Question

Use Maclaurin series to evaluate the limits. \(\lim _{x \rightarrow 0} \frac{\sin x-x}{x^{3}}\)

Short answer

The limit is \ -\frac{1}{6} \.

Step-by-step solution

Write the Maclaurin series for sin(x)

The Maclaurin series expansion for \ \( \sin(x) \ \) is: \ \[ \sin(x) = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \cdots \ \] For small values of \( x \), the higher order terms become negligible.

Substitute the series into the limit expression

Replace \ \( \sin(x) \ \) in the limit expression with its Maclaurin series: \ \[ \lim_{x \rightarrow 0} \frac{x - \frac{x^{3}}{3!} + \cdots - x}{x^{3}} \ \]

Simplify the expression

Combine like terms to simplify the numerator: \ \[ \lim_{x \rightarrow 0} \frac{x - x - \frac{x^{3}}{6}}{x^{3}} = \lim_{x \rightarrow 0} \frac{- \frac{x^{3}}{6}}{x^{3}} \ \] Simplifying further, this becomes: \ \[ \lim_{x \rightarrow 0} \frac{-1}{6} \ \]

Evaluate the limit

Since there are no more variables left to evaluate as \( x \) approaches zero, the limit directly simplifies to: \ \[ \frac{-1}{6} \ \]

Key concepts

Limit Evaluation

Evaluating limits is an essential concept in calculus where we analyze what happens to a function as the input approaches a particular value. Here, we focus on the limit as x approaches 0. In this exercise, we utilize the Maclaurin series expansion of the trigonometric function to simplify our computation. Let's delve into the steps.

When we have a limit expression involving functions like \(\frac{\text{sin(x)} - x}{x^3}\), directly substituting x=0 can lead to an indeterminate form (like \(\frac{0}{0}\)), which is undefined. To resolve this, we expand \(\text{sin(x)}\) as a Maclaurin series, making the expression easier to limit evaluate.

Once the series is substituted, and after like terms are combined, simplifying further becomes straightforward, allowing us to find the limit value.

Series Expansion

Series expansions, like the Maclaurin series, provide a way to represent functions as an infinite sum of terms. For small values of x, we can often approximate functions using only the first few terms of the series. This makes complex functions easier to handle.

The Maclaurin series for \(\text{sin(x)}\) is:
\[ \text{sin(x)} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \text{higher-order terms} \]
For limits near x=0, higher-order terms (terms with x raised to powers greater than 3) can be neglected.

By using this truncated series, we can substitute it into our limit expression and greatly simplify the problem. Here, the key step is to understand how to truncate the series correctly, and then substitute it to simplify the numerator of the fraction.

Trigonometric Functions

Trigonometric functions, like \(\text{sin(x)}\), play a central role in calculus. Understanding their series expansions helps tackle problems involving limits and derivatives.

The function \(\text{sin(x)}\) can be expanded using the Maclaurin series:
\[ \text{sin(x)} = x - \frac{x^3}{6} + \frac{x^5}{120} - \text{higher-order terms} \]
This series gives us a polynomial approximation for \(\text{sin(x)}\) for very small x. When solving limits or manipulating functions close to x=0,accurately using these series allows us to simplify and evaluate expressions that would otherwise be indeterminate.

Remember, the power of using trigonometric series expansions lies in their ability to approximate functions and make complex limit evaluations more approachable and understandable.