Question

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

Short answer

\(\frac{1}{3}\)

Step-by-step solution

Identify the Maclaurin Series for relevant functions

A Maclaurin series is a Taylor series expansion of a function about 0. The first few terms of the Maclaurin series for \(\tan x\) are \[\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + O(x^7)\]. The Maclaurin series for \(x\) is simply \[x = x\].

Substitute the series into the limit expression

Substitute the Maclaurin series for \(\tan x\) and \(x\) into the limit expression: \[\lim_{x \rightarrow 0} \frac{\tan x - x}{x^3} = \lim_{x \rightarrow 0} \frac{\left(x + \frac{x^3}{3} + \frac{2x^5}{15} + O(x^7)\right) - x}{x^3}\].

Simplify the series inside the limit

Simplify the numerator by combining like terms: \[\frac{x + \frac{x^3}{3} + \frac{2x^5}{15} - x + O(x^7)}{x^3} = \frac{\frac{x^3}{3} + \frac{2x^5}{15} + O(x^7)}{x^3}\].

Factor out \(x^3\) from the numerator

Factor \(x^3\) out of the numerator: \[\frac{x^3\left(\frac{1}{3} + \frac{2x^2}{15} + O(x^4)\right)}{x^3} = \frac{1}{3} + \frac{2x^2}{15} + O(x^4)\].

Evaluate the limit as \(x \rightarrow 0\)

As \(x\rightarrow 0\), all terms containing \(x\) vanish: \[\lim_{x \rightarrow 0} \left(\frac{1}{3} + \frac{2x^2}{15} + O(x^4)\right) = \frac{1}{3}\].

Key concepts

Taylor series expansion

A Taylor series expansion allows us to express a function as an infinite sum of terms calculated from the values of its derivatives at a single point. When this expansion is about zero, it's called a Maclaurin series. The general form of a Maclaurin series for a function f(x) is \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + ... \]. For example, the Maclaurin series for \( \tan x \) is \( \tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + O(x^7) \). We use this to simplify functions and solve problems like limit evaluation. This approach transforms complex functions into polynomial approximations that are easier to handle.

Limit evaluation

Evaluating limits, especially as x approaches a specific value, is crucial for understanding function behaviors near that point. In our example, \( \lim_{x \rightarrow 0} \frac{\tan x - x}{x^3} \), direct substitution results in an indeterminate form \( \frac{0}{0} \). Using the Maclaurin series transforms our expression, simplifying the process. By expressing \( \tan x - x \) as a power series, we can examine the term-by-term behavior as x approaches 0. After substituting in the series, simplifying, and factoring, you can see that only constant terms (not involving x) matter in the limit.

Trigonometric functions

Trigonometric functions, such as sine, cosine, and tangent, are frequently used in calculus. They're essential in various applications, including geometry, physics, and engineering. The tangent function, \( \tan x \), is particularly interesting because of its periodicity and behavior near zero. When evaluating limits involving trigonometric functions, converting them into their series form simplifies the process. For instance, the Maclaurin series for \( \tan x \) helps break down complex expressions into polynomials. This transformation is often critical when direct substitution is not feasible due to indeterminate forms.

Series simplification

Series simplification is a powerful technique in calculus, enabling us to work with complex expressions more manageably. By representing functions as series, we can focus on the behavior of each term independently. Consider the expression \( \frac{\tan x - x}{x^3} \). By substituting in the Maclaurin series for \( \tan x \) and simplifying, we combine like terms and factor out common elements: \( \frac{x + \frac{x^3}{3} + \frac{2x^5}{15} - x}{x^3} \rightarrow \frac{\frac{x^3}{3} + \frac{2x^5}{15}}{x^3} \rightarrow \frac{1}{3} + \frac{2x^2}{15} \). Evaluating the limit as x approaches 0, we see that terms containing x vanish, leaving us with \( \frac{1}{3} \). This process showcases the power of series simplification in solving limits and other calculus problems.