Problem 30 - Thomas Calculus

Question

Use series to evaluate the limits. \begin{equation} \lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{x} \end{equation}

Step-by-Step Solution

Answer

The limit is 2.

Step 1: Understand the Problem

We want to find the limit of \( \frac{e^x - e^{-x}}{x} \) as \( x \rightarrow 0 \). This is an indeterminate form \( \frac{0}{0} \), which suggests using series expansion.

Step 2: Taylor Series Expansion

Use the Taylor series for \( e^x \) and \( e^{-x} \). The series for \( e^x \) is \( 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \ldots \). Similarly, \( e^{-x} \) is \( 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \ldots \).

Step 3: Subtract the Series Expansions

Subtract the two series:\[e^x - e^{-x} = (1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \ldots) - (1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \ldots) = 2x + \frac{2x^3}{6} + \ldots\]

Step 4: Simplify the Expression

Focus on the lowest order of \( x \):\[e^x - e^{-x} \approx 2x\] as higher order terms vanish when dividing by \( x \) in the limit as \( x \rightarrow 0 \).

Step 5: Evaluate the Limit

Now consider the limit:\[\lim_{x \rightarrow 0} \frac{e^x - e^{-x}}{x} = \lim_{x \rightarrow 0} \frac{2x}{x} = 2.\] As \( x \rightarrow 0 \), higher-order terms in \( x \) vanish, leaving only the constant.

Key Concepts

Series Expansion

A series expansion breaks down a function into an infinite sum of terms. It allows us to approximate complex functions as simple polynomials centered around a specific point.
In calculus, series expansions are particularly useful for analyzing behavior near that point. We often focus on the Taylor series, one of the most common forms of series expansion.
With a series expansion, complex functions like exponential functions, trigonometric functions, or logarithms can be represented in a simpler polynomial form.

• This simplification becomes crucial when evaluating limits of functions that result in indeterminate forms.
• The terms of the series typically become smaller as we consider higher orders, allowing us to ignore them for an approximation.

Series expansion is a powerful mathematical tool, helping to solve problems where direct computation could be complex or impossible.

Taylor Series

The Taylor series approximates functions as an infinite sum of terms calculated from the function's derivatives at a single point. The general form of a Taylor series about a point is: \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \]
In the exercise, we used the Taylor series of the exponential functions at the expansion point (0).

• The series for \(e^x\) near 0 is \(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \ldots\).
• Similarly, \(e^{-x}\) is \(1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \ldots\).

We subtracted these two series to eliminate constant terms, honing in on the expression involving \(x\).The Taylor series provides not just an approximation but also reveals the rate and nature of a function's changes around a point.

Indeterminate Forms

Indeterminate forms often arise in calculus when the limits result in undefined expressions like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These forms suggest a potential to simplify the expression further before evaluating the limit.
In the given problem, the expression \( \frac{e^x - e^{-x}}{x} \) results in the indeterminate form \(\frac{0}{0}\) as \(x\) approaches zero.

• This indicates that without simplification, the expression is undefined at this point.
• Using series expansions enables the simplification by finding a common factor to cancel out, resolving the indeterminate form smoothly.

Because indeterminate forms indicate an inconclusive limit, resolving them through methods like series expansion is essential for accurate limit evaluation.

Exponential Functions

Exponential functions are characterized by a constant base raised to a variable exponent. They are used to model rapid growth or decay in natural phenomena.
The continuous development of compound interest, population growth, or radioactive decay are typical applications.
In calculus, the function \(e^x\) simplifies differentiation and integration processes due to its unique property:

• The derivative and integral of \(e^x\) are both \(e^x\).

When dealing with limits and series, the exponential function becomes invaluable.The Taylor series provides a straightforward approximation of \(e^x\) around any given point, such as 0.This convenience makes exponential functions favored in solving differential equations, evaluating complex limits, and in numerous asymptotic analyses.