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 non-removable. $$ f(x)=\frac{\sin x}{x} ; c=0 $$

Step-by-Step Solution

Answer

The function is not continuous at \( c = 0 \); the discontinuity is removable.

Step 1: Determine the function value at the point

First, we evaluate the function \( f(x) = \frac{\sin x}{x} \) at the point \( c = 0 \). Since direct substitution leads to the expression \( \frac{\sin(0)}{0} = \frac{0}{0} \), which is undefined, we cannot determine \( f(0) \) directly. Instead, we consider the limit as \( x \) approaches 0 to evaluate the function's behavior at the given point.

Step 2: Determine the limit of the function as x approaches c

To analyze the limit, we use the fact that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), which is a well-known limit result in calculus. Thus, the limit of \( f(x) \) as \( x \to 0 \) is \( 1 \).

Step 3: Check for continuity at the point

A function \( f(x) \) is continuous at a point \( c \) if \( f(c) \) is defined, \( \lim_{x \to c} f(x) \) exists, and \( \lim_{x \to c} f(x) = f(c) \). In this case, \( f(0) \) is not defined, but \( \lim_{x \to 0} f(x) = 1 \), showing a difference between the limit and actual function value. Therefore, \( f(x) \) is not continuous at \( x = 0 \).

Step 4: Determine the type of discontinuity

Since \( \lim_{x \to 0} f(x) \) exists and equals 1, but \( f(0) \) is not defined, the discontinuity at \( x = 0 \) can be "fixed" by redefining \( f(0) \) to be 1. Thus, the discontinuity is removable. This means we can make the function continuous by defining \( f(0) = 1 \).

Key Concepts

Limits

In calculus, limits are a fundamental concept that help us understand the behavior of functions as they approach a specific point. For the function \( f(x) = \frac{\sin x}{x} \), as \( x \) approaches 0, the direct substitution results in an undefined form \( \frac{0}{0} \). This is where limits come into play, allowing us to evaluate the approach of the function to a point without actually reaching it. Using known calculus results, the limit \( \lim_{x \to 0} \frac{\sin x}{x} \) is 1. Limits are essential for checking continuity and solving calculus problems.

Removable Discontinuity

A function can have different types of discontinuities. A removable discontinuity occurs when a function is not defined at a point, but the limit exists, suggesting it can be "fixed." For \( f(x) = \frac{\sin x}{x} \), at \( c = 0 \), though \( f(0) \) is undefined, the limit \( \lim_{x \to 0} f(x) = 1 \) exists. Thus, by setting \( f(0) = 1 \), we can "remove" the discontinuity, making the function continuous. Understanding removable discontinuities helps improve function behavior, especially in cases with undefined points.

Trigonometric Functions

Trigonometric functions are vital in calculus, often appearing in functions that need analysis. One critical property of the sine function, \( \sin x \), is its behavior around 0, such as the well-known limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). This property allows us to deal with certain undefined forms in calculus problems, helping to determine continuity and behavior of functions around specific points. Trigonometric limits are powerful tools when dealing with oscillatory functions or simplifying complex expressions.

Calculus Problem-Solving

Solving calculus problems often involves a structured approach. First, evaluate the function at the point of interest. If it's undefined, as is often the case in indeterminate forms like \( \frac{0}{0} \), consider the limit. Next, analyze if the function is continuous by checking if the limit equals the function value at that point. If not, identify the type of discontinuity. In cases of removable discontinuities, redefine the function at the point for continuity. This methodical process applies to checking continuity and solving a wide range of calculus problems, ensuring accurate solutions.