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Magazine Chapter 1: Problem 29
Evaluate the given limit. $$ \lim _{x \rightarrow 6} \frac{x^{2}-4 x-12}{x^{2}-13 x+42} $$
Short Answer
Expert verified
The limit evaluates to \(-8\).
Step by step solution
01
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.
02
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: \((x - 6)(x + 2)\).
03
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: \((x - 6)(x - 7)\).
04
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: \[ \frac{x + 2}{x - 7} \].
05
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, \[ \lim _{x \rightarrow 6} \frac{x^{2}-4 x-12}{x^{2}-13 x+42} = -8 \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
Techniques such as factoring, algebraic manipulation, or applying L'H么pital's Rule might resolve and simplify such expressions, helping them to be evaluated at a certain limit.
- 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.
Techniques such as factoring, algebraic manipulation, or applying L'H么pital'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 \( b \)
- Factors of \( -12 \) that add up to \( -4 \) are \(-6\) and \(2\).
- Therefore, we factor it as \( (x - 6)(x + 2) \).
- Here, factors of \( 42 \) that add up to \( -13 \) are \(-6\) and \(-7\).
- Thus, it factors to \( (x - 6)(x - 7) \).
Simplifying Expressions
After factoring both the numerator and the denominator in the given problem, simplifying the expression becomes much easier:Having factored forms:
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.
- Numerator: \((x - 6)(x + 2)\)
- Denominator: \((x - 6)(x - 7)\)
- Both contain the factor \( (x - 6) \).
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.
