined at some points because of zero denominators.

In the exercise, the expression \( \lim _{x \rightarrow 3} \frac{x^{2}-16}{x-4} \) is undefined at \( x = 4 \) because the denominator becomes zero there. However, after simplifying the expression, it's possible to redefine the function without the point causing the discontinuity.

• Identify the problematic point: Check where the denominator is zero.
• Determine if the numerator can be factored in a way that cancels with the denominator.
• Simplify the expression by cancelling out common factors to "remove" the discontinuity.

By cancelling out the \( x - 4 \) term in both the numerator and denominator, \( \frac{(x - 4)(x + 4)}{x - 4} \) simplifies to \( x + 4 \), allowing the calculation of the limit. Removing the discontinuity helps understand the function's behavior at the point of concern.

Algebraic Simplification

Algebraic simplification plays a crucial role in solving calculus problems involving limits. When confronting complex algebraic expressions, simplification can reveal hidden trends and make limit calculations more straightforward.

The simplification process involves:

• Rewriting expressions using algebraic identities, such as factoring and expanding.
• Cancelling out common terms between the numerator and denominator.
• Testing if further simplification is possible based on algebraic rules and practices.

In our exercise, we identified that \( x^2 - 16 \) can be factored using the difference of squares into \( (x - 4)(x + 4) \). This algebraic step breaks down the problem into more manageable parts, allowing us to cancel out the common factor and evaluate the expression as \( x + 4 \). Simplification not only clarifies the expression but also directly impacts results, as seen in determining the limit.

Difference of Squares

The difference of squares is a fundamental concept in algebra and is often used to simplify quadratic expressions. It refers to expressions of the form \( a^2 - b^2 \), which can be factored into \( (a - b)(a + b) \).

This factorization is useful for limit problems that involve expressions leading to zero denominators. Here's how it works:

• Identify expressions that meet the difference of squares format.
• Write them as a product of two binomials, where each factors differ by a sum and difference.
• Use the factored form to simplify expressions where terms cancel in the limit calculation.

In the given problem, \( x^2 - 16 \) matches the difference of squares format, because \( 16 = 4^2 \). The transformation from \( x^2 - 16 \) to \( (x - 4)(x + 4) \) reveals the flexibility of expressions that appear complex at first glance. This concept assists in uncovering concluded values in limits and removes discontinuities effectively.