Problem 20 - Precalculus: Mathematics for Calculus

Question

Finding Limits Evaluate the limit if it exists. $$\lim _{x \rightarrow-4} \frac{x^{2}+5 x+4}{x^{2}+3 x-4}$$

Step-by-Step Solution

Answer

The limit is \( \frac{3}{5} \).

Step 1: Identify the Limit Function

We are tasked with evaluating the limit \( \lim_{x \rightarrow -4} \frac{x^2 + 5x + 4}{x^2 + 3x - 4} \). The first step is to look at the function whose limit we are trying to find: \( \frac{x^2 + 5x + 4}{x^2 + 3x - 4} \). Here, we recognize that it is a rational function, which means we have a polynomial divided by another polynomial.

Step 2: Check for Direct Substitution

Substitute \( x = -4 \) into the function: \( \frac{(-4)^2 + 5(-4) + 4}{(-4)^2 + 3(-4) - 4} \). This simplifies to \( \frac{16 - 20 + 4}{16 - 12 - 4} \), which gives \( \frac{0}{0} \). This indicates a \( \frac{0}{0} \) indeterminate form, prompting us to find another method to evaluate the limit.

Step 3: Factorization

Factor the numerator \( x^2 + 5x + 4 \) as \( (x + 4)(x + 1) \) and the denominator \( x^2 + 3x - 4 \) as \( (x + 4)(x - 1) \). Thus, the function can be rewritten as \( \frac{(x + 4)(x + 1)}{(x + 4)(x - 1)} \).

Step 4: Cancel Common Factors

Cancel the common factor \( (x + 4) \) from both the numerator and the denominator. The expression simplifies to \( \frac{x + 1}{x - 1} \).

Step 5: Evaluate the Simplified Limit

Now, solve \( \lim_{x \rightarrow -4} \frac{x + 1}{x - 1} \) by direct substitution. Replace \( x \) with \( -4 \) to obtain \( \frac{-4 + 1}{-4 - 1} = \frac{-3}{-5} \). This simplifies to \( \frac{3}{5} \).

Key Concepts

Rational Function

A rational function is simply a fraction where both the numerator and the denominator are polynomials. Polynomials consist of terms that are combined using addition, subtraction, and multiplication by numbers or variables raised to whole number exponents. The rational function in our exercise is \( \frac{x^2 + 5x + 4}{x^2 + 3x - 4} \). Here, both the numerator \( x^2 + 5x + 4 \) and the denominator \( x^2 + 3x - 4 \) meet the criteria of being polynomials.

Understanding that you are working with a rational function helps guide the process of solving. Rational functions are noteworthy because they can often be simplified if they contain common factors in the numerator and denominator, making it easier to evaluate their limits.

Factoring Polynomials

Factoring polynomials involves breaking down a polynomial into simpler 'factor' polynomials which, when multiplied together, return the original polynomial. This technique is crucial when dealing with rational functions, especially when you encounter an indeterminate form.

For instance, in our problem, the polynomial \( x^2 + 5x + 4 \) breaks down into \((x + 4)(x + 1)\). Similarly, \( x^2 + 3x - 4 \) factors into \( (x + 4)(x - 1) \).

• Start by looking for numbers that multiply to give the constant term (the '4' in \( x^2 + 5x + 4 \)), while adding to the linear coefficient (the '5' in \( x^2 + 5x + 4 \)).
• Apply similar logic to the denominator.

Factoring simplifies the expression and helps identify common factors which may be cancelled, as seen in the subsequent solution steps.

Indeterminate Forms

When substituting a particular value into a rational function leads to an expression like \( \frac{0}{0} \), this is termed an indeterminate form. This does not mean the limit does not exist; it means more investigation is needed.

In our example, plugging \( x = -4 \) directly into the function yields \( \frac{0}{0} \). It signals that further simplification, often through factoring, is necessary to evaluate the limit correctly. Identifying an indeterminate form alerts you to re-examine the expression instead of concluding immediately.

Direct Substitution

Direct substitution, the simplest method to find a limit, involves plugging the value of \( x \) into the function directly. In cases where this doesn’t create an indeterminate form, it gives the limit right away.

• If plugging in the value works, the solution is straightforward.
• In our case, after factoring and cancelling, substituting \( x = -4 \) into \( \frac{x + 1}{x - 1} \) was achievable, leading directly to the solution \( \frac{3}{5} \).

Direct substitution is a speedy option when the expression allows, highlighting its practicality when an expression has been simplified properly.