Problem 32 - Thomas Calculus

Question

Use series to evaluate the limits. \begin{equation} \lim _{\theta \rightarrow 0} \frac{\sin \theta-\theta+\left(\theta^{3} / 6\right)}{\theta^{5}} \end{equation}

Step-by-Step Solution

Answer

The limit evaluates to \( \frac{1}{120} \).

Step 1: Taylor Series Expansion for Sine

We begin by using the Taylor series expansion for \( \sin \theta \) around 0: \[ \sin \theta = \theta - \frac{\theta^3}{6} + \frac{\theta^5}{120} - \ldots \] We will expand to at least the cubic term since we are given \( \theta^3/6 \) in the expression and need terms up to \( \theta^5 \) in the denominator to match the order of the given limit.

Step 2: Substitute Expansion into Expression

Substitute the Taylor expansion into the given limit expression:\[ \lim_{\theta \rightarrow 0} \frac{\sin \theta - \theta + \frac{\theta^3}{6}}{\theta^5} = \lim_{\theta \rightarrow 0} \frac{\left( \theta - \frac{\theta^3}{6} + \frac{\theta^5}{120} \right) - \theta + \frac{\theta^3}{6}}{\theta^5} \]

Step 3: Simplify the Expression

Simplify the expression by canceling terms:- The \( \theta \) terms cancel because \( \theta - \theta = 0 \).- The \( \theta^3/6 \) terms also cancel each other out.The expression simplifies to:\[ \lim_{\theta \rightarrow 0} \frac{\frac{\theta^5}{120}}{\theta^5} \]

Step 4: Evaluating the Limit

Cancel \( \theta^5 \) in the numerator and denominator:\[ \lim_{\theta \rightarrow 0} \frac{\frac{\theta^5}{120}}{\theta^5} = \lim_{\theta \rightarrow 0} \frac{1}{120} \]Since the limit is now a constant, simply evaluate to get:\[ \frac{1}{120} \]

Key Concepts

Understanding Limits in Calculus

The concept of limits is a foundational element in calculus. It helps us understand the behavior of functions as they approach a specific point. Limits are crucial when we deal with functions that might not even be defined at that particular point.

In calculus, we use limits to analyze how a function behaves near a certain input, usually as the input either gets very small or very large. The notation for a limit, such as \( \lim_{x \to a} f(x) = L\), means that as \(x\) gets closer and closer to \(a\), the value of \(f(x)\) approaches \(L\).

Here are some facts about limits:

• Existence: For a limit to exist at a point \(a\), both the left and right-hand limits must exist and be equal.
• Infinite Limits: When a function's value increases or decreases without bound as it approaches a specific point.
• One-Sided Limits: Limits can be taken from either the left (denoted \( \lim_{x \to a^-}\)) or the right (denoted \( \lim_{x \to a^+}\)).

Series Expansion as a Tool in Calculus

Series expansion is a powerful technique in calculus that allows us to represent functions as infinite sums of terms. These expansions make complex functions easier to work with, especially near a specific point.

The most common type of series expansion used is the Taylor series, which expresses functions as an infinite sum of terms calculated from the values of its derivatives at a single point.

Here’s an overview of Taylor series:

• Formula: The Taylor series for a function \(f\) at \(x = a\) is given by \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \]
• Maclaurin Series: A special case of the Taylor series where \(a = 0\), useful for approximating functions around zero.
• Convergence: A Taylor series converges to the function within a radius of convergence, which depends on the function’s nature.

In practice, Taylor series are invaluable for evaluating limits, as seen with expansions like \( \sin \theta \), making it easier to handle expressions that arise from complex functions like sines and cosines.

Incorporating Limits and Series Expansion in Calculus Problems

Calculus often requires combining different concepts to solve a problem, a good example being the use of series expansion to evaluate limits.

This method involves breaking down a function into a simpler form using series expansion, as seen with the Taylor series for \( \sin \theta \). Once simplified, we can then apply the limit to find the value that the function approaches, as we did when simplifying \( \frac{\sin \theta - \theta + \left(\theta^{3} / 6\right)}{\theta^{5}} \).

Steps to follow include:

• Expand: Use series expansions to express terms in a simpler form.
• Substitute and Simplify: Perform algebraic operations to cancel out terms and simplify the expression.
• Evaluate the Limit: After simplification, evaluate the limit of the simplified expression.

By mastering such techniques, tackling calculus problems involving complex functions can become more approachable. Moreover, it highlights the elegance and interconnectedness of calculus concepts, making seemingly challenging problems solvable.