Problem 40 - Thomas Calculus

Question

Use series to evaluate the limits. \begin{equation} \lim _{x \rightarrow 0} \frac{\ln \left(1+x^{3}\right)}{x \cdot \sin x^{2}} \end{equation}

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Series Expansion of ln(1+x³)

Apply the Taylor series expansion for \( \ln(1+y) \) around 0, which is \( \ln(1+y) = y - \frac{y^2}{2} + \frac{y^3}{3} - \cdots \). Substituting \( y = x^3 \), the series becomes \( \ln(1+x^3) = x^3 - \frac{(x^3)^2}{2} + \frac{(x^3)^3}{3} - \cdots = x^3 - \frac{x^6}{2} + \frac{x^9}{3} - \cdots \). For small x, we approximate this as \( \ln(1+x^3) \approx x^3 \).

Step 2: Series Expansion of sin(x²)

Use the Taylor series expansion for \( \sin z \) around 0: \( \sin z = z - \frac{z^3}{6} + \frac{z^5}{120} - \cdots \). Substituting \( z = x^2 \), the expansion becomes \( \sin(x^2) = x^2 - \frac{(x^2)^3}{6} + \cdots = x^2 - \frac{x^6}{6} + \cdots \). For small x, we approximate this as \( \sin(x^2) \approx x^2 \).

Step 3: Simplifying the Fraction

Insert the approximations from the series expansions into the limit expression. We have: \( \frac{\ln(1 + x^3)}{x \cdot \sin(x^2)} \approx \frac{x^3}{x \cdot x^2} = \frac{x^3}{x^3} = 1 \).

Step 4: Calculate the Limit

Now that the expression simplifies to 1, the limit as \( x \rightarrow 0 \) of this expression is simply \( 1 \).

Key Concepts

Limits

Limits are a fundamental concept in calculus used to understand the behavior of functions as they approach a specific point. Think of a limit as a way to predict the value a function gets closer to as the input approaches a certain value. For example, in the problem given, we are looking at what happens to the function \( \frac{\ln(1+x^3)}{x \cdot \sin x^2} \) as \( x \) approaches 0.

• In such cases, directly substituting the value can lead to indeterminate forms like \( \frac{0}{0} \).
• This is where clever algebra or series expansions become handy.
• Understanding limits is crucial for techniques such as finding derivatives and integrals.

By using limits, you gain deeper insights into how functions behave infinitely close to certain points, bolstering your overall mathematical intuition.

Series Expansion

Series expansion is a powerful tool that involves expressing a function as an infinite sum of terms calculated from the values of its derivatives at a particular point. In our problem, series expansions help break down complex functions into simpler polynomial forms. For the natural logarithm \( \ln(1+y) \), the series expansion around 0 is given by:

• \( \ln(1+y) = y - \frac{y^2}{2} + \frac{y^3}{3} - \cdots \)

By substituting \( y = x^3 \), we obtain the series for \( \ln(1+x^3) \). The beauty of series expansion is its ability to approximate complex functions with manageable terms, especially when \( x \) is small. This makes it easier to compute limits and solve indeterminate forms.

Taylor Polynomial

A Taylor polynomial is a finite portion of a Taylor series, which approximates a function near a particular point using a polynomial. It leverages derivatives of the function to form this approximation. The more terms we include from the series, the closer our approximation becomes to the actual function. For example, if we consider up to the first few terms of \( \sin(x^2) \), we use:

• \( \sin z = z - \frac{z^3}{6} + \cdots \)
• Substitute \( z = x^2 \), resulting in \( \sin(x^2) \approx x^2 \).

Using the first term is often enough for approximations when \( x \) is very small. Taylor polynomials hence provide a straightforward way to estimate functions in calculus, playing a key role in computing limits.

Trigonometric Series

Trigonometric series, like the Taylor series for trigonometric functions, are important for simplifying complex expressions, especially when variables are near zero. In this exercise, the trigonometric function \( \sin(x^2) \) is expanded using:

• \( \sin z = z - \frac{z^3}{6} + \frac{z^5}{120} - \cdots \)

This allows for the approximation \( \sin(x^2) \approx x^2 \) when dealing with small \( x \). These kinds of series help transform difficult-to-handle functions into more manageable polynomial approximations:

• Such approximations make solving limits practical and less cumbersome.
• They avoid dealing with complex forms at zero, overcoming the hurdles of indeterminate expressions.

The ability to approximate trigonometric functions accurately is essential in calculus, providing crucial insight for solving a variety of mathematical problems.