Problem 16 - Contemporary Precalculus

Question

Find the limit if it exists. If the limit does not exist, explain why. $$\lim _{x \rightarrow-2} \frac{x^{2}+5 x+6}{x^{2}-x-6}$$

Step-by-Step Solution

Answer

Answer: The limit of the function as x approaches -2 is $\frac{1}{-5}$.

Step 1: 1. Factor the numerator and denominator

To find the limit of the function, we first need to factor the numerator and denominator. The given function is: $$\frac{x^{2}+5x+6}{x^{2}-x-6}$$ The numerator can be factored as \((x+2)(x+3)\). The denominator can be factored as (x+2)(x-3). So we have: $$\frac{(x+2)(x+3)}{(x+2)(x-3)}$$

Step 2: 2. Simplify the expression

As we can see, there is a common factor in the numerator and the denominator – (x+2). We can simplify the expression by canceling out this common factor: $$\frac{(x+2)(x+3)}{(x+2)(x-3)} = \frac{x+3}{x-3}$$

Step 3: 3. Evaluate the limit

Now, let's evaluate the limit as x approaches -2: $$\lim_{x \rightarrow -2} \frac{x+3}{x-3} = \frac{-2+3}{-2-3} = \frac{1}{-5}$$

Step 4: 4. Check if the limit exists

Since the function is now a simplified rational function with no indeterminacy at x = -2, the limit exists and is equal to its current value: $$\lim_{x \rightarrow -2} \frac{x^{2}+5x+6}{x^{2}-x-6} = \lim_{x \rightarrow -2} \frac{x+3}{x-3} = \frac{1}{-5}$$

Key Concepts

Factoring Polynomials

When dealing with limits in calculus, especially those involving rational functions, factoring polynomials is a crucial skill. Polynomials can often be expressed in a factored form, which makes complex expressions easier to handle. Consider a polynomial like \(x^2 + 5x + 6\). To factor it, you need to find two numbers that multiply to 6 (the constant term) and add up to 5 (the coefficient of the linear term).

For \(x^2 + 5x + 6\):

• The two numbers are 2 and 3, since \(2 \times 3 = 6\) and \(2 + 3 = 5\).
• This results in the factors \((x+2)\) and \((x+3)\).

This kind of factoring helps simplify the rational function into a form that is easier to analyze or compute, particularly when exploring limits.

Rational Functions

Rational functions are expressions formed by dividing two polynomials. They can be written in the form \(\frac{P(x)}{Q(x)}\), where both \(P(x)\) and \(Q(x)\) are polynomials. These functions often need simplification, especially when evaluating limits, because they might have terms that appear in both the numerator and denominator.

In the exercise, the rational function was \(\frac{x^2 + 5x + 6}{x^2 - x - 6}\), which can be simplified by factoring:

• This becomes \(\frac{(x+2)(x+3)}{(x+2)(x-3)}\).
• Cancel out the common factor \((x+2)\) to get \(\frac{x+3}{x-3}\).

Such simplification is vital as it allows us to focus on the remaining factors, making it easier to determine the behavior of the function as \(x\) approaches a particular value.

Indeterminate Forms

In calculus, when evaluating limits, you may encounter indeterminate forms. These are expressions that do not initially provide enough information to determine the limit. One of the common indeterminate forms is \(\frac{0}{0}\), which often arises when trying to evaluate limits directly.

When \(x\) approaches -2 in our example, substituting directly into the unsimplified function initially gives \(\frac{0}{0}\). This signals an indeterminate form and suggests the need for algebraic manipulation such as factoring.

• By factoring and simplifying, you can evaluate the limit of the simplified form \(\frac{x+3}{x-3}\) without encountering any indeterminacy.
• This results in a clear and computable limit, as demonstrated: \(\frac{1}{-5}\) when \(x\) approaches -2.

Understanding and resolving indeterminate forms is key to successfully finding limits in calculus problems.