Problem 43 - Precalculus: Mathematics for Calculus

Question

Does the Limit Exist? Let $$f(x)=\left\\{\begin{array}{ll} x-1 & \text { if } x<2 \\ x^{2}-4 x+6 & \text { if } x \geq 2 \end{array}\right.$$ (a) Find \(\lim _{x \rightarrow 2^{-}} f(x)\) and \(\lim _{x \rightarrow 2^{+}} f(x)\) (b) Does \(\lim _{x \rightarrow 2} f(x)\) exist? (c) Sketch the graph of \(f\)

Step-by-Step Solution

Answer

(a) \( \lim_{x \to 2^-} f(x) = 1 \), \( \lim_{x \to 2^+} f(x) = 2 \). (b) The limit does not exist because the left and right limits are not equal. (c) Sketch two lines for different intervals with a discontinuity at \( x = 2 \).

Step 1: Analyze Left-Hand Limit

To find the left-hand limit \( \lim_{x \to 2^-} f(x) \), we need to consider the expression for \( f(x) \) when \( x < 2 \). This is \( f(x) = x - 1 \). To find the limit, substitute \( x = 2 \) into this expression: \( \lim_{x \to 2^-} f(x) = 2 - 1 = 1 \).

Step 2: Analyze Right-Hand Limit

To find the right-hand limit \( \lim_{x \to 2^+} f(x) \), we consider the expression for \( f(x) \) when \( x \geq 2 \). This is \( f(x) = x^2 - 4x + 6 \). Substitute \( x = 2 \) into this expression to find that \( \lim_{x \to 2^+} f(x) = 2^2 - 4(2) + 6 = 4 - 8 + 6 = 2 \).

Step 3: Determine the Existence of the Limit at x = 2

For the overall limit \( \lim_{x \to 2} f(x) \) to exist, the left-hand and right-hand limits must be equal. From Step 1, we found \( \lim_{x \to 2^-} f(x) = 1 \) and from Step 2, \( \lim_{x \to 2^+} f(x) = 2 \). Since \( 1 eq 2 \), the overall limit does not exist.

Step 4: Sketch the Graph of f(x)

To sketch the graph of \( f(x) \), plot the piecewise functions:1. For \( x < 2 \), graph the line \( f(x) = x - 1 \); the line is continuous and has a slope of 1.2. For \( x \geq 2 \), plot the parabola \( f(x) = x^2 - 4x + 6 \). Include the point at \( x = 2 \) by calculating \( f(2) = 2 \). This will create a discontinuity at \( x = 2 \) since the two pieces do not connect.

Key Concepts

Limits

In calculus, limits are a fundamental concept used to analyze functions as they approach a particular point. When dealing with piecewise functions, it's important to understand both the left-hand limit and the right-hand limit. The left-hand limit, denoted as \( \lim_{x \to c^-} f(x) \), refers to the value that the function approaches as \( x \) approaches \( c \) from the left (or smaller than \( c \)).
On the other hand, the right-hand limit, \( \lim_{x \to c^+} f(x) \), represents the value the function approaches as \( x \) approaches \( c \) from the right (or greater than \( c \)).
In our exercise, we calculated:

• \( \lim_{x \to 2^-} f(x) = 1 \)
• \( \lim_{x \to 2^+} f(x) = 2 \)

These differing values indicate the function's behavior as \( x \) approaches 2 from each side. A function's limit at a point only exists if both one-sided limits are equal. So here, the limit \( \lim_{x \to 2} f(x) \) does not exist because 1 is not equal to 2.

Continuity

Continuity in a function means that there are no sudden jumps or breaks as you trace the graph. A function is continuous at a point if three conditions are met:
1. The function is defined at the point.
2. The limit of the function exists at that point.
3. The function's value at that point equals that limit.
If any of these conditions are not satisfied, the function is not continuous at that point.
In the case of our piecewise function at \( x=2 \), there is a discontinuity because the left-hand and right-hand limits do not match. Although the function is defined at this point, the overall limit doesn't exist. Therefore, the function is not continuous at \( x=2 \). Understanding continuity is crucial for graphing as it helps identify where breaks or jumps may occur.

Graphing Piecewise Functions

Graphing a piecewise function involves plotting each piece accurately and considering their domains. For our function:

• For \( x < 2 \), the function is \( f(x) = x - 1 \), which is a simple linear function. You can graph it as a line with a slope of 1, continuing up to \( x = 2 \) but not including it.
• For \( x \geq 2 \), the function is \( f(x) = x^2 - 4x + 6 \), which forms a parabola. Start at \( x = 2 \) and continue for greater values of \( x \).

When the piecewise function changes its formula at a point, it can create a potential discontinuity, especially if the formula values aren't equal at that point, as seen in this exercise. Plotting both sections separately is essential, creating a complete graph for viewers to interpret, noting any gaps or jumps.

Discontinuity in Functions

A discontinuity in a function appears as a break, gap, or jump on its graph. In our exercise, the function shows a clear discontinuity at \( x=2 \). Discontinuities occur when conditions for continuity aren't met, often due to differing left-hand and right-hand limits or if the function is undefined at that point.
There are different types of discontinuities:

• Jump Discontinuity: Seen in our function where \( \lim_{x \to 2^-} f(x) \) and \( \lim_{x \to 2^+} f(x) \) differ.
• Infinite Discontinuity: The function approaches infinity at a point.
• Removable Discontinuity: Occurs when a point is undefined but easily fixable by redefining the function's value there.

Recognizing discontinuity types can guide adjustments in solving equations or tweaking function definitions for specific applications. In this graph, the jump discontinuity visually stands out and is crucial in conveying the true nature of this piecewise relationship.