Problem 1 - University Calculus: Early Transcendentals

Question

Use I'Hópital's Rule to evaluate the limit. Then evaluate the limit using a method studied in Chapter 2. $$\lim _{x \rightarrow-2} \frac{x+2}{x^{2}-4}$$

Step-by-Step Solution

Answer

The limit is \(-\frac{1}{4}\).

Step 1: Identify the Indeterminate Form

First, substitute the value of the limit, which is \(x = -2\), into the expression \( \frac{x+2}{x^2-4} \). This gives \( \frac{-2+2}{(-2)^2-4} = \frac{0}{4-4} = \frac{0}{0} \). This shows that the expression is an indeterminate form and L'Hôpital's Rule can be applied.

Step 2: Apply L'Hôpital's Rule

L'Hôpital's Rule states that if the limit of \( \frac{f(x)}{g(x)} \) as \(x\) approaches \(c\) results in an indeterminate form \( \frac{0}{0} \), then it can be evaluated as follows: \( \lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \lim_{x \rightarrow c} \frac{f'(x)}{g'(x)} \). Take the derivative: \( f(x) = x + 2 \) and \( f'(x) = 1 \); \( g(x) = x^2 - 4 \) and \( g'(x) = 2x \). Substitute in to get \( \lim_{x \rightarrow -2} \frac{1}{2x} \).

Step 3: Calculate the New Limit

Substitute \( x = -2 \) into the new expression \( \frac{1}{2x} \) to find \( \frac{1}{2(-2)} = \frac{1}{-4} = -\frac{1}{4} \). This is the limit using L'Hôpital's Rule.

Step 4: Evaluate the Limit Using Factorization

To evaluate the limit using methods studied earlier, factor the denominator: \( x^2 - 4 = (x+2)(x-2) \). This means the original limit \( \lim_{x \rightarrow -2} \frac{x+2}{(x+2)(x-2)} \) simplifies to \( \lim_{x \rightarrow -2} \frac{1}{x-2} \) after canceling \(x+2\). Substitute \( x = -2 \), leading to \( \frac{1}{-2-2} = \frac{1}{-4} = -\frac{1}{4} \).

Key Concepts

Understanding Indeterminate Forms

When evaluating limits, we may encounter expressions like \( \frac{0}{0} \), which are known as indeterminate forms. These forms arise when both the numerator and the denominator of a fraction approach zero, making it impossible to determine the limit through direct substitution.
Identifying an indeterminate form is essential because it acts as a signal that special mathematical rules or techniques are needed. In calculus, the most common approach to handle these situations is by using L'Hôpital's Rule. This rule helps us resolve these types of limits by differentiating the numerator and the denominator.

Exploring Limits

The concept of a limit is central in calculus and refers to the value that a function approaches as the input approaches some value. For rational functions, limits are often calculated directly by substitution, but it's not always that simple.
When substitution leads to an indeterminate form, we have alternatives:

• Using algebraic manipulation, like factoring or simplifying the expression.
• Applying L'Hôpital's Rule, which involves derivatives, as explored in the given problem.

L'Hôpital's Rule is especially useful when substitution yields \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). It simplifies the evaluation of limits by transforming the problem into one involving derivatives instead of the original functions.

Derivatives and Their Role

A derivative measures how a function changes as its input changes, essentially representing the slope of the tangent line at any point on the function. Derivatives provide a powerful tool in calculus, used in various applications such as optimization and understanding rates of change.
In the context of evaluating limits, derivatives become essential when applying L'Hôpital's Rule. The rule states that if a limit results in an indeterminate form, the limit of the original functions can be replaced by the limit of their derivatives:\[\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}\]This transformation sometimes turns a seemingly unsolvable problem into a straightforward one, demonstrating the power and convenience of using derivatives in calculus.