Problem 21 - A Graphical Approach to Precalculus with Limits

Question

Determine each limit, if it exists. $$\lim _{x \rightarrow-1} \frac{2 x+3}{3 x+4}$$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Understand the Limit Expression

The problem is to find the limit \( \lim_{x \to -1} \frac{2x + 3}{3x + 4} \). Simplifying this rational function can help avoid division by zero when \( x \to -1 \).

Step 2: Direct Substitution Check

First, try substituting \( x = -1 \) directly into the function \( \frac{2x + 3}{3x + 4} \). Substitute and get: \( \frac{2(-1) + 3}{3(-1) + 4} \).

Step 3: Compute the Numerator

Calculate the numerator: \( 2(-1) + 3 = -2 + 3 = 1 \).

Step 4: Compute the Denominator

Calculate the denominator: \( 3(-1) + 4 = -3 + 4 = 1 \).

Step 5: Substitute and Simplify

After substituting and simplifying, the expression becomes \( \frac{1}{1} = 1 \).

Step 6: Conclusion

Since the direct substitution does not result in an indeterminate form, the limit exists and is \( 1 \).

Key Concepts

Direct Substitution

When evaluating a limit using direct substitution, you simply plug the value to which the variable is approaching directly into the function. This is often the quickest method for finding limits, provided the function is continuous at the point you're investigating. In the given problem, we're looking at a limit as \( x \) approaches \(-1\).

• First, substitute \( x = -1 \) in place of \( x \) in the function \( \frac{2x + 3}{3x + 4} \).
• Calculate \( \frac{2(-1) + 3}{3(-1) + 4} \).

If this substitution results in a valid numerical value, meaning no division by zero or other undefined behavior occurs, the process has successfully determined the limit. Direct substitution shows that the expression evaluates to \( 1 \), as it does in this exercise.

Rational Functions

Rational functions are formed by the quotient of two polynomials. Understanding how they work is crucial in limit computation. In this exercise, we have the rational function \( \frac{2x + 3}{3x + 4} \), which is a simple expression where:

• The numerator is \( 2x + 3 \).
• The denominator is \( 3x + 4 \).

Key aspects of rational functions include their potential to have vertical asymptotes, holes, or removable discontinuities depending on the behavior of the denominator. We avoid these issues by performing a direct substitution, since the denominator does not become zero as \( x \to -1 \), meaning the function is continuous at \( x = -1 \). Understanding these properties helps us make quick decisions about whether direct substitution will succeed.

Numerator and Denominator Evaluation

Evaluating both the numerator and the denominator separately can provide insights into the behavior of the entire rational function. This step involves performing basic arithmetic operations:

• For the numerator \( 2x + 3 \), substituting \( x = -1 \) yields \( 2(-1) + 3 = 1 \).
• For the denominator \( 3x + 4 \), substituting \( x = -1 \) results in \( 3(-1) + 4 = 1 \).

Neither calculation results in a zero or undefined state, so placing these values back into the rational function confirms the limit. Recognizing the need for these evaluations is key, especially in cases where the denominator may influence the limit's existence.

Limits of Functions

The concept of limits is foundational in calculus. The limit examines the behavior of a function as the input approaches a particular value. It does not necessarily indicate the function's behavior exactly at that point, but rather its tendency.

• In this case, we are determining how the function \( \frac{2x + 3}{3x + 4} \) behaves as \( x \) gets very close to \(-1\).
• A successful substitution with no indeterminate forms implies the limit exists.

Since our calculations show a result after substitution, it tells us that the function is "behaving" predictably near \(-1\). Thus, we confidently state the limit is \( 1 \). This illustrates the elegant concept of limits smoothing out potentially complex behavior at specific points.