Problem 20 - Calculus - AP Edition

Question

Find the limits. $$\lim _{x \rightarrow 0} \frac{x-\tan ^{-1} x}{x^{3}}$$

Step-by-Step Solution

Answer

The limit is \(\frac{1}{3}\).

Step 1: Identify the Type of Limit Problem

The given limit problem is\[\lim _{x \rightarrow 0} \frac{x-\tan ^{-1} x}{x^{3}}.\]This is an indeterminate form of type \(\frac{0}{0}\), which suggests that we cannot directly substitute \(x = 0\). Instead, we'll use Taylor expansion for \(\tan^{-1} x\) to tackle this limit.

Step 2: Apply Taylor Expansion to \(\tan^{-1} x\)

The Taylor series expansion of \(\tan^{-1} x\) around \(x = 0\) is given by\[\tan^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots\]For our problem, accuracy up to \(x^3\) term is sufficient since the denominator contains \(x^3\). Thus, we can approximate \(\tan^{-1} x\) as\[\tan^{-1} x \approx x - \frac{x^3}{3}.\]

Step 3: Substitute the Expansion into the Limit

Substitute the expression for \(\tan^{-1} x\) from the Taylor expansion into the limit expression:\[\lim _{x \rightarrow 0} \frac{x - \left(x - \frac{x^3}{3}\right)}{x^3} = \lim _{x \rightarrow 0} \frac{x - x + \frac{x^3}{3}}{x^3}.\]

Step 4: Simplify the Expression

Simplify the expression inside the limit:\[= \lim _{x \rightarrow 0} \frac{\frac{x^3}{3}}{x^3} = \lim _{x \rightarrow 0} \frac{1}{3} = \frac{1}{3}.\]As \(x^3\) cancels out from the numerator and the denominator, leaving us with \(\frac{1}{3}\).

Step 5: Conclude the Limit

Therefore, the limit of the given expression as \(x\) approaches 0 is \(\frac{1}{3}\).

Key Concepts

Indeterminate Forms

In calculus, indeterminate forms are expressions that arise when evaluating limits and cannot be directly computed. They include types like \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\). These forms are so-called because they do not immediately suggest a value—hence, they are considered 'indeterminate'.

In our problem, the expression is \(\lim _{x \rightarrow 0} \frac{x-\tan^{-1} x}{x^{3}}\). At \(x = 0\), the numerator becomes \(0 - \tan^{-1}(0) = 0\), and the denominator is \(0^3 = 0\). Thus, the expression is in the \(\frac{0}{0}\) form.

To deal with indeterminate forms, we often use algebraic manipulation or series expansions to simplify the expression. This allows us to evaluate the limit accurately. Techniques like L'Hôpital's Rule can also be used but may not always be applicable depending on the function. In this exercise, a Taylor series expansion is used to resolve the indeterminate form.

Taylor Series

The Taylor series is a powerful tool in calculus that allows us to represent functions as infinite sums of polynomial terms. Each term in the series is derived from the function's derivatives evaluated at a particular point. For any function that can be expanded, this series provides a means to approximate functions to any degree of accuracy.

For the function \(\tan^{-1} x\), its Maclaurin series (a Taylor series centered at zero) is:

• \(\tan^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots\)

To solve our problem, we only need terms up to \(x^3\) since higher-order terms vanish when \(x\) approaches 0. By substituting \(\tan^{-1} x\) with \(x - \frac{x^3}{3}\), this approximation helps simplify the indeterminate form to allow an easy evaluation of the limit.

Taylor expansions not only help in finding limits, but are also useful in approximating complex functions and solving differential equations.

Arctan Function

The arctan function, also known as the inverse tangent function, is the inverse of the tangent function. It is denoted as \(\tan^{-1} x\) or \(\arctan x\). This function provides the angle whose tangent is \(x\), typically resulting in values in the interval \(-\frac{\pi}{2} \) to \(\frac{\pi}{2}\).

Key properties of \(\tan^{-1} x\) include:

• It is an odd function, meaning \(\tan^{-1}(-x) = -\tan^{-1}(x)\).
• The derivative of \(\tan^{-1} x\) is \(\frac{1}{1+x^2}\), which is useful in calculus for integration and Taylor series.

In our example, the behavior of the arctan function around zero is particularly important, as it allows expanding via Taylor series for limit evaluation. The approximation using Taylor terms highlights how minor differences between functions can impact solving limits. Recognizing these patterns is essential in understanding both simple and complex mathematical models.