Problem 62 - Calculus: Early Transcendentals

Question

Use series to evaluate the limit. \( \lim_{x \to 0} \frac {1 - \cos x}{1 + x - e^x} \)

Step-by-Step Solution

Answer

The limit is -1.

Step 1: Identifying Relevant Series Expansions

The problem involves evaluating a limit with functions that are often expressed in terms of their series expansions. The functions are \( \cos x \) and \( e^x \). The series expansion for \( \cos x \) is \( 1 - \frac{x^2}{2} + \frac{x^4}{24} - \ldots \) and for \( e^x \) is \( 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \ldots \).

Step 2: Expanding Cosine Function

For small values of \( x \), \( 1 - \cos x \approx \frac{x^2}{2} \) using the series expansion of the cosine function \( \cos x = 1 - \frac{x^2}{2} + \ldots \), thus the first few terms suffice for \( \cos x \).

Step 3: Expanding Exponential Function

Similarly, we expand \( e^x \) using its series expansion: \( e^x = 1 + x + \frac{x^2}{2} + \ldots \). The terms relevant for small \( x \) up to \( x^2 \) are enough for evaluation.

Step 4: Creating Simplified Numerator and Denominator

Using the series expansions, the numerator becomes \( 1 - \cos x = \frac{x^2}{2} \). The denominator becomes \( 1 + x - e^x \approx 1 + x - (1 + x + \frac{x^2}{2}) = -\frac{x^2}{2} \).

Step 5: Forming the Limit Expression

The limit expression becomes \( \lim_{x \to 0} \frac{\frac{x^2}{2}}{-\frac{x^2}{2}} = \lim_{x \to 0} -1 \).

Step 6: Evaluating the Limit

Since the \( x^2 \) terms cancel out, the limit simplifies to \( -1 \), which is independent of \( x \) as \( x \to 0 \).

Key Concepts

Cosine Series

One of the key elements in evaluating the given limit is understanding the cosine series. The cosine series is a representation of the cosine function as an infinite sum of terms. For small angle approximations, this is particularly useful. The standard series expansion for cosine is:

• \( \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots \)

For very small values of \( x \), higher order terms (such as \( x^4 \) and above) contribute insignificantly, so they can often be ignored in calculations.
In the original problem, this series allows us to approximate \( \cos x \) such that the expression \( 1 - \cos x \) becomes manageable. Specifically, for \( x \to 0 \), we get:

• \( 1 - \cos x \approx \frac{x^2}{2} \)

This simplification is fundamental to evaluating the limit in a straightforward manner. Recognizing when these series expansions help simplify complex expressions is a crucial skill in calculus.

Exponential Series

The exponential series deals with the expansion of the exponential function into a power series. This is another powerful tool for handling limits, especially when dealing with functions involving \( e^x \). The series expansion for \( e^x \) is given by:

• \( e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots \)

The exponential series is invaluable because it provides a way to express exponential growth in terms that are easier to manipulate, particularly for small values of \( x \).
In the context of limits as in the original exercise, these lower order terms are typically sufficient, as higher order terms become negligible when \( x \to 0 \). Using the series expansion:

• \( e^x \approx 1 + x + \frac{x^2}{2} \)

This simplification aids in forming expressions that are more tractable. For this problem, we saw that the difference \( 1 + x - e^x \) led to a cancellation of the constant and linear terms, leaving \( -\frac{x^2}{2} \). Mastering series expansions can greatly enhance your ability to tackle limit problems and gain deeper insights into calculus concepts.

Calculus Limits

Limits are foundational in calculus as they enable the understanding of behavior of functions at points of interest, often where direct substitution isn't feasible. In this exercise, the goal was to evaluate the limit as \( x \to 0 \). By applying series expansions, we converted complex expressions into forms where limits are simpler to compute.
To evaluate the given limit, the important steps were:

• **Express** both the numerator and denominator using series expansions for easy simplification.
• **Simplify** the resulting expression as terms cancel out or become negligible.
• **Compute** the limit, which in this problem led to a constant expression of \(-1\).

The result is independent of \( x \), showing that approaching zero, the expression stabilizes to a constant value. Understanding and applying limits with series expansions illuminate the behavior of complex functions, and using calculus techniques like these makes solving such problems efficient and elegant.