Problem 33 - Thomas Calculus

Question

Use series to evaluate the limits. \begin{equation} \lim _{y \rightarrow 0} \frac{y-\tan ^{-1} y}{y^{3}} \end{equation}

Step-by-Step Solution

Answer

The limit evaluates to \(\frac{1}{3}\).

Step 1: Taylor Series Expansion for arctan(y)

The function \(-\tan^{-1}(y)\) can be expanded using the first few terms of its Taylor series at \(-y = 0\).\ The series expansion is given by:\[\tan^{-1}(y) = y - \frac{y^3}{3} + \frac{y^5}{5} - \frac{y^7}{7} + \cdots\]For small values of \(y\), we can approximate this as:\[\tan^{-1}(y) \approx y - \frac{y^3}{3}\]

Step 2: Substitute the Series in the Limit Expression

Substitute the approximate series expansion from Step 1 into the limit expression:\[\lim_{y \to 0} \frac{y - \left(y - \frac{y^3}{3}\right)}{y^3}\]This simplifies to:\[\lim_{y \to 0} \frac{\frac{y^3}{3}}{y^3}\]

Step 3: Simplify the Expression

Simplify the expression by dividing \(y^3\) in the numerator and the denominator:\[\lim_{y \to 0} \frac{\frac{y^3}{3}}{y^3} = \lim_{y \to 0} \frac{1}{3}\]Since \(-y^3\) cancels out, the expression simplifies to \(y \rightarrow 0\), which is a constant \(\frac{1}{3}\).

Step 4: Evaluate the Limit

Since the simplification in Step 3 results in a constant value and \(y\rightarrow 0\) doesn't change the constant, the limit evaluates directly to:\[\frac{1}{3}\]

Key Concepts

Limits

Understanding limits is crucial in calculus as they are foundational to concepts like derivatives and integrals. In simple terms, a limit is the value that a function approaches as the input approaches a particular point. In the context of our problem, we are interested in how the function behaves as the variable \( y \) approaches zero.To evaluate limits, especially involving indeterminate forms like \( \frac{0}{0} \), we often use techniques such as:

• Substitution: directly substitute the value into the function, if it immediately resolves to a determinate form.
• Algebraic manipulation: simplify the expression, which can lead to canceling terms and finding the limit.
• Series expansion: use a power series to rearrange the expression, often solving more complex limits simply.

In our problem, the approach involves using the Taylor series expansion to simplify the given limit expression into a simple constant \( \frac{1}{3} \), as the terms involving \( y \) effectively cancel out.

Series Expansion

The concept of series expansion is at the heart of many calculus problems. Series expansions allow us to represent functions as infinite sums of terms, and they are especially handy when evaluating functions at specific points or discovering limits like in our current problem.A Taylor series is a type of series expansion used to approximate functions around a small region close to a specific point. When we expand \( \tan^{-1}(y) \) using its Taylor series at \( y = 0 \), we get:\[ \tan^{-1}(y) = y - \frac{y^3}{3} + \frac{y^5}{5} - \frac{y^7}{7} + \cdots \] For small values of \( y \), terms become negligible as they rapidly decrease in size. Therefore, a simple approximation like \( \tan^{-1}(y) \approx y - \frac{y^3}{3} \) is sufficient for evaluating the limit. This simplifies our calculations greatly, as seen in the solution steps, giving us insights into the function's behavior near \( y = 0 \).

Arctan Function

The arctan function, or inverse tangent function, represented as \( \tan^{-1} \), is an important trigonometric function in calculus. It provides the angle whose tangent is the given number, essentially reversing the "tangent" operation.The Taylor series expansion of \( \tan^{-1}(y) \) is especially useful for evaluating limits involving this function. The series allows us to express the function as an infinite sum of polynomial terms, which can be truncated for practical purposes to simplify calculations.For our specific exercise, the arctan function serves as the main function within our limit expression. By expanding \( \tan^{-1}(y) \) into a series, we transformed a potentially complex limit problem into a much simpler one, allowing us to find the solution efficiently by examining only the first few terms of its expansion.

Calculus

Calculus is the mathematical study of continuous change and is essential in understanding how systems evolve over time. It provides tools like differentiation and integration to solve complex problems, especially those involving limits and change. In the context of this limit problem, calculus helps us handle expressions where direct substitution doesn't work, often due to indeterminate forms or complexities in the function itself. We utilize techniques like:

• Series expansions, which simplify functions near certain points,
• Understanding the behavior of functions as inputs tend towards specific values,
• Simplifying complex mathematical expressions through calculus operations.

Overall, calculus provides the framework and tools needed to solve limits efficiently, enabling us to simplify and evaluate expressions that would otherwise be challenging to compute directly.