Problem 13

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

Verified
Answer
The limit is 8.
1Step 1: Analyze the Function at the Limit Point
First, we need to analyze the expression \( \frac{x^2 - 16}{x-4} \) as \( x \) approaches 4. Substitute \( x = 4 \) into the expression to determine if it leads to an undefined form. We get \( \frac{4^2 - 16}{4-4} = \frac{0}{0} \), which is an indeterminate form.
2Step 2: Simplify the Expression
To simplify the expression, we factor the numerator. Notice that \( x^2 - 16 \) is a difference of squares, which can be factored as \((x-4)(x+4)\). So, the expression becomes: \( \frac{(x-4)(x+4)}{x-4} \).
3Step 3: Cancel Common Factors
After factoring, we see the common factor \( x-4 \) in both the numerator and the denominator. Cancel these common factors, leading to the simplified expression \( x + 4 \). Note that this simplification is valid as long as \( x eq 4 \).
4Step 4: Evaluate the Limit of the Simplified Expression
Now that the expression is simplified to \( x + 4 \), calculate the limit as \( x \) approaches 4: \( \lim_{x \to 4} (x + 4) \). Substituting \( x = 4 \), we find \( 4 + 4 = 8 \). Thus, \( \lim_{x \rightarrow 4} \frac{x^{2}-16}{x-4} = 8 \).

Key Concepts

Indeterminate FormsSimplifying ExpressionsFactoring Polynomials
Indeterminate Forms
In calculus, an indeterminate form arises when substituting a limit value results in expressions that are unclear without further analysis. When you encounter situations like \( \frac{0}{0} \), it's called an indeterminate form. This occurs in our problem when substituting \( x = 4 \) into \( \frac{x^2 - 16}{x-4} \), giving us \( \frac{0}{0} \).

To resolve an indeterminate form, we need to take further steps, such as simplifying the expression or using techniques like L'Hôpital's Rule, if applicable. In our particular exercise, the form becomes clear through simplification efforts. But why is it important? Because it helps us find the actual value of a limit or confirm if a limit doesn't exist. Recognizing and addressing indeterminate forms is a vital skill in calculus, helping you understand the behavior of complex functions near certain points.
Simplifying Expressions
Simplifying expressions is the process of reducing complexity in mathematical equations. This is crucial in solving calculus limits, especially when you encounter indeterminate forms. In our example, the original expression \( \frac{x^2 - 16}{x-4} \) becomes undefined at \( x=4 \).

To simplify, you take advantage of algebraic manipulations, such as breaking down the numerator \( x^2 - 16 \). Factoring it reveals the difference of squares structure, making the expression easier to handle. Simplification not only provides a clearer path to finding limits but also enhances understanding by stripping away unnecessary components.
  • Reduces potential calculation errors in lengthy expressions.
  • Opens paths to alternative solving techniques, like substitution or graphical analysis.
Successful simplification can turn an indeterminate form into a straightforward evaluation, ensuring the right limit is determined.
Factoring Polynomials
Factoring polynomials is a method used to break down a polynomial into simpler 'factors' that, when multiplied together, give back the original polynomial. In the context of limits, like in our exercise, factoring is essential. It's because identifying common factors allows us to simplify the expression effectively.

The polynomial in our problem is \( x^2 - 16 \), a classic 'difference of squares'. Recognizable by the format \( a^2 - b^2 \), it can be factored into \((x-4)(x+4)\). This step is crucial for eliminating the part of the expression that causes the indeterminate form.
  • It reveals hidden simplifications in complex algebraic expressions.
  • Enables us to cancel terms, clearing indeterminate forms in rational functions.
Through factoring, you can better analyze the behavior of the function near specific points and ensure accurate computations. Understanding this concept aids in smoother problem-solving across calculus topics.