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: Expand the Numerator Using Series

The numerator of the given expression is \( \ln(1+x^3) \). Use the series expansion for \( \ln(1+u) \) around \( u = 0 \):\[\ln(1+u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \cdots\]For \( u = x^3 \), this becomes:\[\ln(1+x^3) = x^3 - \frac{(x^3)^2}{2} + \frac{(x^3)^3}{3} - \cdots = x^3 - \frac{x^6}{2} + \frac{x^9}{3} - \cdots\]We need only the first term for our calculations as \( x \rightarrow 0 \):\[ \ln(1+x^3) \approx x^3 \]

Step 2: Expand the Denominator Using Series

The denominator of the given expression is \( x \sin(x^2) \). Use the series expansion for \( \sin(v) \) around \( v = 0 \):\[\sin(v) = v - \frac{v^3}{6} + \frac{v^5}{120} - \cdots\]For \( v = x^2 \), this becomes:\[\sin(x^2) = x^2 - \frac{(x^2)^3}{6} + \frac{(x^2)^5}{120} - \cdots = x^2 - \frac{x^6}{6} + \frac{x^{10}}{120} - \cdots\]Thus:\[ x \sin(x^2) = x(x^2 - \frac{x^6}{6} + \cdots) = x^3 - \frac{x^9}{6} + \cdots \]We need only the first term for our calculations as \( x \rightarrow 0 \):\[ x \sin(x^2) \approx x^3 \]

Step 3: Calculate and Simplify the Limit

Substitute the series approximations from Steps 1 and 2 into the limit:\[\lim_{x \to 0} \frac{\ln(1+x^3)}{x \sin(x^2)} = \lim_{x \to 0} \frac{x^3}{x^3}\]This simplifies to:\[\lim_{x \to 0} 1 = 1\]

Key Concepts

Series Expansion

Series expansions allow us to express functions in terms of an infinite sum of terms calculated from the values of their derivatives at a single point.
The series expansion is especially useful for approximating functions for small values of the variable.
In calculus, when evaluating limits or simplifying complex expressions, series expansion is a great tool.
For example, the natural logarithm function can be expanded using its Taylor Series:

• For \( \ln(1+u) \), the expansion around \( u = 0 \) is given by: \[ \ln(1+u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \cdots \]

Applying this to \( \ln(1+x^3) \), only the first term \( x^3 \) is significant for values of \( x \) approaching zero because higher-order terms become insignificant.
Similarly, trigonometric functions can be expanded using series to solve limits and other problems where direct computation becomes difficult.

Logarithmic Functions

Logarithmic functions are vital in calculus and algebra.
They provide us with a way to transform multiplicative relationships into additive ones.
This quality is especially useful in solving limit problems as it simplifies expressions by converting products into sums.For small values of \( x \), the function \( \ln(1+x) \) can be expanded using:

• \( \ln(1+x) \approx x - \frac{x^2}{2} + \frac{x^3}{3} \)

This approximation simplifies the evaluation of limits because it reduces complex expressions into a more manageable form.
The natural logarithm approximates well for values near zero, which helps in applying limits.
In our exercise, using the series expansion of \( \ln(1+x^3) \) simplifies it to \( x^3 \) for very small \( x \), significantly aiding in reaching the solution.

Trigonometric Series

Trigonometric series are expansions of trigonometric functions into infinite sums that simplify their computations, especially useful when working with limits and calculus problems.The sine function, \( \sin(v) \), can be expanded in a series form, which is:

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

For the problem at hand, using \( \sin(x^2) \), we substitute \( x^2 \) into the sine series:

• \( \sin(x^2) = x^2 - \frac{x^6}{6} + \cdots \)

This conversion becomes especially useful for evaluating limits, as complex expressions can be handled more easily by truncating to the most significant term.
In our example, considering \( x \to 0 \), it is enough to keep the leading term \( x^2 \), simplifying our denominator to \( x^3 \) when multiplied by \( x \).
By utilizing trigonometric series, the original problem becomes straightforward, allowing limits to be calculated with much less complexity.