Problem 47 - Calculus Early Transcendentals

Question

Let \(F(x)=\frac{x^{2}-1}{|x-1|}.\) (a) Find (i) $$\lim _{x \rightarrow 1^{+}} F(x)$$ (ii) $$\lim _{x \rightarrow 1^{-}} F(x)$$ (b) Does lim \(_{x \rightarrow 1}\) F(x) exist? (c) Sketch the graph of F.

Step-by-Step Solution

Answer

(i) 2, (ii) -2; (b) No; (c) Discontinuous at x = 1.

Step 1: Analyze the Function for Positive Limit

Firstly, consider the limit as \( x \to 1^+ \). For \( x > 1 \), the expression \( |x-1| = x-1 \). Thus, the function simplifies to: \[ F(x) = \frac{x^2 - 1}{x - 1} \]Next, factor the numerator: \( x^2 - 1 = (x - 1)(x + 1) \). Therefore, \[ F(x) = \frac{(x - 1)(x + 1)}{x - 1} = x + 1 \text{ for } x > 1 \]. Now calculate:\[ \lim _{x \to 1^+} F(x) = \lim _{x \to 1^+} (x + 1) = 2 \].

Step 2: Analyze the Function for Negative Limit

Consider the limit as \( x \to 1^- \). For \( x < 1 \), we have \(|x-1| = -(x-1) \). Thus, the function becomes:\[ F(x) = \frac{x^2 - 1}{-(x - 1)} = \frac{(x - 1)(x + 1)}{-(x - 1)} = -(x + 1) \text{ for } x < 1 \]. Now, calculate:\[ \lim _{x \to 1^-} F(x) = \lim _{x \to 1^-} -(x + 1) = -2 \].

Step 3: Determine the Existence of Limit

To determine if the limit as \( x \to 1 \) exists for \( F(x) \), compare the limits from the right and left. We found:\[ \lim _{x \to 1^+} F(x) = 2 \]and\[ \lim _{x \to 1^-} F(x) = -2 \].Since these one-sided limits are not equal, \( \lim_{x \to 1} F(x) \) does not exist.

Step 4: Sketch the Graph of F(x)

The function \( F(x) = x + 1 \) for \( x > 1 \) is a line with slope 1 intersecting the x-axis at -1. For \( x < 1 \), \( F(x) = -(x + 1) \) is a line with slope -1 intersecting the x-axis at -1. There is a jump discontinuity at \( x = 1 \), where \( F(x) \) shifts from 2 to -2. Plot these two lines on a graph, ensuring to indicate the point of discontinuity at \( x = 1 \).

Key Concepts

One-sided limits

When investigating the behavior of a function as it approaches a particular point, we often examine the one-sided limits. In the given exercise, we analyze the function \( F(x)=\frac{x^2-1}{|x-1|} \) as \( x \) approaches 1 from both the right and left sides.

• The right-hand limit or \( \lim_{x \rightarrow 1^+} F(x) \) considers values of \( x \) slightly greater than 1. For these values, the absolute value \( |x-1| \) simplifies to \( x-1 \), and therefore: \( F(x) = \frac{(x - 1)(x + 1)}{x - 1} = x + 1 \). Evaluating the limit, we find \( \lim_{x \rightarrow 1^+} F(x) = 2 \).
• The left-hand limit or \( \lim_{x \rightarrow 1^-} F(x) \) considers values of \( x \) slightly less than 1. Here, \( |x-1| = -(x-1) \), which changes the sign and leads us to \( F(x) = -(x + 1) \). This gives \( \lim_{x \rightarrow 1^-} F(x) = -2 \).

Understanding one-sided limits is critical for determining whether a general limit exists, which involves comparing these two results.

Discontinuity

Discontinuity in a function refers to a point where a function is not continuous—it may "jump" from one value to another or have a gap. In our exercise, the point of interest is \( x = 1 \). For a limit to exist at a given point, the one-sided limits must be equal. This, however, is not the case here.

From the solution:

• The right-hand limit at \( x = 1 \) is 2, meaning the function approaches 2 as \( x \to 1^+ \).
• On the left hand, the limit is -2, indicating the function approaches -2 as \( x \to 1^- \).

Because these values are different (2 and -2), there is a "jump" discontinuity at \( x = 1 \). This kind of discontinuity is characterized by the sudden shift in function value from one side of \( x=1 \) to the other, making \( \lim_{x \to 1} F(x) \) nonexistent.

Graph Sketching

Sketching the graph of a function allows us to visualize its behavior and understand the presence of any discontinuities or other interesting features. To sketch the graph of \( F(x) = \frac{x^2-1}{|x-1|} \), we consider each piece of the function separately:

• For \( x > 1 \), we have \( F(x) = x + 1 \). This is a straight line with a slope of 1, crossing the x-axis at \( x = -1 \) and passing through \( (1, 2) \).
• For \( x < 1 \), the function is \( F(x) = -(x + 1) \), a line with a slope of -1 that also crosses the x-axis at \( x = -1 \) and passes through \( (1, -2) \).

As you sketch, ensure the lines reflect the difference in slopes and intersect correctly on the x-axis. A key part of the sketch is marking the discontinuity between these lines at \( x = 1 \), where there is no connecting point. Visualizing the graph with these jumps helps in better understanding the function's structure and behavior around the point of interest.