Problem 29 - APEX Calculus

Question

Evaluate the given limit. $$ \lim _{x \rightarrow 6} \frac{x^{2}-4 x-12}{x^{2}-13 x+42} $$

Step-by-Step Solution

Answer

The limit evaluates to \(-8\).

Step 1: Substitute and Evaluate

First, substitute the value of the limit, 6, into the expression to check if the formula becomes indeterminate. \[ f(x) = \frac{x^2 - 4x - 12}{x^2 - 13x + 42} \] When \( x = 6 \), \( f(6) = \frac{6^2 - 4\cdot6 - 12}{6^2 - 13\cdot6 + 42} = \frac{36 - 24 - 12}{36 - 78 + 42} = \frac{0}{0} \). This is an indeterminate form, meaning we need to factor the expression to simplify it.

Step 2: Factor the Numerator

Factor the quadratic expression in the numerator \(x^2 - 4x - 12\). Look for two numbers that multiply to \(-12\) and add to \(-4\). The numbers \(-6\) and \(2\) work. Thus, the factorization is: \((x - 6)(x + 2)\).

Step 3: Factor the Denominator

Factor the quadratic expression in the denominator \(x^2 - 13x + 42\). Look for two numbers that multiply to \(42\) and add to \(-13\). The numbers \(-6\) and \(-7\) work. Thus, the factorization is: \((x - 6)(x - 7)\).

Step 4: Simplify by Canceling Common Factors

Now that you have factored both the numerator and the denominator, the expression is: \[ \frac{(x - 6)(x + 2)}{(x - 6)(x - 7)} \] Cancel the common factor \((x - 6)\) from both the numerator and the denominator to get: \[ \frac{x + 2}{x - 7} \].

Step 5: Evaluate the Simplified Limit

Now substitute \(x = 6\) into the simplified expression: \[ \frac{x + 2}{x - 7} = \frac{6 + 2}{6 - 7} = \frac{8}{-1} = -8 \]. Thus, \[ \lim _{x \rightarrow 6} \frac{x^{2}-4 x-12}{x^{2}-13 x+42} = -8 \].

Key Concepts

Indeterminate Forms

In calculus, an indeterminate form is a mathematical expression that becomes ambiguous or undefined when direct substitution is applied to it. Consider the expression from the original problem, when we tried substituting \( x = 6 \) directly:

• In the numerator, \( f(6) = 36 - 24 - 12 = 0 \)
• In the denominator, \( f(6) = 36 - 78 + 42 = 0 \)
• This results in \( \frac{0}{0} \), which is undefined.

0 divided by 0 is specifically known as an indeterminate form. This is not just a numerical undefined situation, it indicates that further analysis is needed to evaluate the limit.
Techniques such as factoring, algebraic manipulation, or applying L'Hôpital's Rule might resolve and simplify such expressions, helping them to be evaluated at a certain limit.

Factoring Quadratics

Factoring quadratics is a crucial skill while dealing with algebraic expressions, especially when tackling indeterminate forms in limits. Here's how it is applied:When you have a quadratic in the form of \( ax^2 + bx + c \), the goal of factoring is to express it as \((px + q)(rx + s)\). To do so, look for pairs of factors:

• Find two numbers whose product is equal to \( a \cdot c \)
• These same two numbers should add up to \( b \)

For example, let's examine the numerator \( x^2 - 4x - 12 \):

• Factors of \( -12 \) that add up to \( -4 \) are \(-6\) and \(2\).
• Therefore, we factor it as \( (x - 6)(x + 2) \).

Now consider the denominator \( x^2 - 13x + 42 \):

• Here, factors of \( 42 \) that add up to \( -13 \) are \(-6\) and \(-7\).
• Thus, it factors to \( (x - 6)(x - 7) \).

This process reveals common factors between the numerator and denominator, simplifying indeterminate forms.

Simplifying Expressions

After factoring both the numerator and the denominator in the given problem, simplifying the expression becomes much easier:Having factored forms:

• Numerator: \((x - 6)(x + 2)\)
• Denominator: \((x - 6)(x - 7)\)

Identify and cancel the common terms present in both. In this case:

• Both contain the factor \( (x - 6) \).

By canceling these factors, you simplify the expression to \( \frac{x + 2}{x - 7} \).
This step of canceling allows the limit to be evaluated without running into the indeterminate form while substituting back:When we plug \( x = 6 \) into the simplified function, it gives \( \frac{6 + 2}{6 - 7} = \frac{8}{-1} = -8 \).The simplification of expressions by canceling is vital for finding proper solutions without undefined outcomes.