Problem 41 - Calculus: Early Transcendentals

Question

Determine the infinite limit. \( \displaystyle \lim_{x \to 2^+}\frac{x^2 - 2x -8}{x^2 -5x +6} \)

Step-by-Step Solution

Answer

The limit is \(-\infty\).

Step 1: Investigate the Limit Approach

We are asked to find the limit of a function as x approaches 2 from the right (indicated by the '+'). This means we will be examining the values of \( x \) that are slightly greater than 2.

Step 2: Identify the Form of the Function

The given function is \( \frac{x^2 - 2x - 8}{x^2 - 5x + 6} \). We need to factor both the numerator and the denominator to see if the limit can be simplified.

Step 3: Factor the Numerator

Factor \( x^2 - 2x - 8 \). We look for two numbers that multiply to -8 and add to -2. These numbers are -4 and +2, so it factors into \((x-4)(x+2)\).

Step 4: Factor the Denominator

Factor \( x^2 - 5x + 6 \). We look for two numbers that multiply to 6 and add to -5. These numbers are -3 and -2, so it factors into \((x-3)(x-2)\).

Step 5: Simplify the Function

Substitute the factored forms into the original expression: \( \frac{(x-4)(x+2)}{(x-3)(x-2)} \). Notice that the denominator becomes zero if \( x = 2 \), indicating a potential vertical asymptote, and cannot be simplified further.

Step 6: Determine Behavior Around Vertical Asymptote

As \( x \to 2^+ \), \( (x-2) \) in the denominator is a small positive number approaching zero from the positive side. Therefore, the entire function approaches \(-\infty\) given that \( (x-4) \approx -2 \) and \( (x-3) \approx -1 \) at \( x=2^+ \).

Step 7: Conclusion of the Limit

We've determined that the limit as \( x \to 2^+ \) results in an infinite decrease. Therefore, the infinite limit is \(-\infty\).

Key Concepts

Factoring Polynomials

Factoring polynomials is a key algebraic skill that helps simplify expressions and solve equations. When you see a polynomial, the goal is to break it down into simpler polynomials, or "factors," that multiply together to make the original expression. This process helps in determining the behavior of a function, especially when analyzing limits or solving equations.

Here's how you can factor a quadratic polynomial like the ones in our problem:

• Look for two numbers that multiply to the constant term (the last number) and add to the coefficient of the linear term (the middle number).
• For the numerator, we factored \(x^2 - 2x - 8\), finding numbers -4 and +2 that satisfy the conditions. This results in \((x-4)(x+2)\).
• For the denominator, \(x^2 - 5x + 6\) breaks down into \((x-3)(x-2)\) using numbers -3 and -2.

The ability to smoothly factor polynomials allows for simplification of expressions and greater insight into the function's behavior.

Vertical Asymptotes

Vertical asymptotes are an important concept in calculus and understanding them helps reveal crucial behaviors of functions. They occur when the denominator of a rational function approaches zero, causing the function to tend towards infinity or negative infinity. A vertical asymptote indicates a point where a function is undefined and typically leads to a dramatic change in the function's graph.

To identify vertical asymptotes, examine the denominator of a fraction and determine values of \(x\) that make it zero.

• In the exercise, the denominator \((x-3)(x-2)\) reveals that at \(x=2\) and \(x=3\), the function is undefined.
• The vertical asymptote at \(x=2\) denotes that as \(x\) approaches 2 from the right \((x \to 2^+)\), the function becomes unbounded, thus going to \(-\infty\).

Vertical asymptotes don't always imply limits of infinity; assessing each case is essential.

Limit Behavior Analysis

Limit behavior analysis involves examining how a function behaves as it approaches a certain value. In the context of infinite limits, we explore how the function value increases or decreases as \(x\) approaches a specific point, often indicating the presence of vertical asymptotes.

In our example, the goal is to analyze how the function \(\frac{(x-4)(x+2)}{(x-3)(x-2)}\) behaves as \(x\) approaches 2 from the right.

• We simplify the function after factoring and notice that near \(x=2^+\), \(x-2\) is a small positive number close to zero.
• The components \((x-4) \approx -2\) and \((x-3) \approx -1\) contribute to an overall negative value, moving towards \(-\infty\).
• Understanding these behaviors allows one to correctly predict that the limit of the function tends to \(-\infty\) as \(x \to 2^+\).

Grasping limit behavior allows for deeper insights into the nature of functions and predicts their tendencies around certain points.