Problem 19 - Calculus Early Transcendentals: Pearson New International Edition

Question

Plot the functions \(u(x), l(x)\), and \(f(x)\). Then use these graphs along with the Squeeze Theorem to determine \(\lim _{x \rightarrow 0} f(x)\). $$ u(x)=2, l(x)=2-x^{2}, f(x)=1+\frac{\sin x}{x} $$

Step-by-Step Solution

Answer

The limit of \( f(x) \) as \( x \rightarrow 0 \) is 2.

Step 1: Plot the Functions

To solve this exercise, we first plot the functions. Start by plotting \( u(x) = 2 \), which is a horizontal line at \( y = 2 \) for all values of \( x \). Next, plot \( l(x) = 2 - x^2 \), which is a downward-opening parabola with its vertex at \( (0, 2) \). Finally, plot \( f(x) = 1 + \frac{\sin x}{x} \), noting that this function has a removable discontinuity at \( x = 0 \) since \( \frac{\sin x}{x} \rightarrow 1 \) as \( x \rightarrow 0 \).

Step 2: Understand the Squeeze Theorem

The Squeeze Theorem states that if \( l(x) \leq f(x) \leq u(x) \) for all \( x \) in some open interval containing \( c \), except possibly at \( c \) itself, and if \( \lim_{x \to c} l(x) = \lim_{x \to c} u(x) = L \), then \( \lim_{x \to c} f(x) = L \).

Step 3: Apply the Squeeze Theorem

Check that \( l(x) \leq f(x) \leq u(x) \) around the neighborhood of \( x = 0 \):1. For \( u(x) = 2 \), since \( f(x) \) adds a term \( \frac{\sin x}{x} \), and knowing that \( -1 \leq \sin x \leq 1 \), \( 1 - 1 = 0 \leq 1 + \frac{\sin x}{x} \leq 1 + 1 = 2 \).2. For \( l(x) = 2 - x^2 \), near \( x = 0 \), \( 2 - x^2 \approaches 2 \), ensuring \( f(x) = 1 + \frac{\sin x}{x} \geq 2 - x^2 \).Therefore, \( l(x) \leq f(x) \leq u(x) \).

Step 4: Determine the Limit Using Squeeze Theorem

Based on the fact that \( \lim_{x \to 0} l(x) = 2 \) and \( \lim_{x \to 0} u(x) = 2 \), the Squeeze Theorem confirms that \( \lim_{x \to 0} f(x) = 2 \).

Key Concepts

Limits in Calculus

Limits are a fundamental concept in calculus. They describe how a function behaves as the input approaches a particular point. In this exercise, we need to find the limit of the function \( f(x) = 1 + \frac{\sin x}{x} \) as \( x \rightarrow 0 \). This function contains a removable discontinuity at \( x = 0 \), which means that although \( f(x) \) is not defined exactly at \( x = 0 \), the function approaches a finite value near this point.Understanding limits:

• Limits are used to analyze the behavior of functions as they approach specific points or infinity.
• If a function approaches a specific value as \( x \) approaches a certain point, that value is the limit.

In this case, by using the Squeeze Theorem and evaluating the limits of \( l(x) \) and \( u(x) \), we can deduce the limit of \( f(x) \) as \( x \rightarrow 0 \) is 2.

Function Behavior Near Discontinuities

Understanding function behavior near discontinuities is essential for solving problems using the Squeeze Theorem, like in our exercise. A removable discontinuity occurs where a function does not have a defined value but tends towards a limit. For \( f(x) = 1 + \frac{\sin x}{x} \), there is a removable discontinuity at \( x = 0 \). Key points about discontinuities:

• Discontinuities occur when a function is not continuous at a certain point.
• There are several types of discontinuities: removable, jump, and infinite, among others.
• A removable discontinuity is where the limit exists, although the function value at that point is undefined or different.

For \( f(x) \), as \( x \rightarrow 0 \), the quotient \( \frac{\sin x}{x} \) approaches 1 because "\( \sin x \) approximates \( x \) for small \( x \)". This approximation helps us apply the Squeeze Theorem to determine the limit.

Graphing Functions

Graphing functions can provide valuable visual insights into their behavior, especially near points of interest, such as discontinuities. In this exercise, graphing \( u(x) = 2 \), \( l(x) = 2 - x^2 \), and \( f(x) = 1 + \frac{\sin x}{x} \) helps us visualize their relationships as \( x \rightarrow 0 \).Steps to graph these functions:

• \( u(x) = 2 \) is a constant function—a horizontal line at \( y = 2 \) across all \( x \).
• \( l(x) = 2 - x^2 \) is an upside-down parabola with a peak at \( x = 0 \).
• \( f(x) = 1 + \frac{\sin x}{x} \) behaves like a curved path that dips near \( x = 0 \).

Using a graph:Visual cues from such plots enable us to understand how \( l(x) \leq f(x) \leq u(x) \) within a range around \( x = 0 \), justifying why \( \lim_{x \to 0} f(x) \) using the Squeeze Theorem is consistent with \( \lim_{x \to 0} l(x) = \lim_{x \to 0} u(x) = 2 \).