Problem 1 - Thomas Calculus

Question

In Exercises \(1-6,\) use 1'Hopital'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 Indeterminate Form

First, substitute the limit value into the expression to see if it results in an indeterminate form. Substitute \( x = 2 \) into \( \frac{x-2}{x^2-4} \). You get \( \frac{0}{0} \), an indeterminate form.

Step 2: Apply L'Hopital's Rule

Since the limit results in an indeterminate form \( \frac{0}{0} \), we can apply L'Hopital's Rule. Differentiate the numerator and the denominator separately. The derivative of the numerator \( x - 2 \) is \( 1 \), and the derivative of the denominator \( x^2 - 4 \) is \( 2x \).

Step 3: Evaluate the New Limit

Using L'Hopital's Rule, the new limit becomes: \( \lim_{x \rightarrow 2} \frac{1}{2x} \). Substitute \( x = 2 \) into \( \frac{1}{2x} \) to get \( \frac{1}{4} \).

Step 4: Verify by Factoring

Another way to find the limit is by factoring the original expression. The denominator \( x^2 - 4 \) can be factored as \((x-2)(x+2) \). The expression becomes \( \frac{x-2}{(x-2)(x+2)} \). Simplify by canceling \( x-2 \), which leaves \( \frac{1}{x+2} \).

Step 5: Evaluate Simplified Expression

Now, substitute \( x = 2 \) into the simplified expression \( \frac{1}{x+2} \). This gives \( \frac{1}{4} \).

Key Concepts

Indeterminate Forms

In calculus, we often encounter unexpected results when calculating limits. One common scenario is known as an "indeterminate form," which arises when both the numerator and denominator of a fraction approach zero as the variable approaches a certain value. In our example, when substituting \( x = 2 \) into the limit \( \frac{x-2}{x^2-4} \), both the numerator and denominator become zero, leading to \( \frac{0}{0} \). This is an indeterminate form and means more work is needed to evaluate the limit. Indeterminate forms are not just limited to \( \frac{0}{0} \); other forms, like \( \frac{\infty}{\infty} \), \( 0 \times \infty \), and \( \infty - \infty \), are also common in different scenarios.

Factoring

Factoring is a method used to simplify expressions, and it's particularly useful for addressing limits that result in indeterminate forms. In our problem, the expression \( x^2 - 4 \) can be factored into \((x-2)(x+2)\). This factorization allows us to simplify the expression \( \frac{x-2}{x^2-4} \) to \( \frac{x-2}{(x-2)(x+2)} \).
Once factored, the \((x-2)\) terms in the numerator and denominator are common and can be canceled, provided \( x eq 2 \). This leaves us with \( \frac{1}{x+2} \). By simplifying, it becomes much more straightforward to evaluate the limit as \( x \) approaches 2. This process of canceling terms is essential in resolving indeterminate forms without necessarily needing more complex calculus techniques.

Limits

The concept of a limit is central in calculus. A limit describes the value that a function approaches as the input approaches a certain point. Understanding limits is crucial for concepts like continuity, derivatives, and integration.

• In our exercise, we are tasked with finding \( \lim_{x \to 2} \frac{x-2}{x^2-4} \).
• Initially, substituting \( x = 2 \) leads to the indeterminate form \( \frac{0}{0} \).

This suggests that we need to analyze the behavior of the function as x gets very close to 2.
By applying techniques like factoring or using advanced rules such as L'Hopital's Rule, we can transform this complicated problem into a simpler one, finding the limit accurately.

Derivatives

Derivatives are tools in calculus that measure how a function changes as its input changes. Essentially, a derivative represents the "rate of change" or "slope" of a function at a particular point.
Derivatives come in handy for problems involving limits that result in indeterminate forms. Specifically, L'Hopital's Rule uses derivatives to evaluate these limits.

• For the exercise given, we find that the original limit \( \frac{x-2}{x^2-4} \) reaches \( \frac{0}{0} \).
• According to L'Hopital's Rule, we differentiate both the numerator (\(x-2\)) and the denominator (\(x^2-4\)).

The derivatives are \(1\) for the numerator and \(2x\) for the denominator.
We then evaluate the limit \( \lim_{x \to 2} \frac{1}{2x} \), which simplifies to \( \frac{1}{4} \).
By understanding and using derivatives in this way, evaluating limits becomes a systematic process.