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Chapter 30: Problem 33

Solve the given problems. $$\text { Evaluate } \lim _{x \rightarrow 0} \frac{\sin x-x}{x^{3}} \text { by using the expansion for } \sin x$$

Short Answer

Expert verified
The limit is \(-\frac{1}{6}\).

Step by step solution

01

Understanding the Problem

We need to evaluate the limit \( \lim_{x \rightarrow 0} \frac{\sin x - x}{x^3} \). To do this, we'll use the Taylor series expansion for \( \sin x \). Recall that the Taylor series for \( \sin x \) around \( x = 0 \) is \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \).
02

Substitute the Series into the Expression

Substitute the Taylor series expansion of \( \sin x \) into the expression: \( \frac{\sin x - x}{x^3} = \frac{x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots - x}{x^3} \). Simplifying inside the numerator gives \( \frac{-\frac{x^3}{3!} + \frac{x^5}{5!} - \cdots}{x^3} \).
03

Cancel and Simplify the Expression

Simplify the expression by canceling the common factor of \( x^3 \) in the numerator and the denominator: \( \frac{-\frac{x^3}{6} + \frac{x^5}{120} - \cdots}{x^3} = -\frac{1}{6} + \frac{x^2}{120} - \cdots \). As \( x \rightarrow 0 \), the higher powers of \( x \) approach zero.
04

Evaluate the Limit

Take the limit as \( x \rightarrow 0 \): \( \lim_{x \rightarrow 0} \left(-\frac{1}{6} + \frac{x^2}{120} - \cdots\right) = -\frac{1}{6} \). The terms involving \( x^2 \) and higher powers vanish as \( x \rightarrow 0 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Taylor Series
In calculus, the Taylor series is a powerful tool used to approximate functions. It provides a series expansion of a function around a specific point. For instance, the Taylor series expansion of a function \( f(x) \) around \( x = a \) is given by:\[f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots\]The series continues indefinitely, capturing more characteristics of the original function with each additional term.
  • Important Points:
    • Convergence: A Taylor series might not converge everywhere but usually does near the point \( a \).
    • General Use: Most useful when functions are hard to compute directly, but you want to evaluate them around a small interval.
    • Calculus Application: Provides a simple way to evaluate limits, derive differential equations, and more.
Sine Function Expansion
The sine function \( \sin x \) can be expanded as a Taylor series around 0, known as the Maclaurin series. The Maclaurin series takes the form:\[\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots\]This series is very handy when calculating limits or integrals involving \( \sin x \).
  • Utility:
    • Evaluating Limits: When working with indeterminate forms like \( \lim_{x \to 0} \frac{\sin x}{x} \).
    • Approximation: Offers a way to compute \( \sin x \) quickly when \( x \) is small.
  • Characteristics: The series alternates and decreases in magnitude, which helps with convergence.
Evaluating Limits
Evaluating limits is a fundamental concept in calculus that often involves simplifying an expression as a variable approaches a certain value. When direct substitution in a limit results in an indeterminate form like \( \frac{0}{0} \), special techniques such as L'H么pital's Rule or series expansion can be used.
For example, in the problem \( \lim_{x \rightarrow 0} \frac{\sin x - x}{x^3} \):
  • Use the expansion \( \sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots \)
  • Simplify: Substitute and simplify the expression to \( -\frac{1}{6} + \frac{x^2}{120} \).
  • Limit Evaluation: As \( x \to 0 \), higher power terms vanish, resulting in the limit value \( -\frac{1}{6} \).

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Most popular questions from this chapter

Find the indicated series by the given operation. Find the first four nonzero terms of the expansion of the function \(f(x)=\frac{1}{2}\left(e^{x}-e^{-x}\right)\) by subtracting the terms of the appropriate series. The result is the series for \(\sinh x\). (See Exercise 55 of Section \(27.6 .)\)

Find the indicated series by the given operation. By using the properties of logarithms and the series for \(\ln (1+x)\) find the series for \(\ln \frac{1+x}{1-x}\)

Calculate the value of each of the given functions. Use the expansion for \(\sqrt{1+x},\) and in Exercises 23 and \(24,\) use the expansion for \(\sqrt[3]{1+x} .\) Use three terms. $$\sqrt[3]{1.1392}$$

Find the indicated series by the given operation. Show that by differentiating term by term the expansion for \(e^{x}\) the result is also the expansion for \(e^{x}\)

Calculate the value of each of the given functions. Use the expansion for \(\sqrt{1+x},\) and in Exercises 23 and \(24,\) use the expansion for \(\sqrt[3]{1+x} .\) Use three terms. $$\sqrt{1.1076}$$
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