Problem 44 - 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. $$ f(x)=\frac{\cos x}{x} ; c=0 $$

Step-by-Step Solution

Answer

The function is discontinuous at \( c = 0 \) with a nonremovable discontinuity.

Step 1: Analyze Function at Point c

First, observe the function \( f(x) = \frac{\cos x}{x} \) and the point \( c = 0 \). Notice that directly substituting \( c = 0 \) will result in division by zero, which indicates that the function is not defined at \( c = 0 \).

Step 2: Check the Limit from Both Sides

Next, find the limit of \( f(x) = \frac{\cos x}{x} \) as \( x \) approaches 0 from both sides. Calculate \( \lim_{x \to 0} \frac{\cos x}{x} \). Since \( \cos x \) is continuous and approaches \( \cos(0) = 1 \), the limit overall does not exist as \( x \to 0 \), as the denominator approaches zero faster than the numerator changes.

Step 3: Determine the Type of Discontinuity

Since \( f(x) \) is not defined at \( x = 0 \) and the limit \( \lim_{x \to 0} \frac{\cos x}{x} \) does not exist, the discontinuity is nonremovable. A removable discontinuity occurs when the limit exists but is not equal to the function's value at that point (or the function needs redefining at that point).

Key Concepts

Limits

Limits are fundamental to understanding continuity. When we talk about limits, we refer to the value that a function approaches as the input (or variable) gets closer to a certain point.
If we have a function \( f(x) \) and a point \( c \), the limit \( \lim_{x \to c} f(x) \) describes what value \( f(x) \) approaches as \( x \) nears \( c \).
This concept is crucial for determining if a function is continuous at a point. A key requirement for continuity at \( c \) is that this limit actually exists and equals the function's value at \( c \), i.e., \( \lim_{x \to c} f(x) = f(c) \).
In the context of the function \( f(x) = \frac{\cos x}{x} \) at \( c = 0 \), the limit \( \lim_{x \to 0} \frac{\cos x}{x} \) does not exist due to division by zero, indicating a potential discontinuity.

Removable Discontinuity

A removable discontinuity occurs at a point on a function where the limit exists, but is not equal to the function's value, often because the function is not defined there.
This type of discontinuity suggests that we can "fix" the function's continuity by redefining it at the discontinuous point.
For example, if \( \lim_{x \to c} f(x) \) exists but \( f(c) \) is undefined or not equal to this limit, we could make the function continuous by setting \( f(c) \) to be equal to the limit.
In our exercise, \( f(x) = \frac{\cos x}{x} \) at \( x = 0 \) would not be a removable discontinuity because the limit itself does not exist.

Nonremovable Discontinuity

A nonremovable discontinuity is more severe than a removable one. It occurs when the limit of a function does not exist at a specific point or even at all around that point.
This means that there is no way to redefine the function at the point of discontinuity to make it continuous.
In the case where \( f(x) = \frac{\cos x}{x} \) and \( c = 0 \), we face a nonremovable discontinuity.
The limit \( \lim_{x \to 0} \frac{\cos x}{x} \) does not exist because of the undefined behavior caused by division by zero.
This scenario is typical of nonremovable discontinuities, where sudden jumps or undefined values occur that cannot be patched by a simple function redefinition.