Problem 22 - APEX Calculus

Question

Evaluate the limit: \(\lim _{x \rightarrow-1} \frac{x^{2}+5 x+4}{x^{2}-3 x-4}\).

Step-by-Step Solution

Answer

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

Step 1: Identify Direct Substitution

First, substitute \(x = -1\) directly into the function \( \frac{x^2 + 5x + 4}{x^2 - 3x - 4}\) to see if we can evaluate the limit directly.

Step 2: Apply Direct Substitution

Substituting \(x = -1\), we have the expression: \( \frac{(-1)^2 + 5(-1) + 4}{(-1)^2 - 3(-1) - 4} = \frac{1 - 5 + 4}{1 + 3 - 4} = \frac{0}{0}.\) Since we get an indeterminate form \(\frac{0}{0}\), we cannot find the limit by direct substitution.

Step 3: Factor the Numerator and Denominator

Factor the numerator \(x^2 + 5x + 4\) and the denominator \(x^2 - 3x - 4\). We begin by factoring: \(x^2 + 5x + 4 = (x + 4)(x + 1)\) and \(x^2 - 3x - 4 = (x - 4)(x + 1)\).

Step 4: Simplify the Expression

The expression \( \frac{(x + 4)(x + 1)}{(x - 4)(x + 1)} \) allows us to cancel out the common factor \((x + 1)\) from both numerator and denominator, which simplifies to \( \frac{x + 4}{x - 4} \), with \(x eq -1\).

Step 5: Re-evaluate Limit

Now that the expression is simplified to \( \frac{x + 4}{x - 4} \), substitute \(x = -1\) to find the limit: \( \frac{-1 + 4}{-1 - 4} = \frac{3}{-5} = -\frac{3}{5}.\) So, \(\lim _{x \rightarrow-1} \frac{x^{2}+5 x+4}{x^{2}-3 x-4} = -\frac{3}{5}.\)

Key Concepts

Direct Substitution

When trying to find the limit of a function as it approaches a specific point, direct substitution is often the first technique to try. It involves directly plugging in the value of the limit into the function. For example, given the function \( \frac{x^2 + 5x + 4}{x^2 - 3x - 4} \), and the limit point \( x = -1 \), we substitute \( x = -1 \) directly into the function.By substituting \( x = -1 \), we obtain:

• Numerator: \( (-1)^2 + 5(-1) + 4 = 0 \)
• Denominator: \( (-1)^2 - 3(-1) - 4 = 0 \)

Resulting in \( \frac{0}{0} \), which is an indeterminate form and requires further simplification through more advanced algebra methods.

Factoring Polynomials

Factoring polynomials is essential when dealing with indeterminate forms encountered in limits. It helps simplify expressions into more manageable forms. Given the polynomial expressions in both the numerator and the denominator of our function:\[ \begin{align*} \text{Numerator: } & x^2 + 5x + 4 \\text{Denominator: } & x^2 - 3x - 4 \end{align*} \]We factor:

• Numerator: \( x^2 + 5x + 4 = (x + 4)(x + 1) \)
• Denominator: \( x^2 - 3x - 4 = (x - 4)(x + 1) \)

The common factor \((x + 1)\) can now be canceled from both terms, which simplifies the function and aids in reevaluating the limit.

Indeterminate Form

An indeterminate form occurs when substituting a value into a limit gives a result such as \( \frac{0}{0} \). This indicates that the limit cannot be derived through direct substitution alone. Common indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \times \infty \), among others.In our exercise, we encountered an indeterminate form when evaluating the limit by direct substitution. At this stage, further manipulation of the function is necessary. Methods like factoring, using L'Hôpital's Rule, or algebraic simplification are often employed to resolve such forms and find the actual limit value.

Rational Functions

Rational functions are composed of two polynomials: a numerator and a denominator. These functions often involve more complex limits due to the potential for expressions in both the numerator and denominator to approach zero or infinity as variables change.In our given problem, we examined the rational function \( \frac{x^2 + 5x + 4}{x^2 - 3x - 4} \). The process of finding limits for rational functions often includes:

• Checking for points of discontinuity (where denominator is zero).
• Performing algebraic manipulations, like factoring, to simplify or eliminate terms.
• Evaluating simplified expressions at the intended limit point to find the actual limit value.

This problem demonstrated these steps effectively, leading us to the final limit \(-\frac{3}{5}\).