Problem 50 - Calculus: Early Transcendentals

Question

Let \( f(x) = \left\\{ \begin{array}{ll} x^2 + 1 & \mbox{if \) x < 1 \(}\\\ (x - 2)^2 & \mbox{if \) x \ge 1 \(} \end{array} \right.\) (a) Find \( \displaystyle \lim_{x \to 1^-}f(x) \) and \( \displaystyle \lim_{x \to 1^+}f(x) \). (b) Does \( \displaystyle \lim_{x \to 1}f(x) \) exist? (c) Sketch the graph of \( f \).

Step-by-Step Solution

Answer

The limits \( \lim_{x \to 1^-} f(x) = 2 \) and \( \lim_{x \to 1^+} f(x) = 1 \); \( \lim_{x \to 1} f(x) \) does not exist.

Step 1: Identify the Limit Type for Part (a)

To find the left-hand limit \( \lim_{x \to 1^-}f(x) \) and the right-hand limit \( \lim_{x \to 1^+}f(x) \), we need to identify which expression for \( f(x) \) is used for each case. For \( x < 1 \), use \( f(x) = x^2 + 1 \). For \( x \ge 1 \), use \( f(x) = (x - 2)^2 \).

Step 2: Calculate the Left-Hand Limit

For \( \lim_{x \to 1^-} f(x) \): Since \( x < 1 \), use the expression \( f(x) = x^2 + 1 \).\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2 + 1) = 1^2 + 1 = 2. \]

Step 3: Calculate the Right-Hand Limit

For \( \lim_{x \to 1^+} f(x) \): Since \( x \ge 1 \), use the expression \( f(x) = (x - 2)^2 \).\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x - 2)^2 = (1 - 2)^2 = 1. \]

Step 4: Check Limit Existence for Part (b)

To determine if \( \lim_{x \to 1} f(x) \) exists, check if \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) \). Since these limits are 2 and 1, respectively, they are not equal. Thus, \( \lim_{x \to 1} f(x) \) does not exist.

Step 5: Sketch the Graph for Part (c)

For \( x < 1 \), plot \( f(x) = x^2 + 1 \) as a parabola with vertex at (0, 1), approaching the point (1, 2). For \( x \ge 1 \), plot \( f(x) = (x - 2)^2 \) as a parabola opening upwards with vertex at (2, 0), starting at (1, 1). The two parts should connect but not have the same value at \( x = 1 \).

Key Concepts

Limits

When dealing with piecewise functions, understanding limits is crucial. Limits help describe the behavior of a function as the input approaches a certain value. In this exercise, we focused on the limits of the function \( f(x) \) as \( x \) approaches 1 from both sides.

• The left-hand limit \( \lim_{x \to 1^-} f(x) = 2 \) tells us what value \( f(x) \) approaches as \( x \) gets closer to 1 from values less than 1.
• The right-hand limit \( \lim_{x \to 1^+} f(x) = 1 \) tells us the value that \( f(x) \) approaches as \( x \) comes from values greater than or equal to 1.

These calculations are essential to understand whether the function has a limit at a certain point. If the left-hand and right-hand limits do not agree, the overall limit does not exist. This occurrence is exactly what we found with \( f(x) \). Since the left and right limits weren't equal, \( \lim_{x \to 1} f(x) \) doesn't exist.

Continuity

Continuity in a function means there are no breaks, jumps, or holes in the graph at the point in question. For a function to be continuous at a point, three conditions must be met: the function must be defined at the point, the limit must exist at that point, and the value of the function at that point must equal the limit value.
In the given exercise, the limit \( \lim_{x \to 1} f(x) \) does not exist due to differing left-hand and right-hand limits, indicating a discontinuity at \( x = 1 \). This means there is a jump in the graph of \( f(x) \) from the value 2 (coming from the left) to the value 1 (coming from the right).
Discontinuities are important to note as they can affect the interpretation and application of mathematical models, especially when considering real-world scenarios modeled by such piecewise functions.

Graph Sketching

Graph sketching involves plotting a visual representation of mathematical functions. With piecewise functions, it's important first to pay attention to the different expressions defining the function across its domain. This exercise involves two different quadratic functions for \( x < 1 \) and \( x \ge 1 \).

• For \( x < 1 \), the function is \( f(x) = x^2 + 1 \), which is a parabola with its vertex at (0, 1).
• For \( x \ge 1 \), the function is \( f(x) = (x - 2)^2 \), another parabola represented by a vertex at (2, 0).

During sketching, you should illustrate these parts until they meet visually at \( x = 1 \), showing that at \( x = 1 \), the graph jumps from point (1, 2) to (1, 1). Although these two parts of the graph get close, they do not connect directly, emphasizing the discontinuity discussed earlier.
By sketching the graph accurately, you can better visualize how the behavior of each piece changes around these key points, making it easier to understand the overall structure of the function.