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Chapter 9: Problem 36

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

Short Answer

Expert verified
The limit is 1.

Step by step solution

01

Rewrite Using the Series Expansion

The expression involves \(\sin \frac{1}{x+1}\), which can be expanded using the Taylor series for sine. Recall that the Taylor series for \(\sin u\) is \(u - \frac{u^3}{3!} + \frac{u^5}{5!} - \cdots\). For very large \(x\), this becomes:\[\sin \frac{1}{x+1} \approx \frac{1}{x+1}\].
02

Simplify the Expression

Substituting the series expansion into the original limit gives us:\[\lim_{x \to \infty} (x+1) \sin \frac{1}{x+1} \approx \lim_{x \to \infty} (x+1) \left( \frac{1}{x+1} \right) = \lim_{x \to \infty} 1\].
03

Evaluate the Limit

Since the expression simplifies to \(1\), the limit becomes:\[\lim_{x \to \infty} 1 = 1\]. The result shows that as \(x\) approaches infinity, the original expression approaches 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Taylor Series
The Taylor series is a mathematical tool used to express functions as infinite sums of terms calculated from the values of their derivatives at a single point. It acts like a "recipe" to approximate more complex functions by a polynomial, making them easier to analyze or compute.
  • The formula for a Taylor series about zero (also known as a Maclaurin series) for a function \( f(x) \) is expressed as:
\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \]
  • For the sine function, \( \sin(u) \), the series becomes \( u - \frac{u^3}{3!} + \frac{u^5}{5!} - \cdots \).
  • This is particularly useful when \( u \) is small, allowing higher-order terms to become negligible, simplifying calculations.
When evaluating limits, the Taylor series helps approximate functions close to points of interest, such as zero or infinity. This process simplifies expressions, making it feasible to find limits that might otherwise be challenging to evaluate directly.
Sine Function
The sine function, noted as \( \sin(x) \), is a fundamental trigonometric function known for its wave-like pattern.
  • Its values oscillate between -1 and 1, making it periodic with a period of \( 2\pi \).
  • Key properties include its symmetry and its odd function characteristic: \( \sin(-x) = -\sin(x) \).
For small angles, such as those described by \( \sin(\frac{1}{x+1}) \) in the original problem, the sine function can be approximated using its first Taylor series term:
  • Hence, \( \sin(\frac{1}{x+1}) \approx \frac{1}{x+1} \) when \( x \) is large.
  • This approximation is crucial when evaluating limits, as it simplifies the terms involved.
This function is integral in various fields of mathematics and physics, especially when modeling cyclical phenomena.
Limit Evaluation
Limit evaluation is a method used to determine the behavior of a function as its input approaches a particular value. This is integral in calculus and helps to solve indeterminate forms or to evaluate the behavior of functions at boundaries.
  • One relies on algebraic manipulation, L'H么pital's rule, or series expansion to resolve these limits.
In the given exercise, limit evaluation is applied after substituting the sine's series expansion for a simplified form:
  • The expression \( \lim_{x \to \infty} (x+1) \sin(\frac{1}{x+1}) \) transforms into \( \lim_{x \to \infty} 1 \).
  • This simplification was possible because the product \( (x+1)\frac{1}{x+1} \) intuitively resolves to 1 as \( x \) approaches infinity.
Understanding limits is crucial for mathematical modeling, calculus problems, and continuous function analysis.

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Most popular questions from this chapter

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