Problem 11 - 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 of the function is 1.

Step 1: Substitute the Limit into the Function

Let's directly substitute \( x = 0 \) into the function \( \lim_{x \to 0} \frac{e^x - e^{-x}}{2\sin x} \). This gives us \( \frac{e^0 - e^0}{2\sin 0} = \frac{0}{0} \), which is indeterminate, so we cannot evaluate it directly.

Step 2: Recognize the Form for L'Hôpital's Rule

Since the limit is in an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule. This rule states that if a limit leads to an indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), you can take the derivative of the numerator and the derivative of the denominator separately until you no longer have an indeterminate form.

Step 3: Differentiate the Numerator and Denominator

Find the derivative of the numerator and denominator. The derivative of \( e^x - e^{-x} \) is \( e^x + e^{-x} \), and the derivative of \( 2\sin x \) is \( 2\cos x \). Thus, the limit transforms to \( \lim_{x \to 0} \frac{e^x + e^{-x}}{2\cos x} \).

Step 4: Substitute the Limit into the New Function

Substitute \( x = 0 \) into the new function from Step 3: \( \frac{e^0 + e^{-0}}{2\cos 0} = \frac{1 + 1}{2 \cdot 1} = \frac{2}{2} = 1 \).

Step 5: Verify and Conclude

Re-evaluate each step to ensure that no calculation errors occurred. The function has been simplified appropriately using L'Hôpital's Rule, and the substitution yielded a determinate form.

Key Concepts

Indeterminate Form

When you encounter the challenge of finding the limit in calculus, sometimes you might come across forms that don't allow for straightforward evaluation. These are called indeterminate forms. A common example is the

• \( \frac{0}{0} \) form
• \( \frac{\infty}{\infty} \) form

Such forms make it impossible to simply substitute the value into the function to find a result. Instead, additional methods, like algebraic manipulation or specialized rules such as L'Hôpital's Rule, are needed.
In our example, when substituting \( x = 0 \) into \( \lim_{x \to 0} \frac{e^x - e^{-x}}{2\sin x} \), you obtain \( \frac{0}{0} \), highlighting the need for an alternative method to solve it.

Derivative

The derivative is a fundamental concept in calculus that reflects how a function changes. To overcome the indeterminate form, we employ a mathematical tool called the derivative, which helps us approximate the slope of a tangent line to a function at any point. When applying L'Hôpital's Rule to the problem, we take the derivatives of both the numerator and the denominator:

• For \( e^x - e^{-x} \), the derivative turns into \( e^x + e^{-x} \).
• The derivative of \( 2\sin x \) becomes \( 2\cos x \).

The application of these derivatives transforms the expression, aiding in resolving the previously indeterminate form.

Exponential Functions

Exponential functions are functions of the form \( e^x \), where \( e \) is the base of natural logarithms approximately equal to 2.718. They are powerful in modeling growth and decay patterns due to their constant rate of change. In the given limit problem, we have the

• \( e^x \)
• \( e^{-x} \)

Both share the same base but have different exponents, making them crucial in the numerator. Understanding the role of exponential functions is key, especially since they yield differentials that are relatively simple, such as \( e^x + e^{-x} \), to help solve the limit problem effectively using L'Hôpital's Rule.

Trigonometric Functions

Trigonometric functions like sine and cosine are vital in calculus due to their periodic properties and wide-ranging applications in modeling waves and oscillations. In this context, the denominator \( 2\sin x \) involves the sine function. This function, when differentiated, becomes \( 2\cos x \), which is essential in simplifying limits that present indeterminate forms.
These functions exhibit unique properties that allow derivatives to exist at crucial points, such as \( x = 0 \). In our problem, replacing \( 2\sin x \) with \( 2\cos x \) through differentiation was pivotal in resolving the indeterminate form and simplifying the limit expression successfully.