Problem 41 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \lim _{x \rightarrow 0} \frac{e^{x}-1}{e^{-x}-1} $$

Step-by-Step Solution

Answer

The limit is -1.

Step 1: Identify the Indeterminate Form

First, let's substitute the value of \(x = 0\) into the function \( \frac{e^{x}-1}{e^{-x}-1} \). Doing this yields \( \frac{e^0 - 1}{e^{-0} - 1} = \frac{0}{0} \). This is an indeterminate form, so we need a different approach to solve the limit.

Step 2: Apply L'Hôpital's Rule

Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule. This rule states that: \[ \lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)} \] when \( \frac{f(x)}{g(x)} \) is of the form \( \frac{0}{0} \). So, we need to find the derivatives of the numerator and the denominator.

Step 3: Differentiate the Numerator

Differentiate the numerator \( e^x - 1 \). The derivative of \( e^x \) is \( e^x \), and the derivative of \(-1\) is \(0\). Thus, the derivative of the numerator is \( f'(x) = e^x \).

Step 4: Differentiate the Denominator

Now, differentiate the denominator \( e^{-x} - 1 \). The derivative of \( e^{-x} \) is \( -e^{-x} \), and the derivative of \(-1\) is \(0\). So, the derivative of the denominator is \( g'(x) = -e^{-x} \).

Step 5: Apply L'Hôpital's Rule Again

Substitute the derivatives we found into L'Hôpital's Rule: \[ \lim_{x \to 0} \frac{e^x}{-e^{-x}} = \lim_{x \to 0} \frac{e^x}{- \frac{1}{e^x}} \] Simplifying this expression, it becomes \[ \lim_{x \to 0} -e^{2x} \].

Step 6: Evaluate the Simplified Limit

Now, evaluate the expression \( -e^{2x} \) as \( x \) approaches \(0\). Substitute \( x = 0 \) into the expression to get \( -e^{0}^2 = -1 \), because \( e^{0} = 1 \). Therefore, the limit is \(-1\).

Key Concepts

Indeterminate Forms

When working with limits in calculus, you might encounter expressions that are undefined or unclear, known as indeterminate forms. These forms arise when you directly substitute a value into a function and obtain an ambiguous result such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms indicate that the limit cannot be determined without further analysis or additional techniques. L'Hôpital's Rule becomes particularly useful in resolving these indeterminate forms, especially the \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \) cases.

• The exercise: \( \lim _{x \rightarrow 0} \frac{e^{x}-1}{e^{-x}-1} \) produces an indeterminate form \( \frac{0}{0} \) through direct substitution.
• This signals the need to employ a method like L'Hôpital's Rule to properly evaluate the limit.

Understanding these forms and knowing when to apply different calculus techniques is essential for solving complex limits.

Derivatives in Calculus

Derivatives are fundamental in calculus. They measure the rate at which a function changes at any given point. In the context of limits and L'Hôpital's Rule, derivatives help simplify complex expressions that are in indeterminate form. Calculating the derivative involves using rules and formulas to derive new functions from original ones.

• For the numerator \( e^x - 1 \), the derivative \( e^x \) is computed using the rule that the derivative of \( e^x \) is simply \( e^x \).
• Likewise, for the denominator \( e^{-x} - 1 \), the derivative is \( -e^{-x} \) due to the chain rule, which handles the negative exponent.

When applying L'Hôpital's Rule, these derivatives replace the original functions to create a new limit that is easier to evaluate. Mastery of derivative rules, such as the product rule, quotient rule, and chain rule, is essential for tackling these types of problems in calculus.

Limits in Calculus

Limits are at the heart of calculus, describing how a function behaves as it approaches a specific point. The concept of limits is crucial for understanding continuity, series, and the foundational theories of derivatives and integrals. Evaluating limits can sometimes become challenging, especially with indeterminate forms or complex functions.

• The exercise in question reduces to evaluating \( \lim_{x \to 0} -e^{2x} \) after applying L'Hôpital's Rule and simplifying the expression.
• Substituting \( x = 0 \) in this simplified expression yields \( -1 \), as \( e^0 = 1 \).

By understanding how to manipulate and simplify expressions using limits, and then applying appropriate techniques like L'Hôpital's Rule, complex problems can become manageable and solvable.