Problem 41 - Calculus for Biology and Medicine

Question

In Problems 29-48, find the limits. $$ \lim _{x \rightarrow 0} \frac{e^{2 x}-1}{e^{x}-1} $$

Step-by-Step Solution

Answer

The limit is 2.

Step 1: Understand the Problem

The problem asks for the limit of the expression \( \frac{e^{2x} - 1}{e^x - 1} \) as \( x \) approaches 0. This means we want to find the value the expression gets closer to as \( x \) becomes very small, approaching zero.

Step 2: Apply L'Hôpital's Rule

Since direct substitution of \( x = 0 \) results in the indeterminate form \( \frac{0}{0} \), we apply L'Hôpital's Rule, which states that for limits resulting in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), you can differentiate the numerator and the denominator separately: \[ \lim_{x \rightarrow 0} \frac{f(x)}{g(x)} = \lim_{x \rightarrow 0} \frac{f'(x)}{g'(x)} \]Here, \( f(x) = e^{2x} - 1 \) and \( g(x) = e^x - 1 \).

Step 3: Differentiate the Numerator and Denominator

Differentiate \( f(x) = e^{2x} - 1 \) with respect to \( x \) to get \( f'(x) = 2e^{2x} \). Differentiate \( g(x) = e^x - 1 \) with respect to \( x \) to get \( g'(x) = e^x \).

Step 4: Evaluate the Limit with the New Expressions

Substitute the derivatives into the limit:\[ \lim_{x \rightarrow 0} \frac{2e^{2x}}{e^x} \]Now, simplify the expression inside the limit:\[ \frac{2e^{2x}}{e^x} = 2e^{x} \]Now evaluate the limit: \[ \lim_{x \rightarrow 0} 2e^{x} = 2e^{0} = 2 \]

Step 5: Interpret the Result

The limit evaluates to \( 2 \), meaning that as \( x \) approaches 0, the ratio \( \frac{e^{2x} - 1}{e^x - 1} \) approaches the value 2.

Key Concepts

L'Hôpital's Rule

When you encounter a limit problem that results in an indeterminate form like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), L'Hôpital's Rule comes to the rescue. This rule provides a method to simplify these tricky limit expressions.
The essence of L'Hôpital's Rule is that it allows you to take derivatives of the numerator and the denominator separately, replacing the problematic indeterminate form with a potentially simpler expression.

• Identify the indeterminate form: If substituting \(x\) yields \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), L'Hôpital's Rule can be applied.


• Differentiate both the numerator and the denominator: Only the first derivatives are necessary for the initial application.


• Evaluate the limit again: Substitute the new expressions resulting from the derivatives into the limit equation.

In the exercise, the limit \(\lim \limits_{x \rightarrow 0} \frac{e^{2x} - 1}{e^x - 1}\) originally results in \(\frac{0}{0}\) upon substitution. By applying L'Hôpital's Rule, you differentiate to get \(\frac{2e^{2x}}{e^x}\), which can then easily be evaluated.

Indeterminate Forms

Indeterminate forms are situations in calculus where direct substitution in a limit problem gives you an undefined or ambiguous scenario. These forms generally include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0\cdot \infty\), \(\infty - \infty\), \(0^0\), \(\infty^0\), and \(1^\infty\).
They occur because the standard arithmetic rules don't work, making it necessary to employ special techniques like L'Hôpital's Rule to resolve them. Here's how you recognize a need for L'Hôpital's Rule:

• Check if substitution results in \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).


• Attempt to find derivatives of the numerator and denominator.


• Apply L'Hôpital's Rule iteratively if necessary until a determinate form appears.

In simple terms, despite their intimidating appearance, indeterminate forms are a rule’s way of telling you that there’s more work to be done to find a clear result.

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. A classic example is the natural exponential function \(e^x\), where \(e\) is Euler's number (approximately 2.71828).
These functions are fundamental in describing growth and decay processes, appearing in a variety of fields like finance, biology, and physics.
Key points regarding exponential functions:

• The derivative of \(e^x\) is \(e^x\), and similarly, \(e^{2x}\) differentiates to \(2e^{2x}\), as used in the problem.


• The exponential function \(e^x\) possesses the unique property that its rate of increase is proportional to its current value.


• Exponential expressions are powerful tools for simplifying calculations in calculus, especially when dealing with growth-related problems or limits.

In the provided exercise, exponential functions as \(e^x\) and \(e^{2x}\) play a key role in obtaining the necessary derivatives to apply L'Hôpital's Rule, leading to the simplified limit solution.