Problem 53 - Essential Calculus Early Transcendentals

Question

Use series to cvaluate the limit. $$\lim _{x \rightarrow 0} \frac{\sin x-x+\frac{1}{6} x^{3}}{x^{5}}$$

Step-by-Step Solution

Answer

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

Step 1: Recognize the Series Expansion

We know that the Maclaurin series expansion for \( \sin x \) is \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \). Since this series is centered at 0, it helps us approximate \( \sin x \) near \( x = 0 \).

Step 2: Substitute the Series into the Expression

Substitute \( \sin x \) with its series representation in the limit expression: \[ \lim_{x \rightarrow 0} \frac{(x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots) - x + \frac{x^3}{6}}{x^5} \].

Step 3: Simplify the Expression

Simplify the numerator: \( x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots - x + \frac{x^3}{6} = \frac{x^5}{120} - \cdots \). This simplifies further as higher order terms become negligible as \( x \rightarrow 0 \).

Step 4: Evaluate the Limit

Substitute back into the limit and simplify: \[ \lim_{x \rightarrow 0} \frac{\frac{x^5}{120}}{x^5} = \lim_{x \rightarrow 0} \frac{1}{120} = \frac{1}{120} \]. Since all higher order terms do not affect the outcome in this limit scenario, the solution is correctly evaluated.

Key Concepts

Limit of a Function

Understanding the concept of a limit is fundamental in calculus, as it helps us determine the behavior of functions as they approach a specific point. In this exercise, we want to find the limit as \(x\) approaches zero for the given expression.

The expression involves the sine function, which we approximate using the Maclaurin series to simplify the calculation.
When we're looking for the limit, our primary focus is often on what happens to our function values as we get very close to the point of interest, not necessarily at the point.

In general, the limit \(\lim_{x \to a} f(x) = L\) implies that as \(x\) gets closer and closer to \(a\), the values of \(f(x)\) get closer and closer to \(L\). In this context, as \(x\) approaches zero, the original function can initially seem complex. However, by substituting with a series expansion, we simplify to terms that make direct evaluation possible, leading us to the solution of \(\frac{1}{120}\).

Sinusoidal Series Expansion

The Maclaurin series is the key to solving this exercise, as it allows us to represent a function as an infinite sum of terms calculated from the values of its derivatives at a single point.

The Maclaurin series expansion for the sine function is particularly useful because it provides an approximation that is highly accurate near \(x = 0\). For \(\sin x\), the expansion is given by:

• \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \)

In our task, we substituted the series expansion for \(\sin x\) into the limit expression. This substitution allowed us to replace potentially complicated trigonometric terms with simple polynomial terms. By working with polynomials, we gain the ability to better manipulate and simplify the overall expression. This approach is efficient because as \(x\) approaches zero, higher-order terms of \(x\) have negligible impact, leaving us a clean path to our answer.

Calculus Step by Step

The step-by-step breakdown of the given problem illustrates how calculus techniques can simplify seemingly complex expressions.

Starting with the Maclaurin series expansion, we translated the problem into a polynomial form that is easier to handle.

• **Recognize the Series Expansion:** Identify the relevant series to express \(\sin x\).
• **Substitute and Simplify:** Replace \(\sin x\) with its series form, making sure to simplify the expression by canceling out terms.
• **Evaluate the Limit:** Pay close attention to about how terms behave as \(x\) approaches zero, concentrating only on impactful terms.

Each step builds upon the previous one, guiding us through a logical progression from recognizing the series expansion, substituting it into our expression, and finally evaluating the simplified limit. By breaking down the problem into these distinct phases, it becomes much more approachable and highlights the power of calculus in solving limits through series expansions.