Problem 61 - Calculus for Biology and Medicine

Question

Let \(f(x)=\ln x .\) We know that \(f^{\prime}(x)=\frac{1}{x} .\) We will use this fact and the definition of derivatives to show that $$\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n}=e$$ (a) Use the definition of the derivative to show that $$f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{\ln (1+h)}{h}$$ (b) Show that (a) implies that $$\ln \left[\lim _{h \rightarrow 0}(1+h)^{1 / h}\right]=1$$ (c) Set \(h=\frac{1}{n}\) in (b) and let \(n \rightarrow \infty\). Show that this implies that $$\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n}=e$$

Step-by-Step Solution

Answer

\(\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e\) using the derivative of \(\ln x\).

Step 1: Understanding the Derivative at a Point

To find the derivative of a function at a point using the definition, we use \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}. \]For our function \( f(x) = \ln x \), we evaluate the derivative at \( x = 1 \), so it becomes \[ f'(1) = \lim_{h \to 0} \frac{\ln(1+h) - \ln 1}{h}. \]Since \( \ln 1 = 0 \), this simplifies to \[ f'(1) = \lim_{h \to 0} \frac{\ln(1+h)}{h}. \]

Step 2: Show Derivative Equals 1

Since \(f'(x) = \frac{1}{x}\), substituting \(x = 1\) gives us \(f'(1) = 1\). This means that \[ \lim_{h \to 0} \frac{\ln(1+h)}{h} = 1. \]

Step 3: Expressing in Terms of Limit

Using the result from Step 2, we rewrite it in a form that involves limits:\[ \ln \left( \lim_{h \to 0} (1+h)^{1/h} \right) = 1. \]This shows that the natural logarithm of the limit expression equals 1, suggesting a base of \(e\).

Step 4: Substitution and Simplification

In Step 3, set \(h = \frac{1}{n}\). As \(h \to 0\), it implies that \(n \to \infty\). Therefore, the expression becomes:\[ \ln \left( \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \right) = 1. \]Taking the exponential of both sides gives:\[ \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e. \]

Key Concepts

Derivative of Natural Logarithm

When we talk about the natural logarithm, we often refer to its derivative properties. The natural logarithm, denoted as \( \ln x \), is a fundamental function in calculus, particularly when exploring exponential growth and decay. The derivative of \( \ln x \) is \( \frac{1}{x} \), which reflects how the function changes as \( x \) increases or decreases.

This derivative tells us the rate of change of \( \ln x \) at any point \( x \). For instance, evaluating the derivative at \( x = 1 \) gives us \( f'(1) = \frac{1}{1} = 1 \). In simple terms, this means that at \( x = 1 \), the slope of the tangent to the curve of \( \ln x \) is exactly 1.

This property is pivotal when applying the limit definition of exponential functions, as it helps us establish fundamental constants such as \( e \). The derivative of the natural logarithm gives us a foundation for proving how expressions like \( \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e \) behave.

Definition of Derivative

The definition of a derivative is a classical concept in calculus. It's centered around understanding the instant rate of change of a function at a specific point. To find the derivative of a function \( f(x) \), we use the limit definition:

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

This expression looks at the difference between the function value at \( x+h \) and the value at \( x \), divided by the small increment \( h \). As \( h \) approaches zero, this quotient gives us the slope of the tangent line at point \( x \), which is the derivative.

For the natural logarithm, \( f(x) = \ln x \), applying this formula at \( x = 1 \) looks like this:

\[ f'(1) = \lim_{h \to 0} \frac{\ln(1+h) - \ln 1}{h} \]

Knowing \( \ln 1 = 0 \), simplifies the expression to:

\[ f'(1) = \lim_{h \to 0} \frac{\ln(1+h)}{h} \]

This neat simplification underpins a crucial step in determining how \( e \), the base of natural logarithims, is defined through limits and exponential function expressions.

Limit Definition

A limit is a fundamental concept in calculus used to find the value that a function approaches as the input approaches a particular point. The limit definition becomes essential when exploring expressions like \( \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e \).

In this context, consider setting \( h = \frac{1}{n} \), which means as \( h \to 0 \), \( n \) will be heading towards infinity. This transforms the limit to:

• The expression \( (1+h)^{1/h} \)
• When \( \to \), converges to \( e \)

This particular limit is the foundation for defining the number \( e \), which is approximately equal to 2.718. By showing that:

\[ \ln \left( \lim_{h \to 0} (1+h)^{1/h} \right) = 1 \]

We confirm that:

\[ \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e \]

This representation, both intuitive and precise, reveals how the concept of limits is crucial for understanding exponential growth, logarithms, and the natural base \( e \). It further demonstrates the profound interconnections between calculus concepts.