Question

Determine if \(f\) is continuous at the indicated values. If not, explain why. \(f(x)=\left\\{\begin{array}{cl}1 & x=0 \\ \frac{\sin x}{x} & x>0\end{array}\right.\) (a) \(x=0\) (b) \(x=\pi\)

Short answer

\(f\) is continuous at both \(x = 0\) and \(x = \pi\).

Step-by-step solution

Analyze Continuity Definition

To determine if a function is continuous at a point, we need to check if the limit of the function as it approaches the point equals the function's value at that point. For a function \(f\) to be continuous at \(x = a\), \(\lim_{x \to a} f(x) = f(a)\) must hold.

Check Continuity at \(x=0\)

For \(x = 0\), the function \(f\) is defined as \(f(0) = 1\). We need to find \(\lim_{x \to 0} f(x)\), which is \(\lim_{x \to 0} \frac{\sin x}{x}\) for \(x > 0\). Using the standard limit result, \(\lim_{x \to 0} \frac{\sin x}{x} = 1\). Therefore, the limit equals the function value, \(\lim_{x \to 0} f(x) = 1 = f(0)\). Thus, \(f\) is continuous at \(x = 0\).

Check Continuity at \(x=\pi\)

For \(x = \pi\), \(f(\pi)\) is not directly defined in the piecewise function, but we can assume it uses the piece \(x > 0\), so \(f(\pi) = \frac{\sin \pi}{\pi} = 0\). The limit as \(x\) approaches \(\pi\) from either side is \(\lim_{x \to \pi} f(x) = \frac{\sin \pi}{\pi} = 0\). Since \(\lim_{x \to \pi} f(x) = 0\) and \(f(\pi) = 0\), \(f\) is continuous at \(x = \pi\).

Key concepts

Limits of Functions

In calculus, limits help us understand the behavior of functions as they approach a certain point. The limit of a function as it approaches a particular value is one of the core concepts in calculus that lays the foundation for understanding continuity and differentiability.

When determining if a function is continuous at a point, limits are crucial. A function is considered continuous at a point, say \(x = a\), if the following conditions are met:

• The function \(f(x)\) is defined at \(x = a\).
• The limit \(\lim_{x \to a} f(x)\) exists.
• The limit equals the function's value at that point, i.e., \(\lim_{x \to a} f(x) = f(a)\).

If all these conditions are satisfied, the function is continuous at \(x = a\). If any of these are missing, the function is discontinuous. Understanding limits not only aids in determining continuity but also paves the way for exploring properties like differentiability.

Piecewise Functions

Piecewise functions are functions that have different definitions for different intervals of the input variable. This means the rule that defines the function changes depending on the value of \(x\). They are particularly useful to model situations where a rule or trend is not uniformly followed over the entire input range.

Consider the piecewise function \(f(x)\):
\[f(x) = \begin{cases} 1 & \text{if } x = 0 \ \frac{\sin x}{x} & \text{if } x > 0 \end{cases}\]

Here, \(f(x)\) takes the value 1 when \(x = 0\), and follows a different rule, \(\frac{\sin x}{x}\), for all \(x > 0\). Such functions often require careful evaluation of limits to check their continuity, especially at the boundaries where the rules change.

For instance, checking continuity at \(x = 0\) for \(f(x)\) involves ensuring that the limit from the defined piece matches the value at \(x = 0\). These types of functions aid in modeling real-world scenarios where a condition suddenly changes, providing a more accurate representation than a uniform model.

Trigonometric Limits

Trigonometric limits are specific types of limits that involve trigonometric functions like sine, cosine, and tangent. They can often appear tricky due to the oscillatory nature of these functions as they approach 0 or other critical angles.

One of the most well-known trigonometric limits is \(\lim_{x \to 0} \frac{\sin x}{x} = 1\). This limit plays a key role when dealing with many types of problems, including those involving piecewise functions.

In the piecewise function example of \(f(x)\) given above, checking continuity at \(x=0\) requires the use of this limit to find \(\lim_{x \to 0} \frac{\sin x}{x}\). Since this limit equals 1, it ensures the continuity of \(f(x)\) at \(x = 0\) because the limit matches the function value at that point.

Trigonometric limits also help simplify problems and find exact values of otherwise complicated expressions, making them integral to calculus and analysis. Understanding and remembering the critical trigonometric limits can save time and effort in solving a wide range of mathematical problems.