Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the indicated limit. Make sure that you have an indeterminate form before you apply l'Hôpital's Rule. $$ \lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{2 \sin x} $$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Verify the form at the point of interest

First, substitute \( x = 0 \) into the function \( \frac{e^{x} - e^{-x}}{2 \sin x} \) to determine if it results in an indeterminate form. In the numerator, \( e^0 - e^0 = 0 \). In the denominator, \( 2 \sin(0) = 0 \). Therefore, the expression is of the \( \frac{0}{0} \) indeterminate form, which allows us to apply l'Hôpital's Rule.

Step 2: Apply l'Hôpital's Rule

Since the limit is in the \( \frac{0}{0} \) form, we can apply l'Hôpital's Rule, which requires differentiating the numerator and the denominator separately. The derivative of the numerator \( e^{x} - e^{-x} \) is \( e^{x} + e^{-x} \). The derivative of the denominator \( 2 \sin x \) is \( 2 \cos x \). This gives us a new limit: \( \lim_{x \to 0} \frac{e^{x} + e^{-x}}{2 \cos x} \).

Step 3: Evaluate the limit after applying l'Hôpital's Rule

Substitute \( x = 0 \) into the new expression \( \frac{e^{x} + e^{-x}}{2 \cos x} \):- Numerator: \( e^0 + e^0 = 2 \).- Denominator: \( 2 \cos(0) = 2 \).Thus, the limit simplifies to \( \frac{2}{2} = 1 \).

Key Concepts

l'Hôpital's Rule

When solving limits in calculus, you might come across expressions that take an indeterminate form, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). This is where l'Hôpital's Rule is particularly useful. This rule helps us find limits that seem unsolvable at first glance.

Here is how l'Hôpital's Rule works:

• Confirm the limit results in an indeterminate form like \( \frac{0}{0} \).
• Differentiation: Compute the derivative of the numerator and the denominator separately.
• Re-evaluate the limit with these new functions.

For example, if we have \( \lim_{x \to a} \frac{f(x)}{g(x)} \), and \( f(a) = 0 \) and \( g(a) = 0 \), you differentiate both \( f(x) \) and \( g(x) \) to solve the limit. Overlaying these steps provides a streamlined solution for what initially seems perplexing.

Limit Evaluation

The evaluation of limits is a fundamental concept in calculus. Limits help us understand the behavior of functions as they approach specific points, even if they do not exactly reach those points.

When evaluating a limit, you need to determine what value the function approaches. Sometimes this involves straightforward substitution. Other times, as with indeterminate forms, special tactics like l'Hôpital's Rule are necessary. Here’s a step-by-step approach:

• Direct Substitution: Start by plugging the point into the function as a first check.
• Determine Indeterminate Form: Identify if the substitution leads to a form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
• Apply Special Techniques: Use rules like l'Hôpital's Rule or algebraic manipulation to resolve the limit.

Grasping this process is crucial for solving complex calculus problems, making it a centerpiece in the field of mathematical analysis.

Indeterminate Forms

Indeterminate forms are expressions that arise in limit evaluation, where conventional calculation methods do not provide a clear answer. They signal a need for advanced techniques to find the true limit.

There are several common indeterminate forms:

• \( \frac{0}{0} \)
• \( \frac{\infty}{\infty} \)
• \( 0 \cdot \infty \)
• \( \infty - \infty \)
• \( 0^0 \), \( 1^\infty \), and \( \infty^0 \)

For the limit \( \lim_{x \to 0} \frac{e^{x}-e^{-x}}{2 \sin x} \), substituting \( x = 0 \) yields \( \frac{0}{0} \), which is a classic indeterminate form. It is these scenarios where rules like l'Hôpital's become invaluable for resolution.By identifying indeterminate forms, you can determine the next step in your calculus computations, which often includes the application of calculus principles like derivatives and simplifying expressions.