Problem 37 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The limit is 2.

Step 1: Identify Series Expansion for the Numerator

The numerator is \( \ln(1+x^2) \). For small \( x \), use the series expansion \( \ln(1+u) \approx u - \frac{u^2}{2} + \frac{u^3}{3} - \ldots \). Substituting \( x^2 \) for \( u \), we get \( \ln(1+x^2) \approx x^2 - \frac{x^4}{2} + \frac{x^6}{3} - \ldots \).

Step 2: Identify Series Expansion for the Denominator

The denominator is \( 1 - \cos x \). The series expansion for \( \cos x \) is \( 1 - \frac{x^2}{2} + \frac{x^4}{24} - \ldots \). Thus, \( 1 - \cos x \approx \frac{x^2}{2} - \frac{x^4}{24} + \ldots \).

Step 3: Simplify the Series Expressions

From the previous steps, we have \( \ln(1+x^2) \approx x^2 - \frac{x^4}{2} + \ldots \) and \( 1 - \cos x \approx \frac{x^2}{2} - \ldots \). Focus on leading terms, where \( \ln(1+x^2) \) is \( x^2 \) and \( 1 - \cos x \) is \( \frac{x^2}{2} \).

Step 4: Form the Limit Expression

Substitute the simplified series expressions into the limit: \[ \lim_{x \to 0} \frac{x^2 - \frac{x^4}{2} + \ldots}{\frac{x^2}{2} - \ldots} \]

Step 5: Cancel Dominant Terms and Evaluate the Limit

Cancel \( x^2 \) from the numerator and \( \frac{x^2}{2} \) from the denominator to simplify the fraction: \[ \lim_{x \to 0} \frac{2(x^2 - \frac{x^4}{2} + \ldots)}{x^2} = \lim_{x \to 0} \frac{2 - x^2 + \ldots}{1} \]. Since higher-order terms go to zero, the limit is 2.

Key Concepts

Series Expansion

Series expansion is a powerful tool for approximating functions around a particular point. It allows complex functions to be expressed in a sum of terms, which are polynomials of different degrees. This is particularly useful for calculating limits, especially when the limit takes the form of an indeterminate form like \(\frac{0}{0}\).

For instance, the function \(\ln(1+x^2)\) can be expanded using the known series expansion for \(\ln(1+u)\), which is \(u - \frac{u^2}{2} + \frac{u^3}{3} - \ldots\). By replacing \(u\) with \(x^2\), we approximate the function as \(x^2 - \frac{x^4}{2} + \frac{x^6}{3} - \ldots\).

Similarly, for \(\cos x\), the series expansion is \(1 - \frac{x^2}{2} + \frac{x^4}{24} - \ldots\). The denominator, \(1 - \cos x\), becomes \(\frac{x^2}{2} - \frac{x^4}{24} + \ldots\). By focusing on the leading terms, we simplify the expressions, crucially aiding the limit evaluation process.

Limit Evaluation

Evaluating limits using series expansion involves simplifying the expression by canceling out dominant terms. This method is especially useful in cases where direct substitution results in an indeterminate form, such as \(\frac{0}{0}\).

In the given exercise, the limit \(\lim _{x \rightarrow 0} \frac{\ln(1+x^2)}{1-\cos x}\) initially results in an indeterminate form. However, with the series expansions \(\ln(1+x^2) \approx x^2 - \frac{x^4}{2} + \ldots\) and \(1 - \cos x \approx \frac{x^2}{2} - \ldots\), we simplify the limit expression to:
\[\lim_{x \to 0} \frac{x^2 - \frac{x^4}{2} + \ldots}{\frac{x^2}{2} - \ldots}\]

By cancelling the \(x^2\) terms, the limit simplifies to \(\lim_{x \to 0} \frac{2 - x^2 + \ldots}{1}\). As \(x\) approaches zero, higher-order terms vanish, giving a definitive limit value of 2.

Indeterminate Forms

Indeterminate forms arise in calculus when the limit evaluation ends up as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), among others. These forms don't provide sufficient information about the behavior of the function close to the point of interest.

In such cases, typical algebraic simplification may not work. That's where series expansion comes into play, effectively transforming the seemingly unsolvable limits into simpler expressions. When evaluating \(\lim _{x \rightarrow 0} \frac{\ln(1+x^2)}{1-\cos x}\), both the numerator \(\ln(1+x^2)\) and the denominator \(1-\cos x\) individually approach zero as \(x\) becomes very small.

By expanding both functions into series, we focus on the dominant term that dictates the limit's behavior. This dominant subtraction leads from the indeterminate \(\frac{0}{0}\) to a clear limit value, revealing the hidden simplicity in the original problem. Such techniques are fundamental in solving complex calculus problems that involve indeterminate forms.