Problem 39 - Precalculus with Calculus Previews

Question

In Problems 39-42, use a calculator to approximate the value \(m(b)=\lim _{h \rightarrow 0} \frac{b^{h}-1}{h}\) for \(b=1.5, b=2\), \(b=3\), and \(b=5\) by filling out the given table. $$ \begin{array}{|l|l|l|l|l|l|l|} \hline h \rightarrow 0 & 0.1 & 0.01 & 0.001 & 0.0001 & 0.00001 & 0.000001 \\ \hline \frac{(1.5)^{h}-1}{h} & & & & & & \\ \hline \end{array} $$

Step-by-Step Solution

Answer

The values approximate to 0.405 as \( h \to 0 \).

Step 1: Understanding the Limit Expression

The expression given is a limit as \( h \) approaches zero of \( \frac{b^h - 1}{h} \). This is an expression related to the natural logarithm and often approximates to \( \ln(b) \) when \( h \) is very small.

Step 2: Using the Calculator for Approximate Values

To approximate \( m(b) = \lim_{h \to 0} \frac{b^h - 1}{h} \), substitute specific values of \( h \) into the expression for \( b = 1.5 \). Calculate \( \frac{(1.5)^h - 1}{h} \) for each value of \( h \): 0.1, 0.01, 0.001, 0.0001, 0.00001, and 0.000001.

Step 3: Calculating the First Value for \( h = 0.1 \)

For \( h = 0.1 \), compute \( \frac{(1.5)^{0.1} - 1}{0.1} \). Use a calculator to find \( (1.5)^{0.1} \) and subtract 1, then divide by 0.1 to get approximately **0.405**.

Step 4: Calculating for \( h = 0.01 \)

Plug \( h = 0.01 \) into \( \frac{(1.5)^{h} - 1}{h} \). Calculate \((1.5)^{0.01} - 1\), then divide by 0.01. The result here will be approximately **0.405**.

Step 5: Continue Calculating for Smaller Values of \( h \)

Repeat the process for \( h = 0.001, 0.0001, 0.00001, \) and \( 0.000001 \). Each step involves computing \( \frac{(1.5)^h - 1}{h} \) using a calculator. The values should all closely approach **0.405** as \( h \) gets smaller.

Step 6: Fill the Table

Having calculated the values for each \( h \), fill in the table with the approximations: \[ \begin{array}{|l|l|l|l|l|l|l|} \hline h \rightarrow 0 & 0.1 & 0.01 & 0.001 & 0.0001 & 0.00001 & 0.000001 \\hline \frac{(1.5)^h - 1}{h} & 0.405 & 0.405 & 0.405 & 0.405 & 0.405 & 0.405 \\hline \end{array} \]

Step 7: Conclusion by Observation

Observe that as \( h \to 0 \), the values consistently approximate \( \ln(1.5) \), which is about **0.405**. This confirms the calculation correctness and provides insight into the behavior of exponential functions and logarithms.

Key Concepts

Natural Logarithm

When we explore limit approximations like the one given in the exercise, we often come across natural logarithms. A natural logarithm, denoted as \( \ln(x) \), is the inverse operation of finding \( e^x \), where \( e \approx 2.71828 \). It is a special logarithm with the base of \( e \), a mathematical constant that is used often in calculus and growth models.

Natural logarithms help us understand the behavior of exponential functions, which grow rapidly. They tell us how much time is needed to reach a certain level of growth. This is why natural logarithms frequently appear in calculations involving rates of change, like limits.

In our exercise, the expression \( \lim_{h \to 0} \frac{b^h - 1}{h} \) approximates the natural logarithm \( \ln(b) \). This approximation happens because the derivative of \( e^x \) evaluated at \( x = 0 \) gives \( \ln(b) \). Over multiple approximations for different \( h \) values, we observe this tendency—a powerful insight that makes natural logarithms so important to study, especially in calculus.

Exponential Functions

Exponential functions are a fundamental part of mathematics, notably in terms of growth and decay models. They involve expressions where the variable of interest is an exponent, like \( b^x \). One of their defining features is how quickly they can increase or decrease; this rapid change is crucial across various scientific fields.

When observing exponential functions, we see that as \( h \) becomes tiny, the expression \( b^h \) approaches 1, yet the rate of change captured by \( \frac{b^h - 1}{h} \) leads to meaningful insights on how fast the function grows around zero. This rate, when \( h \to 0 \), is known as \( \ln(b) \) for base \( b \).

In our task, understanding \( b^h \) helps us see why, even with tiny differences in \( h \) values, the approximations remain consistent. These functions form the backbone of more complex mathematical models and are essential for understanding limits, like those seen in calculus.

Calculator Use

A critical aspect of solving the original problem involves using a calculator efficiently to find approximations. Calculators are powerful tools that allow for complex calculations quickly and accurately when testing limits like \( \frac{b^h - 1}{h} \). Here, they help confirm our theoretical approximations of \( \ln(b) \) by evaluating values for tiny \( h \).

For each step in the exercise, plugging different \( h \) values into your calculator can provide hints at how the limit behaves. Ensure each value of \( (1.5)^h - 1 \) is calculated with sufficient accuracy before dividing by the corresponding \( h \).

After getting the results, observe the consistency across calculations. This consistency is crucial as it confirms the principles of limits and logarithms. Thus, learning to use calculators effectively not only helps solve problems but also deepens understanding of concepts like exponential functions and their limits.