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: Write the Taylor Series of the Exponentials

The Taylor series expansion for the exponential function around 0 is given by \[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \] and the series expansion for \( e^{-x} \) is:\[ e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \ldots \]

Step 2: Substitute and Simplify the Expression

Substitute the Taylor series expansions of \( e^x \) and \( e^{-x} \) into the original limit expression:\[ \lim _{x \rightarrow 0} \frac{(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots) - (1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \ldots)}{x} \]Simplify the numerator:\[ e^x - e^{-x} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \ldots - (1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \ldots) \ = 2x + \frac{2x^3}{6} + \ldots \]

Step 3: Divide by x and Take the Limit

Substitute the simplified expression back into the limit:\[ \lim _{x \rightarrow 0} \frac{2x + \frac{2x^3}{6} + \ldots}{x} \]Cancel out \( x \) in the numerator and denominator:\[ \lim _{x \rightarrow 0} (2 + \frac{2x^2}{6} + \ldots) \]As \( x \) approaches 0, higher order terms vanish, thus:\[ \lim _{x \rightarrow 0} 2 = 2 \]

Key Concepts

Taylor series

When solving problems involving limits, the Taylor series is a fundamental tool. It allows us to express functions like exponentials in an infinite sum of terms that involve powers of a variable. These expansions converge to the original function for values near the point of expansion. Here's a simple breakdown:

• The Taylor series for a function at a particular point averages out its derivatives to approximate the function's value close to that point.
• In our problem, we use the Taylor series of the exponential function, which is given as: \[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \]

By rewriting complicated functions into a sum of simpler terms, you can more easily manipulate and simplify expressions, particularly when you're dealing with limits, as substitutions help reveal hidden simplifications.

exponential functions

Exponential functions, like \( e^x \), play a crucial role in calculus, appearing in many limits and differentials. Here's why they're vital:

• They grow quickly, making their derivatives especially notable, as the derivative of \( e^x \) is \( e^x \) itself.
• These functions are often expanded using series, like the Taylor series, making them easier to handle analytically.

For our specific limit, we need the expansions of both \( e^x \) and \( e^{-x} \), the latter being:\[ e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \ldots \]This results from the property that replacing \( x \) with \( -x \) changes the sign of all odd terms. This clever technique allows us to construct a manageable difference by simplifying across the series.

symbolic limits

Symbolic limits involve analyzing expressions as one variable approaches a specific value, often zero or infinity. Consider these key points:

• When a limit includes a function, representing it using series expansions can reveal simplifications.
• In our exercise, after substituting Taylor series expansions for \( e^x \) and \( e^{-x} \) and simplifying, we find the expression boils down to eliminated terms where possible.

The limit \( \lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{x} \) simplified into\[ \lim _{x \rightarrow 0} (2 + \frac{2x^2}{6} + \ldots) \]. As \( x \) becomes very small, all terms with \( x \) in the numerator vanish, leaving:\[ \lim _{x \rightarrow 0} 2 = 2 \].By understanding these steps, you can approach limits systematically using series expansions, making them accessible for evaluation.