Problem 23 - APEX Calculus

Question

Evaluate the limit: \(\lim _{x \rightarrow-4} \frac{x^{2}-16}{x^{2}-4 x-32}\)

Step-by-Step Solution

Answer

The limit is \( \frac{2}{3} \).

Step 1: Identify the Structure

The goal is to find the limit as \( x \) approaches -4 of the rational function \( \frac{x^{2}-16}{x^{2}-4x-32} \). Recognize that direct substitution of \( x = -4 \) results in an indeterminate form \( \frac{0}{0} \).

Step 2: Factorize the Numerator

The numerator \( x^2 - 16 \) is a difference of squares and can be factored as \( (x - 4)(x + 4) \). So, \( x^2 - 16 = (x - 4)(x + 4) \).

Step 3: Factorize the Denominator

The denominator \( x^2 - 4x - 32 \) can be factorized by looking for two numbers that multiply to -32 and add to -4. These numbers are -8 and 4. So, \( x^2 - 4x - 32 = (x - 8)(x + 4) \).

Step 4: Simplify the Expression

Now that both the numerator and the denominator are factored, the expression is: \( \frac{(x - 4)(x + 4)}{(x - 8)(x + 4)} \). Cancel out the common factor \( (x + 4) \) from the numerator and the denominator, simplifying the expression to \( \frac{x - 4}{x - 8} \).

Step 5: Evaluate the Simplified Limit

With the expression simplified, evaluate the limit \( \lim_{x \to -4} \frac{x - 4}{x - 8} \) by substitution: \( \frac{-4 - 4}{-4 - 8} = \frac{-8}{-12} = \frac{2}{3} \).

Key Concepts

Indeterminate Forms

When dealing with limits in calculus, an indeterminate form arises when direct substitution into a function results in expressions like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), or others such as \( 0 \cdot \infty \). These forms don't provide a clear answer about the behavior of the function near that point.
To handle them, it's critical to use techniques such as algebraic manipulation to simplify the function or apply L'Hopital's Rule where applicable. In this exercise, substituting \( x = -4 \) directly into \( \frac{x^2 - 16}{x^2 - 4x - 32} \) gives \( \frac{0}{0} \), an indeterminate form, prompting us to find a way to simplify it further.
This simplification often involves factoring polynomials, as shown in our solution. By reducing the expression, we can resolve the indeterminate form and calculate the limit confidently.

Factoring Polynomials

Factoring is a method used to simplify polynomial expressions, making them easier to work with, especially when solving limits and other algebraic problems. The process involves breaking down a polynomial into simpler 'factor' elements whose product is the original polynomial.
In the numerator \( x^2 - 16 \), we recognize it as a "difference of squares," a special pattern that allows us to factor it into \( (x - 4)(x + 4) \).

For the denominator \( x^2 - 4x - 32 \), we need to find values that multiply to -32 and sum to -4. Through this, we find that the required values are -8 and 4, giving us the factored form \( (x - 8)(x + 4) \).

This factoring simplification helps in canceling out terms, simplifying the function to help eliminate indeterminate forms.

Rational Functions

Rational functions are quotients of two polynomials. They are written in the form \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials, and \( Q(x) eq 0 \). These functions can appear complicated, but understanding their behavior is key when determining limits.
In the exercise, \( \frac{x^2 - 16}{x^2 - 4x - 32} \) is a rational function. When it is factorized and simplified by canceling common factors, it shows a clearer picture of how the function behaves near a particular point. No longer indeterminate, it becomes easier to determine the limit through direct substitution.
This simplification process not only aids in solving limits but is fundamental in calculus, offering ways to graph functions, find asymptotes, or analyze continuity and other characteristics of rational functions.