Problem 41 - Calculus: Early Transcendental Functions

Question

In the text, we deferred the proof of \(\lim _{h \rightarrow \infty} \frac{e^{h}-1}{h}=1\) to the exercises. In this exercise, we guide you through one possible proof. (Another proof is given in exercise 42.) Starting with \(h>0,\) write \(h=\ln e^{h}=\int_{1}^{e^{h}} \frac{1}{x} d x .\) Use the Integral Mean Value Theorem to write \(\int_{1}^{e^{h}} \frac{1}{x} d x=\frac{e^{h}-1}{\bar{x}}\) for some number \(\bar{x}\) between 1 and \(e^{h} .\) This gives you \(\frac{e^{h}-1}{h}=\bar{x} .\) Now, take the limit as \(h \rightarrow 0^{+} .\) For \(h<0,\) repeat this argument, with \(h\) replaced with \(-h.\)

Step-by-Step Solution

Answer

The limit \(\lim _{h \rightarrow \infty} \frac{e^{h}-1}{h}=1\) is proved using the integral mean value theorem.

Step 1: Convert h using logarithmic and integral

Start with \( h>0 \), the expression \( h = \ln e^{h} = \int_{1}^{e^{h}} \frac{1}{x} d x \) is written.

Step 2: Apply Mean Value Theorem

By applying the Mean Value Theorem to the integral part of previous expression, it can be shown that \( \int_{1}^{e^{h}} \frac{1}{x} d x = \frac{e^{h}-1}{\bar{x}} \), where \( \bar{x} \) is a number between 1 and \( e^{h} \).

Step 3: Solve the equation for \( \frac{e^{h}-1}{h} \)

By using the previous two steps, we obtain an equation \( \frac{e^{h}-1}{h} = \bar{x} \).

Step 4: Take the limit as \( h \rightarrow 0^{+} \)

When we introduce the limit and \( h \rightarrow 0^{+} \), we have \( \lim _{h \rightarrow 0^{+}} \frac{e^{h}-1}{h} = \bar{x} \). As \( h \rightarrow 0^{+} \), \( e^{h} \rightarrow 1 \) and therefore \( \bar{x} \rightarrow 1 \). Thus, \( \lim _{h \rightarrow 0^{+}} \frac{e^{h}-1}{h} = 1 \).

Step 5: Deal with negative \( h \)

For \( h<0 \), repeat the above steps replacing \( h \) with \( -h \). The result will be same that is, \( \lim_{h \rightarrow 0^-} \frac{e^{-h}-1}{-h} = 1 \). Hence, the statement is proved.

Key Concepts

Integral Mean Value Theorem

The Integral Mean Value Theorem (IMVT) is a handy tool for approximating the average value of a function over an interval. It's a close cousin to the Mean Value Theorem for derivatives. It tells us that for a continuous function \( f \) on an interval \([a, b]\), there is at least one point \( c \) in that interval where:
\[\int_a^b f(x) \, dx = f(c) \, (b-a)\] This means the definite integral of a function can be regarded as if the entire interval had this one average value \( f(c) \). This is especially useful in connecting different parts of calculus.

In our problem, it helps us by showing that the integral of \( \frac{1}{x} \) from \(1\) to \( e^h \) can be expressed as \( \frac{e^h - 1}{\bar{x}} \), where \( \bar{x} \) is some value between \(1\) and \( e^h \). This simplifies handling the integration part when dealing with limits.

Natural Exponential Function

The natural exponential function \( e^x \) is one of the most important functions in mathematics. It has the unique property that its rate of growth is proportional to its current value. This leads to the derivative of \( e^x \) being itself. This makes it ideal for many mathematical models, especially those involving growth and decay.

In the context of our problem, we use this natural function to represent growth, and as \( h \) approaches zero, the behavior of \( e^h \) becomes more evident, approaching 1. This property is crucial for determining limits, such as \( \lim_{h \to 0^+} \frac{e^h - 1}{h} = 1 \). This reflects the underlying principles of calculus where the infinitesimally small changes in \( h \) help us understand the behavior near zero, a common technique in analyzing functions.

Logarithmic Integration

Logarithmic integration involves integrals that include logarithmic functions, commonly arising in calculus when solving for the natural logarithm \( \ln(x) \). To integrate \( \frac{1}{x} \), we recall that it results in \( \ln|x| + C \) where \( C \) is the constant of integration.

In our exercise, integrating \( \frac{1}{x} \) from 1 to \( e^h \) provides us with a natural framework [\( \ln(e^h) - \ln(1) = h \)]. This integral helps to transform multiplication into addition, simplifying how we solve things in calculus.

• This concept shows how logarithms simplify integration tasks.
• It also elegantly connects exponential and logarithmic functions, illustrating their inverse relationship.

Using logarithmic integration alongside the IMVT enables us to connect the value \( \int_{1}^{e^h} \frac{1}{x} \, dx \) to \( \frac{e^h-1}{\bar{x}} \), essential for proving our limit statement.