Problem 31 - Thomas Calculus

Question

Find the limits. Are the functions continuous at the point being approached? $$\lim _{x \rightarrow \pi} \sin (x-\sin x)$$

Step-by-Step Solution

Answer

The limit is 0 and the function is continuous at \(x = \pi\).

Step 1: Understand the Function

The problem asks us to find the limit as \(x\) approaches \(\pi\) for the function \(\sin(x - \sin x)\). This involves evaluating how the function behaves as \(x\) gets closer and closer to \(\pi\).

Step 2: Substitute the Limit Point

Direct substitution is the first method to check if the limit can be easily evaluated. Substitute \(x = \pi\) into the expression: \(\sin(\pi - \sin \pi)\).

Step 3: Simplify After Substitution

Since we know \(\sin \pi = 0\), the expression simplifies to \(\sin(\pi - 0)\). Thus, the expression becomes \(\sin \pi\).

Step 4: Evaluate \(\sin \pi\)

Recall the unit circle properties: \(\sin \pi = 0\).

Step 5: Conclusion on the Limit

Since the direct substitution gave us a real number, the limit is 0. Thus, \(\lim_{x \to \pi} \sin(x - \sin x) = 0\).

Step 6: Determine Continuity

A function is continuous at a point if the limit as \(x\) approaches that point equals the function's value at that point. Evaluate the function \(f(x) = \sin(x - \sin x)\) at \(x = \pi\).

Step 7: Evaluate \(f(\pi)\)

Substitute \(x = \pi\) to find \(f(\pi) = \sin(\pi - \sin \pi) = \sin(\pi) = 0\).

Step 8: Conclusion on Continuity

Since \(f(\pi) = 0\) and \(\lim_{x \to \pi} f(x) = 0\), the function is continuous at \(x = \pi\).

Key Concepts

Continuity of Functions

Continuity is a crucial concept in calculus, depicting how smooth a function behaves at a given point. A function is said to be continuous at a point if there is no interruption, jump, or gap. In simpler terms, if you can draw the function's graph without picking up your pencil, the function is continuous. Mathematically, for a function \( f(x) \) to be continuous at a point \( x = c \), two main conditions must be satisfied:

• The limit of \( f(x) \) as \( x \) approaches \( c \) must exist.
• The limit value \( \lim_{x \to c} f(x) \) must be equal to the function's value at that point, i.e., \( f(c) \).

In our exercise, the function \( f(x) = \sin(x - \sin x) \) was tested for continuity at \( x = \pi \). By utilizing the limit we found previously, and knowing that \( f(\pi) = 0 \), it was confirmed that the function is continuous because both the limit and the function value were equal. Thus, no breaks or jumps exist at that particular point.

Trigonometric Limits

In calculus, trigonometric limits involve finding the limit of a function that includes trigonometric expressions, such as sine, cosine, or tangent. These limits are fundamental when dealing with periodic functions and require understanding properties of trigonometric functions. The unit circle plays a significant role here, providing rapid evaluation of trigonometric values, especially at common angles like \( 0, \pi/2, \pi, 3\pi/2, \) etc.
For example, we exploited the property \( \sin \pi = 0 \) in our step-by-step solution. This property facilitated simplifying the expression \( \sin(\pi - \sin x) \) by direct substitution, which is necessary for evaluating the limit as \( x \to \pi \). Remembering such unit circle values allows you to quickly tackle similar exercises involving trigonometric limits.

Direct Substitution Method

The direct substitution method is the go-to approach for finding limits when the function behaves well at the point of interest. This straightforward technique involves substituting the value that \( x \) approaches directly into the function. If this action doesn't result in an undefined expression, like division by zero, it confirms the limit.
In our exercise, we used the direct substitution method by substituting \( x = \pi \) into \( \sin(x - \sin x) \). Upon simplifying \( \sin(\pi - \sin \pi) \), we realized it equated to \( \sin \pi = 0 \). The whole process was seamless, indicating that the function behaved neatly, thereby providing the limit directly.

• Direct substitution is especially useful when dealing with polynomial functions, rational functions that don’t lead to indeterminate forms, and many trigonometric functions.
• This method simplifies the process significantly and is an efficient first check for a limit problem.

So, always try using direct substitution to find a limit first! If it fails, only then you need to consider other techniques.