Problem 15 - Calculus for Biology and Medicine

Question

Use l'Hospital's rule to find the limits. $$ \lim _{x \rightarrow 0} \frac{2^{x}-1}{3^{x}-1} $$

Step-by-Step Solution

Answer

The limit is \( \frac{\ln(2)}{\ln(3)} \).

Step 1: Identify the Indeterminate Form

First, let's determine whether we have an indeterminate form. Substitute \( x = 0 \) into the expression to get \( \frac{2^0 - 1}{3^0 - 1} = \frac{1-1}{1-1} = \frac{0}{0} \). Since we have the form \( \frac{0}{0} \), l'Hospital's Rule is applicable.

Step 2: Differentiate the Numerator and Denominator

Apply l'Hospital's Rule by differentiating the numerator \( 2^x - 1 \) and the denominator \( 3^x - 1 \). This gives the derivatives: - Numerator's derivative: \( \frac{d}{dx}(2^x - 1) = 2^x \ln(2) \)- Denominator's derivative: \( \frac{d}{dx}(3^x - 1) = 3^x \ln(3) \)

Step 3: Apply l'Hospital's Rule

Substitute these derivatives back into the original limit: \[ \lim_{x \to 0} \frac{2^x \ln(2)}{3^x \ln(3)} \]

Step 4: Evaluate the Limit of the New Expression

Substitute \( x = 0 \) into the expression after differentiating: \[ \frac{2^0 \ln(2)}{3^0 \ln(3)} = \frac{1 \times \ln(2)}{1 \times \ln(3)} = \frac{\ln(2)}{\ln(3)} \]

Step 5: Conclusion

Thus, the original limit \( \lim _{x \rightarrow 0} \frac{2^{x}-1}{3^{x}-1} \) evaluates to \( \frac{\ln(2)}{\ln(3)} \).

Key Concepts

Understanding Limits

Limits are a fundamental concept in calculus. They help us understand the behavior of functions as inputs approach a specific value. In simple terms, a limit examines how a function behaves when its variable gets closer to a particular value, such as zero, infinity, or any other number.

• When we say "the limit of \( f(x) \) as \( x \) approaches \( a \) is \( L \)," it means that as \( x \) gets closer to \( a \), \( f(x) \) gets closer to \( L \).
• Limits are used to define derivatives and integrals, and they are essential in formulating the concept of continuity.

Exploring limits can reveal trends in a function's behavior, allowing us to predict how it behaves as it approaches a critical point. In the given problem, understanding limits lets us evaluate the function's form when both the numerator and denominator approach zero.

Identifying Indeterminate Forms

Indeterminate forms often arise in the study of limits and are essential to identify before using advanced techniques like l'Hospital's Rule.

• An indeterminate form, such as \( \frac{0}{0} \), occurs when both the numerator and denominator of a fraction head towards zero or infinity, making direct evaluation impossible.
• There are other types of indeterminate forms, including \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), \( \infty - \infty \), \( 0^0 \), \( 1^\infty \), and \( \infty^0 \).

Recognizing an indeterminate form signals that standard evaluation won't work, prompting the use of different techniques, such as l'Hospital’s Rule, to find limits. In our exercise, substituting \( x = 0 \) results in \( \frac{0}{0} \), so we apply l'Hospital's Rule.

Basics of Differentiation

Differentiation is a key process in calculus that involves finding the derivative of a function. A derivative represents the rate of change of a function concerning its variable.

• In simple terms, the derivative of a function gives us the slope of the tangent line at any point on its curve.
• Common rules of differentiation include the power rule, product rule, quotient rule, and chain rule.

To apply l'Hospital's Rule, we first need to find the derivatives of both the numerator and the denominator of the given limit expression:

• For \( 2^x - 1 \), the derivative is \( 2^x \ln(2) \).
• For \( 3^x - 1 \), the derivative is \( 3^x \ln(3) \).

Differentiation here allows us to resolve the indeterminate form \( \frac{0}{0} \), leading to a solution.

Exploring Exponential Functions

Exponential functions, like \( 2^x \) and \( 3^x \) in the problem, are vital in mathematics and science.

• These functions involve exponentiation where a constant base is raised to a variable exponent.
• Exponential functions grow quickly and exhibit unique properties, such as constant growth rates proportional to their current value.

When dealing with limits and differentiation, knowing the derivative of an exponential function is important:

• The derivative of \( a^x \), where \( a \) is a constant, is \( a^x \ln(a) \).

In our exercise, acknowledging these derivatives allows us to use l'Hospital's Rule to simplify and compute the limit of an indeterminate form, resulting in an understandable outcome.