Problem 14 - Calculus for Biology and Medicine

Question

In Problems 1-32, use a table or a graph to investigate each limit. $$ \lim _{x \rightarrow 3} \frac{x^{2}-9}{x+3} $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Understanding the Problem

The problem requires us to find the limit of the function \( f(x) = \frac{x^2 - 9}{x + 3} \) as \( x \) approaches 3. Direct substitution results in an indeterminate form \( \frac{0}{0} \), so other methods like factoring or simplifying the expression around \( x = 3 \) are needed.

Step 2: Simplifying the Expression

The expression \( x^2 - 9 \) can be factored as \((x - 3)(x + 3)\). Therefore, \( \frac{x^2 - 9}{x + 3} = \frac{(x - 3)(x + 3)}{x + 3} \). For \( x eq -3 \), this simplifies to \( x - 3 \).

Step 3: Reevaluating the Limit

The simplified expression \( f(x) = x - 3 \) is continuous for all \( x eq -3 \); thus, the limit as \( x \to 3 \) is \( f(3) = 3 - 3 \).

Step 4: Computing the Limit

Substituting \( x = 3 \) into the simplified expression \( x - 3 \), we have \( f(3) = 3 - 3 = 0 \). Hence, the limit is 0.

Key Concepts

Indeterminate Forms

In calculus, an indeterminate form arises when substituting numbers into a function yields an undefined expression. One classic example is the limit form \( \frac{0}{0} \). This often occurs in limit problems where direct substitution doesn't provide a valid result. In our exercise, when we try to evaluate \( \lim_{x \rightarrow 3} \frac{x^2 - 9}{x + 3} \), plugging in \( x = 3 \) directly results in \( \frac{0}{0} \).
To resolve such situations, alternative methods or simplifications need to be employed. Indeterminate forms require us to delve deeper into the function's algebraic properties to find a way to rewrite the function to remove the problematic zero expression. Addressing this through factoring, L'Hôpital's rule, or by deriving equivalent simplified forms becomes essential.

Factoring Polynomials

Factoring is an essential algebraic skill, especially in calculus, to simplify complex expressions and resolve indeterminate forms. In our exercise, we start with the function \( f(x) = \frac{x^2 - 9}{x + 3} \). This expression involves a difference of squares, \( x^2 - 9 \), which can be factored into \( (x - 3)(x + 3) \).

• This allows the expression \( \frac{x^2 - 9}{x + 3} \) to transform as \( \frac{(x - 3)(x + 3)}{x + 3} \).
• The \( x + 3 \) terms can be canceled out, provided \( x eq -3 \), leaving us with \( x - 3 \).

Factoring here is crucial because it simplifies the problem and is a common approach to handle quadratic and higher-order polynomial expressions in limits. Remember, the goal of factoring is to break down expressions into simpler, manageable pieces that can transform or eliminate initial indeterminate results whenever possible.

Continuous Functions

A continuous function is one where small changes in the input result in small changes in the output. For a function to be continuous at a point, the limit as \( x \) approaches the point must equal the function value at that point. This concept is pivotal in evaluating limits.
In our exercise, after simplification, the function \( f(x) = x - 3 \) emerges. This is a linear function, naturally continuous for all real numbers except at \( x = -3 \) where the original expression was undefined.

• For \( x = 3 \), substitution into the simplified function becomes straightforward: \( f(3) = 3 - 3 = 0 \).
• The continuity of \( f(x) = x - 3 \) ensures that the limit is far easier to handle, reinforcing the power of transforming the original expression.

Understanding continuity provides confidence when tackling limits because it guarantees the function behaves predictably around the point of interest.