Problem 61 - Calculus for Biology and Medicine

Question

use l'Hospital'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 the Indeterminate Form

First, we evaluate the limit by substituting the value of \(x\) directly. Substituting \(x = 0\) in the numerator and the denominator, we get \(a^0 - 1 = 0\) and \(b^0 - 1 = 0\). Thus, the expression \(\frac{a^x - 1}{b^x - 1}\) is in the indeterminate form \(\frac{0}{0}\).

Step 2: Apply l'Hospital's Rule

Since we have an indeterminate form \(\frac{0}{0}\), we can apply l'Hospital's Rule, which allows us to differentiate the numerator and the denominator separately. We take the derivative of the numerator \(\frac{d}{dx}(a^x - 1) = a^x \ln(a)\) and the derivative of the denominator \(\frac{d}{dx}(b^x - 1) = b^x \ln(b)\).

Step 3: Evaluate the New Limit

Now we need to evaluate the limit of the new expression:\[\lim_{x \to 0} \frac{a^x \ln(a)}{b^x \ln(b)}.\]At \(x = 0\), the expression simplifies to:\[\frac{a^0 \ln(a)}{b^0 \ln(b)} = \frac{\ln(a)}{\ln(b)}.\]This resolves to a determinate form and can be evaluated directly.

Key Concepts

Indeterminate Forms

When we talk about limits, sometimes we encounter expressions that don't immediately provide a clear outcome when substituting the value towards which the variable is approaching. These cases are known as indeterminate forms. An indeterminate form occurs when substituting the value into a function results in expressions like \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(\infty - \infty\), among others. Such forms suggest that simply plugging in the number won't work, and we need some more powerful mathematical tools to resolve the limit.

• The most common indeterminate forms are \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\).
• They require applying special techniques, like manipulating the algebra or using L'Hospital's Rule, to evaluate properly.
• L'Hospital's Rule specifically helps us handle these forms involving two functions, by letting us take derivatives.

Recognizing an indeterminate form immediately helps us decide the next steps for evaluating a limit, as seen in this exercise.

Limits

A limit helps determine the value a function approaches as the input approaches a particular point. Understanding limits is crucial for solving calculus problems, especially when assessing behavior that is not explicitly clear by direct substitution.Calculating a limit involves observing how values get closer to a specific number, even if that number may not be included in the domain of the function. In the initial problem, we have a limit approaching zero:\[ \lim_{x \rightarrow 0} \frac{a^x - 1}{b^x - 1} \]By directly substituting \(x = 0\), and finding it leads to an indeterminate form \(\frac{0}{0}\), we recognize the need for further evaluation. Limits are often evaluated using various mathematical tools, such as:

• Simple substitution, when direct evaluation does not lead to an indeterminate form.
• Algebraic manipulation to simplify expressions.
• Techniques like L'Hospital's Rule to deal with complex indeterminate forms.

In this case, once L'Hospital's Rule is applied, the limit simplifies and can be resolved to provide a meaningful result.

Derivatives

Derivatives are a central concept in calculus, reflecting how a function changes as its input changes. Simply put, they provide a way to compute the slope or the rate of change. The power of derivatives in limit problems shines through when employing L'Hospital's Rule.To properly use L'Hospital's Rule, we need to differentiate the parts of our indeterminate quotient. For the problem:- Differentiate the numerator: \( \frac{d}{dx}(a^x - 1) = a^x \ln(a) \)- Differentiate the denominator: \( \frac{d}{dx}(b^x - 1) = b^x \ln(b) \)These derivatives transform the original problematic expression into something manageable. The derivatives help convert the indeterminate form \(\frac{0}{0}\) into an expression whose limit can be evaluated straightforwardly:\[ \lim_{x \to 0} \frac{a^x \ln(a)}{b^x \ln(b)} = \frac{\ln(a)}{\ln(b)} \]Through derivatives, we unlock the ability to resolve complicated limits, turning abstract concepts into clear, calculable results.