Problem 36 - Thomas Calculus

Question

Use series to evaluate the limits. \begin{equation} \lim _{x \rightarrow \infty}(x+1) \sin \frac{1}{x+1} \end{equation}

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Understanding the Limit

The given limit is \( \lim _{x \rightarrow \infty} (x+1) \sin \frac{1}{x+1} \). As \( x \) approaches infinity, \( x+1 \) will also tend towards infinity. We need to evaluate how the expression behaves as \( x \) becomes very large.

Step 2: Series Expansion of \( \sin \theta \)

Since \( \frac{1}{x+1} \) becomes very small as \( x \) approaches infinity, we can use the Taylor series expansion for \( \sin \theta \) around 0: \( \sin \theta \approx \theta - \frac{\theta^3}{6} + \cdots \). So \( \sin \frac{1}{x+1} \approx \frac{1}{x+1} - \frac{1}{6(x+1)^3} + \cdots \).

Step 3: Substituting the Series Approximation

Substitute the approximation in the limit: \( (x+1) \sin \frac{1}{x+1} \approx (x+1) \left( \frac{1}{x+1} - \frac{1}{6(x+1)^3} + \cdots \right) \).

Step 4: Simplifying the Expression

Distribute \( (x+1) \) across the series terms: \( = 1 - \frac{1}{6(x+1)^2} + \cdots \).

Step 5: Evaluating the Limit

As \( x \rightarrow \infty \), \( \frac{1}{6(x+1)^2} \rightarrow 0 \). Thus, the approximation \( 1 - \frac{1}{6(x+1)^2} + \cdots \) simplifies to 1 in the limit.

Step 6: Conclusion

Therefore, \( \lim _{x \rightarrow \infty} (x+1) \sin \frac{1}{x+1} = 1 \).

Key Concepts

Taylor Series

The Taylor Series is a powerful tool in calculus used to represent functions as infinite sums of their derivatives at a single point. It can be thought of as a way to approximate a function using polynomials. This series comes in handy especially when dealing with limits and small angle approximations.

For a function \( f(x) \), the Taylor series around the point \( a \) is given by:

• \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots \)

In our problem, we utilize the series expansion of the sine function. Since \( \sin(\theta) \) can be approximated when \( \theta \) is close to zero by \( \theta - \frac{\theta^3}{6} + \cdots \), it allows us to substitute it into expressions where the angle is very small. This helps simplify calculations in limit evaluations and can lead to a clearer insight into the behavior of the function as \( x \) tends to infinity.

Understanding and using the Taylor Series is crucial for evaluating limits, expanding functions, and tackling infinite behaviors in calculus.

Series Expansion

Series expansion refers to expressing a function in a form that is the sum of simpler terms. This technique is useful in calculus to handle complex functions, simplify calculations, and provide approximate solutions to problems.

In the given exercise, we performed a series expansion on \( \sin \frac{1}{x+1} \) because it becomes a small angle as \( x \) approaches infinity. By using the initial terms of the series, \( \sin \theta \approx \theta - \frac{\theta^3}{6} + \cdots \), the computation is simplified. We substitute this into our expression which then is:

• \( (x+1) \left( \frac{1}{x+1} - \frac{1}{6(x+1)^3} + \cdots \right) \)

Expanding functions into series is a fundamental technique for evaluating limits because it helps highlight which components are significant as the variable grows large or small. It reveals the "big picture" of what happens to each term in the expansion as \( x \) tends toward a certain value, such as infinity in this exercise.

Series expansions simplify complex expressions, making them more manageable in rigorous calculus tasks involving limits and approximations.

Limit Evaluation

Limit evaluation is a core aspect of calculus aimed at understanding the behavior of functions as their inputs approach a certain value. This includes handling cases as inputs become extremely large, close to zero, or hit any point where the function might behave unpredictably.

In our specific problem, the goal was to evaluate \( \lim_{x \rightarrow \infty} (x+1) \sin \frac{1}{x+1} \). The handy trick here was recognizing how the expression can be simplified using the series expansion so that evaluating the limit becomes straightforward:

• Substitute \( \sin \frac{1}{x+1} \) with its series expansion.
• Distribute \( (x+1) \) across terms.
• Simplify to end up analyzing the behavior of each term as \( x \to \infty \)
• Acceptably ignore insignificant terms as they tend to zero.

By doing so, we're left with the leading term, which gives us the limit result. Practicing limit evaluation with series expansion helps demystify challenging calculus questions, providing a systematic way to approach seemingly complex limits. It's a skill that uncovers the essence of how functions behave and serves as a building block for more advanced mathematical concepts.