Problem 32 - Calculus Early Transcendentals

Question

Use continuity to evaluate the limit. $$\lim _{x \rightarrow \pi} \sin (x+\sin x)$$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Identifying the Function

The given function is \( f(x) = \sin(x + \sin x) \). Our goal is to evaluate the limit \( \lim_{x \to \pi} f(x) \).

Step 2: Recognizing Continuity

The sine function is continuous everywhere, and the sum \( x + \sin x \) is also continuous being the sum of continuous functions. Therefore, \( f(x) = \sin(x + \sin x) \) is continuous for all real numbers.

Step 3: Applying Continuity Theorem

Because \( f(x) \) is continuous at \( x = \pi \), the limit of \( f(x) \) as \( x \) approaches \( \pi \) can be found by evaluating \( f(\pi) \):\[ f(\pi) = \sin(\pi + \sin \pi) \]

Step 4: Simplifying the Expression

Since \( \sin \pi = 0 \), we simplify the expression:\[ f(\pi) = \sin(\pi + 0) = \sin \pi = 0 \]

Step 5: Final Evaluation

Thus, the limit \( \lim_{x \to \pi} \sin(x + \sin x) = 0 \).

Key Concepts

Continuity

When we talk about continuity in calculus, we mean that a function behaves smoothly without any gaps, jumps, or holes. In other words, you can draw the function's graph without lifting your pencil from the paper. Mathematically, a function is continuous at a point if the limit of the function as it approaches that point equals the function's value at that point.

Continuity is essential because it allows us to evaluate limits more conveniently. If we know that a function is continuous at a certain point, we can directly substitute the point into the function to find the limit. This greatly simplifies the process of limit evaluation, as demonstrated in the original exercise.

Sine Function

The sine function, often represented as \( \sin(x) \), is a fundamental trigonometric function that describes the relationship of an angle within a right triangle to the ratio of the length of the opposite side to the hypotenuse. The sine function is periodic with a period of \( 2\pi \) meaning it repeats its values every \( 2\pi \) radians.

Key properties of the sine function include:

• It is continuous and smooth across its domain of all real numbers.
• The range of the sine function is [-1, 1].
• It has a critical value of 0 at multiples of \( \pi \), such as \( \pi, 2\pi, 3\pi \,... \).

This periodic and predictable behavior of the sine function ensures that calculations involving sine, such as in limits, can be made straightforward through continuity.

Continuous Functions

When dealing with continuous functions, it's helpful to remember that compositional functions—those made up of two or more continuous functions—are also continuous. This principle is crucial when evaluating complex expressions like \( f(x) = \sin(x + \sin x) \).

Since both \( x \) and \( \sin(x) \) are individually continuous, their sum \( x + \sin x \) remains continuous. Applying a continuous function, like sine, to another continuous function results in another continuous function. Thus, \( \sin(x + \sin x) \) is continuous for all real numbers.

This property significantly aids in evaluating limits because for a continuous function, you can simply evaluate the function directly at the point being approached.

Calculus

Calculus is a branch of mathematics focused on change and motion, through the concepts of derivatives and integrals. One of its fundamental ideas is the concept of limits, which are essential in analyzing the behavior of functions as they approach specific points.

Limits help us understand how functions behave near a point and are crucial for defining both derivative and integral concepts.

In the context of the original problem, calculus enables us to use the concept of limits and continuity to evaluate expressions that may initially seem complex. Evaluating \( \lim_{x \rightarrow \pi} \sin(x + \sin x) \) demonstrates how limits and continuity smooth the path towards finding solutions, thus illustrating the elegance and utility of calculus in solving mathematical problems.