Chapter 6: Problem 40
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. $$ \begin{array}{|l|l|l|l|l|l|l|} \hline h \rightarrow 0 & 0.1 & 0.01 & 0.001 & 0.0001 & 0.00001 & 0.000001 \\ \hline \frac{2^{h}-1}{h} & & & & & & \\ \hline \end{array} $$
Short Answer
Step by step solution
Understand the Function m(b)
Substitute Values for b
Calculate for h = 0.1
Calculate for h = 0.01
Calculate for h = 0.001
Calculate for h = 0.0001
Calculate for h = 0.00001
Calculate for h = 0.000001
Complete the Table
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative approximation
Exponential functions
- Base greater than 1: Function grows as \( x \) increases.
- Base equal to 1: Function remains constant.
- Base between 0 and 1: Function decreases as \( x \) increases.
Calculus limits
In our calculation for \( b = 2 \), by substituting even smaller numbers for \( h \), you might notice that the values of \( \frac{2^{h}-1}{h} \) stabilize around 0.72, illustrating the process of approaching the limit. This consistency indicates that our approximation is becoming increasingly precise as \( h \) nears zero, leading us to a clearer understanding of the derivative at that point.