Problem 29 - Precalculus: Mathematics for Calculus

Question

Finding Limits Evaluate the limit if it exists. $$\lim _{x \rightarrow-4} \frac{\frac{1}{4}+\frac{1}{x}}{4+x}$$

Step-by-Step Solution

Answer

The limit is \(-\frac{1}{16}\).

Step 1: Substitution

First, try to directly substitute the value of \( x \) approaching \(-4\) into the expression. Substituting gives:\[\frac{\frac{1}{4} + \frac{1}{-4}}{4 + (-4)} = \frac{\frac{1}{4} - \frac{1}{4}}{0}\]This results in a \(\frac{0}{0}\) indeterminate form.

Step 2: Simplify the Expression

Since direct substitution resulted in an indeterminate form, simplification is needed. Combine the fractions in the numerator:\[\frac{\frac{1}{4} + \frac{1}{x}}{4+x} = \frac{\frac{x+4}{4x}}{4+x}\]Now, the expression becomes:\[\frac{x+4}{4x(4+x)}\]

Step 3: Factor and Cancel

Notice that \(x+4\) appears in the numerator and denominator. We can cancel it out:\[\frac{x+4}{4x(4+x)} = \frac{1}{4x}\]This leaves us with a simpler expression after cancellation.

Step 4: Substitute Again

Substitute \(x = -4\) into the simplified expression:\[\frac{1}{4(-4)} = \frac{1}{-16} = -\frac{1}{16}\]This substitution is valid now as it does not result in an indeterminate form.

Key Concepts

Indeterminate Forms

When working with limits, we often encounter indeterminate forms. These forms happen when substitution into an expression results in situations like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), or other undefined forms. In our exercise, substituting \( x = -4 \) into the expression initially yields \( \frac{0}{0} \), a classic indeterminate form.

Indeterminate forms indicate that our work is not done yet. They tell us that further manipulation, such as algebraic simplification or factoring, is needed. By performing these additional steps, we can potentially rewrite the expression into a new form suitable for limit calculation. Understanding indeterminate forms prepares us for the methods required to resolve them, leading to valid limit evaluations.

Simplifying Expressions

Simplifying expressions is a crucial step when dealing with limits, especially when direct substitution yields an indeterminate form. In our given exercise, the expression initially was \( \frac{\frac{1}{4} + \frac{1}{x}}{4+x} \).

To simplify this, we first combined the fractions in the numerator. The result was \( \frac{\frac{x+4}{4x}}{4+x} \). Further simplification revealed \( \frac{x+4}{4x(4+x)} \) by multiplying through by \( 4x \).

This simplification is essential because it prepares the expression for possible factoring and further manipulations that help evaluate the limit successfully.

Factoring Rational Expressions

Factoring plays a significant role when simplifying or reducing rational expressions in calculus. In our expression, \( \frac{x+4}{4x(4+x)} \), the term \( x+4 \) appears in both the numerator and the denominator. This presence allows us to factor and cancel out \( x+4 \).

Factoring is not only about canceling terms. It is fundamentally about simplifying expressions. This technique often paves the way to eliminating indeterminate forms. In this instance, once \( x+4 \) is canceled, the remaining expression becomes \( \frac{1}{4x} \), a much easier expression to handle and evaluate as \( x \) approaches \(-4\).

Direct Substitution in Limits

With a simplified expression in hand, direct substitution becomes straightforward. Initially, plugging in \( x = -4 \) was problematic. After simplification, the expression \( \frac{1}{4x} \) is left.

Now, substituting with \( x = -4 \) gives \( \frac{1}{4(-4)} \) which evaluates to \( -\frac{1}{16} \). This result comes without the complications of an indeterminate form.

Direct substitution is one of the simplest methods for finding limits. But it only works well when the expression is free from indeterminate forms. Hence, the value of simplifying and factoring beforehand becomes evident, paving the way for effective use of direct substitution.