Problem 61 - Calculus for Biology and Medicine

Question

Use l'Hôpital's rule to find $$ \lim _{x \rightarrow 0} \frac{a^{x}-1}{b^{x}-1} $$ where \(a, b>0\).

Step-by-Step Solution

Answer

The limit is \(\frac{\ln(a)}{\ln(b)}\).

Step 1: Check Indeterminate Form

First, evaluate the limit directly by substituting \(x = 0\) into the expression \(\frac{a^x - 1}{b^x - 1}\). This gives \(\frac{a^0 - 1}{b^0 - 1} = \frac{1 - 1}{1 - 1}\), which is \(\frac{0}{0}\), an indeterminate form.

Step 2: Apply l'Hôpital's Rule

Since the expression is in the form \(\frac{0}{0}\), we can apply l'Hôpital's rule, which states that \(\lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \lim_{x \rightarrow c} \frac{f'(x)}{g'(x)}\) if this limit exists. Thus, we differentiate the numerator and the denominator with respect to \(x\).

Step 3: Differentiate the Numerator

The derivative of the numerator, \(a^x - 1\), with respect to \(x\) is \(a^x \ln(a)\) because the derivative of \(a^x\) is \(a^x \ln(a)\) and the derivative of \(-1\) is 0.

Step 4: Differentiate the Denominator

The derivative of the denominator, \(b^x - 1\), with respect to \(x\) is \(b^x \ln(b)\) because the derivative of \(b^x\) is \(b^x \ln(b)\) and the derivative of \(-1\) is 0.

Step 5: Re-evaluate the Limit Using l'Hôpital's Rule

Re-evaluate the limit by using the derivatives found in the previous steps. So, the limit becomes \(\lim_{x \rightarrow 0} \frac{a^x \ln(a)}{b^x \ln(b)}\).

Step 6: Simplify and Evaluate the New Limit

Substitute \(x=0\) into the expression \(\frac{a^x \ln(a)}{b^x \ln(b)}\), which simplifies to \(\frac{a^0 \ln(a)}{b^0 \ln(b)}\). Since \(a^0 = 1\) and \(b^0 = 1\), the expression simplifies to \(\frac{\ln(a)}{\ln(b)}\).

Key Concepts

Indeterminate Forms

Indeterminate forms arise when you substitute values into a mathematical expression and encounter an ambiguous situation that can take on multiple values. One common type is the \( \frac{0}{0} \) form. This occurs when the numerator and denominator of a fraction both approach zero particularly through the process of evaluating limits.
This scenario happened when trying to compute the limit \( \lim_{x \rightarrow 0} \frac{a^x-1}{b^x-1} \). Direct substitution of \(x = 0\) led to \(\frac{1-1}{1-1}\), which simplifies to \(\frac{0}{0}\).

• Other common forms include \(\frac{\infty}{\infty}\), \(0 \cdot \infty\), and \(\infty - \infty\).
• When an indeterminate form is encountered, further analysis, such as using l'Hôpital's Rule in this case, is required to find a meaningful limit.

Derivatives

Derivatives are the cornerstone of calculus, used to determine the rate of change of one quantity with respect to another. When applying l'Hôpital's Rule, finding derivatives of the functions involved is essential.
In the context of the exercise, we needed to find the derivatives of the numerator \(a^x - 1\) and the denominator \(b^x - 1\):

• The derivative of \(a^x - 1\) is \(a^x \ln(a)\). This follows from the derived formula for the derivative of exponential functions with a base not equal to \(e\): \(a^x\) becomes \(a^x \ln(a)\).
• Similarly, the derivative of \(b^x - 1\) is \(b^x \ln(b)\).

These derivatives provide the functions needed to apply l'Hôpital's Rule effectively, allowing us to evaluate the limit without the troublesome indeterminate form.

Exponential Functions

Exponential functions are mathematical functions of the form \(f(x) = a^x\), where \(a\) is a positive constant and \(x\) is the exponent. These functions are essential for modeling growth processes, natural phenomena, and are widely applicable in calculus.
A critical property of exponential functions is their unique behavior under differentiation, which makes them integral to solving many problems:

• The derivative of \(a^x\) is \(a^x \ln(a)\), showcasing how they differ slightly from the natural exponential function \(e^x\), which differentiates to itself \(e^x\).
• In our problem, we examined functions \(a^x\) and \(b^x\). Their derivatives helped us rewrite the limit expression using l'Hôpital's Rule.

Understanding these basics about exponential functions allows one to tackle a variety of advanced problems in calculus including those needing special techniques like l'Hôpital's Rule.