Problem 43 - Calculus Early Transcendentals: Pearson New International Edition

Question

Determine whether the function is continuous at the given point \(c\). If the function is not continuous, determine whether the discontinuity is removable or nonremovable. $$ f(x)=\frac{\sin x}{x} ; c=0 $$

Step-by-Step Solution

Answer

The function is not continuous at \(x=0\), and the discontinuity is removable.

Step 1: Definition of Continuity at a Point

A function is continuous at a point \(c\) if the following three conditions are met: 1) \(f(c)\) is defined, 2) \(\lim_{{x \to c}} f(x)\) exists, and 3) \(\lim_{{x \to c}} f(x) = f(c)\). First, evaluate \(f(0)\). The function is given by \(f(x) = \frac{\sin x}{x}\), so \(f(0) = \frac{\sin 0}{0}\), which is undefined since division by zero is not possible.

Step 2: Check the Existence of Limit

Since \(f(0)\) is undefined, check if the limit \(\lim_{{x \to 0}} \frac{\sin x}{x}\) exists. It is known from calculus that \(\lim_{{x \to 0}} \frac{\sin x}{x} = 1\) based on the Squeeze Theorem.

Step 3: Conclusion on Continuity

For the function to be continuous at \(x=0\), \(f(0)\) must be defined to equal \(1\). Since \(f(0)\) is currently undefined, \(f(x)\) is not continuous at \(x=0\).

Step 4: Identify Type of Discontinuity

Since the limit exists but \(f(0)\) is undefined, the discontinuity at \(x=0\) is removable. We can define a new function \(g(x)\) such that \(g(x) = \frac{\sin x}{x}\) for \(x eq 0\) and \(g(0) = 1\) for continuity at \(x=0\).

Key Concepts

Discontinuity

In mathematics, a discontinuity occurs when a function is not continuous at a specific point. A function is continuous at a point if there are no abrupt changes or gaps, meaning it can be drawn without lifting the pencil from the paper. For a function to be continuous at a point \(c\), three conditions need to be satisfied:

• The function value \(f(c)\) must exist and should be defined.
• The limit of the function as it approaches \(x = c\) must exist.
• The limit of the function as \(x\) approaches \(c\) should equal the function value \(f(c)\).

For example, the function \(f(x) = \frac{\sin x}{x}\) at \(x = 0\) is not continuous because \(f(0)\) is undefined. But there is more to discover about this type of discontinuity.

Removable Discontinuity

A removable discontinuity is a gap in the graph of a function that can be "removed" by simply redefining the function at a point. This means that a limit exists at this point, but the function is not initially defined there. By adjusting the definition, the function becomes continuous.Consider the function \(f(x) = \frac{\sin x}{x}\). At \(x = 0\), the function is originally undefined because it involves division by zero. However, the limit \(\lim_{{x \to 0}} \frac{\sin x}{x} = 1\) exists, suggesting that the graph approaches a single point.To "remove" the discontinuity, we define a new function \(g(x)\) where \(g(x) = 1\) when \(x = 0\) and \(g(x) = \frac{\sin x}{x}\) elsewhere. By doing this, the continuity is restored at \(x = 0\), creating a seamless graph.

Squeeze Theorem

The Squeeze Theorem is a principle used in calculus to determine limits of a function when direct evaluation is challenging. It is especially useful when a function is trapped between two other functions whose limits at a point are already known.This theorem states that if you have three functions, \(f(x)\), \(g(x)\), and \(h(x)\), where \(f(x) \leq g(x) \leq h(x)\) for values of \(x\) close to a value \(c\) (excluding \(c\)), and if the limits of \(f(x)\) and \(h(x)\) as \(x\) approaches \(c\) are the same, then the limit of \(g(x)\) as \(x\) approaches \(c\) must also exist and be equal to that same limit.For the function \(f(x) = \frac{\sin x}{x}\), as \(x\) approaches zero, we use the Squeeze Theorem to show that \(\lim_{{x \to 0}} \frac{\sin x}{x} = 1\). This clever approach helps us navigate around the issue of direct substitution of \(x = 0\), where the function is originally undefined.