Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{2 \sin x} $$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Recognize Applicable L'Hôpital's Rule

Since the limit presents an indeterminate form \( \frac{0}{0} \), L'Hôpital's Rule can be used to solve this. Recall that L'Hôpital's Rule states that for indeterminate forms \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), the limit \( \lim_{x \rightarrow a} \frac{f(x)}{g(x)} \) can be computed using \( \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)} \) if this latter limit exists.

Step 2: Differentiate the Numerator and Denominator

Differentiate the numerator and the denominator of the given function separately. The derivative of the numerator \( f(x) = e^x - e^{-x} \) is \( f'(x) = e^x + e^{-x} \). The derivative of the denominator \( g(x) = 2 \sin x \) is \( g'(x) = 2 \cos x \).

Step 3: Apply L'Hôpital's Rule

Now that we have the derivatives, apply L'Hôpital's Rule: \[\lim_{x \rightarrow 0} \frac{e^x - e^{-x}}{2 \sin x} = \lim_{x \rightarrow 0} \frac{e^x + e^{-x}}{2 \cos x}.\]

Step 4: Evaluate the New Limit

Evaluate the new limit by substituting \( x = 0 \): \[\frac{e^0 + e^{-0}}{2 \cos 0} = \frac{1 + 1}{2 \times 1} = \frac{2}{2} = 1.\] The value of the limit is 1.

Key Concepts

Indeterminate Forms

In calculus, an indeterminate form arises when substituting in a value results in a form that does not give enough information about the actual limit. Common indeterminate forms include \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\).

These forms occur frequently in problems involving limits, especially when dealing with functions that behave unpredictably at specific points.

• The form \(\frac{0}{0}\) happens when both the numerator and denominator approach zero as \(x\) approaches a certain value.
• This creates uncertainty, which means we cannot determine the limit's value directly from the expression.

To solve indeterminate forms, we often use techniques such as factoring, conjugates, or, as in this case, L'Hôpital's Rule. Recognizing these special forms is crucial, as it guides us toward the correct methods to use.

Understanding when a problem presents an indeterminate form is an important step to solving it accurately.

Differentiation

Differentiation is a fundamental operation in calculus, used to compute the derivative of a function. The derivative represents the rate of change of the function with respect to a variable, typically \(x\).

The process of differentiation provides a way to find the slope of a function at any given point.

• The derivative of \(f(x) = e^x - e^{-x}\) is \(f'(x) = e^x + e^{-x}\).
• For \(g(x) = 2 \sin x\), the derivative is \(g'(x) = 2 \cos x\).

These derivatives are essential in applying L'Hôpital's Rule because they allow us to form a new limit expression that can be easier to evaluate.

Differentiation thus helps simplify complex expressions, making it a vital tool in solving various problems in calculus.

Limits

The concept of limits is central to calculus, providing a way to describe the behavior of functions as they approach a specific point or value. Limits help us understand how functions behave close to a point rather than exactly at it.

In this exercise, the limit \(\lim_{x \rightarrow 0} \frac{e^{x} - e^{-x}}{2 \sin x}\) was indeterminate. We reformulate it using L'Hôpital's Rule as \(\lim_{x \rightarrow 0} \frac{e^x + e^{-x}}{2 \cos x}\).

• By substituting \(x = 0\), we simplified the expression to \(\frac{1 + 1}{2 \times 1}\).
• This evaluation gives a limit value of 1.

When handling limits, determining when and how they exist is critical. A correct approach ensures accurate evaluation and interpretation, making limits a foundational concept in understanding continuity and solving calculus problems.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, play an essential role in calculus, often appearing in limits, derivatives, and integrals. These functions describe the relationships between angles and sides in right triangles and can extend to describe periodic phenomena.

In this problem, we encountered the function \(2 \sin x\).

• The derivative, \(2 \cos x\), was computed to apply L'Hôpital's Rule.
• These trigonometric derivatives are obtained from the basic identities: the derivative of \(\sin x\) is \(\cos x\), and the derivative of \(\cos x\) is \(-\sin x\).

Trigonometry provides tools for resolving specific kinds of limits and understanding cyclical behaviors.

Mastering these functions and their properties is crucial for effectively tackling calculus problems involving trigonometry.