Problem 16 - Calculus for Biology and Medicine

Question

Find the limits in Problems 1-60; not all limits require use of l'Hôpital's rule. $$ \lim _{x \rightarrow 0} \frac{2^{x}-1}{3^{x}-1} $$

Step-by-Step Solution

Answer

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

Step 1: Analyze the Limit Form

Observe the expression \( \lim_{x \rightarrow 0} \frac{2^x - 1}{3^x - 1} \). As \( x \) approaches 0, both the numerator \( 2^x - 1 \) and the denominator \( 3^x - 1 \) approach 0, resulting in the indeterminate form \( \frac{0}{0} \). This indicates that the limit might be solvable using l'Hôpital's Rule.

Step 2: Apply l'Hôpital's Rule

l'Hôpital's Rule can be used when the limit is in the form \( \frac{0}{0} \) or \( \frac{\pm \infty}{\pm \infty} \). To apply it, differentiate the numerator and the denominator with respect to \( x \). The derivative of \( 2^x - 1 \) is \( 2^x \ln(2) \), and the derivative of \( 3^x - 1 \) is \( 3^x \ln(3) \). The limit then becomes \( \lim_{x \rightarrow 0} \frac{2^x \ln(2)}{3^x \ln(3)} \).

Step 3: Evaluate the New Limit Expression

Substitute \( x = 0 \) into the new expression to find the limit: \[ \lim_{x \rightarrow 0} \frac{2^x \ln(2)}{3^x \ln(3)} = \frac{2^0 \ln(2)}{3^0 \ln(3)} = \frac{1 \cdot \ln(2)}{1 \cdot \ln(3)} = \frac{\ln(2)}{\ln(3)}. \]

Step 4: Calculate the Final Value

The expression \( \frac{\ln(2)}{\ln(3)} \) is a constant value that represents the limit of the original function as \( x \) approaches 0. No further simplification is needed as \( \ln(2) \) and \( \ln(3) \) are constants.

Key Concepts

l'Hôpital's Rule

l’Hôpital’s Rule is a handy tool for limits. It helps when you face indeterminate forms like \( \frac{0}{0} \) or \( \frac{\pm \infty}{\pm \infty} \). If you have such a limit, you can apply l’Hôpital’s Rule.
To use it, take the derivatives of the numerator and denominator separately.

• First, ensure the function is in an indeterminate form.
• Differentiate the numerator and denominator.
• Re-evaluate the limit with these new derivatives.

This rule is named after the 17th-century French mathematician Guillaume de l'Hôpital.
In our exercise, \( \lim_{x \rightarrow 0} \frac{2^x - 1}{3^x - 1} \) results in an indeterminate form \( \frac{0}{0} \). Hence, we apply l'Hôpital's rule by differentiating \( 2^x - 1 \) and \( 3^x - 1 \). The derivatives are \( 2^x \ln(2) \) and \( 3^x \ln(3) \), respectively. Re-evaluating with l'Hôpital's Rule simplifies the problem.

Indeterminate forms

An indeterminate form occurs when substituting a value into a limit doesn't yield a clear answer. Common types include \( \frac{0}{0} \) and \( \frac{\pm \infty}{\pm \infty} \). These forms signal that the usual methods for evaluating limits won't work directly.
Indeterminate forms occur because the function's behavior near a point is not straightforward.
In the exercise, both \( 2^x - 1 \) and \( 3^x - 1 \) approach zero as \( x \rightarrow 0 \). This gives a \( \frac{0}{0} \) scenario. But don't worry, indeterminate forms like \( \frac{0}{0} \) are solvable with techniques like l'Hôpital's Rule.

Derivatives

Derivatives are a core component of calculus. They measure how a function changes at any point. When you find a derivative, you're essentially calculating the slope of the tangent line at a point.
In our problem, we need derivatives to use l'Hôpital's Rule. We find the derivative of the numerator \( 2^x - 1 \) to be \( 2^x \ln(2) \). Similarly, the derivative of the denominator \( 3^x - 1 \) is \( 3^x \ln(3) \).

• Exponentials like \( a^x \) have derivatives of \( a^x \ln(a) \).
• This use of derivatives turns an indeterminate form into something manageable.

By differentiating, we could simplify the expression \( \frac{2^x \ln(2)}{3^x \ln(3)} \), allowing us to easily evaluate as \( x \rightarrow 0 \). Now, we can find a clear limit value, \( \frac{\ln(2)}{\ln(3)} \) without confusion.