Problem 46 - Calculus Early Transcendentals

Question

Let $$f(x)=\left\\{\begin{array}{ll}{4-x^{2}} & {\text { if } x \leqslant 2} \\\ {x-1} & {\text { if } x>2}\end{array}\right.$$ (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) \(\lim_{x \rightarrow 2^-} f(x) = 0\), \(\lim_{x \rightarrow 2^+} f(x) = 1\). (b) No, the limit does not exist. (c) A discontinuous piecewise graph.

Step 1: Analyze Left-Hand Limit

To find \(\lim _{x \rightarrow 2^-} f(x)\), we consider the expression of \(f(x)\) when \(x\leq 2\). This is given by \(f(x) = 4-x^2\). Substituting \(x=2\) gives us \(4-2^2 = 4-4 = 0\). Therefore, \(\lim _{x \rightarrow 2^-} f(x) = 0\).

Step 2: Analyze Right-Hand Limit

To find \(\lim _{x \rightarrow 2^+} f(x)\), we use the expression of \(f(x)\) for \(x>2\), which is \(f(x) = x-1\). Substituting \(x=2\) yields \(2-1=1\). Thus, \(\lim _{x \rightarrow 2^+} f(x) = 1\).

Step 3: Determine if Overall Limit Exists

The overall limit \(\lim_{x \rightarrow 2} f(x)\) exists if and only if the left-hand limit \(\lim_{x \rightarrow 2^-} f(x)\) and the right-hand limit \(\lim_{x \rightarrow 2^+} f(x)\) are equal. Since the left-hand limit is 0 and the right-hand limit is 1, they are not equal. Therefore, \(\lim_{x \rightarrow 2} f(x)\) does not exist.

Step 4: Sketch the Graph

The graph of \(f(x)\) consists of two parts. For \(x \leq 2\), the function is a downward-opening parabola segment given by \(y = 4 - x^2\). For \(x > 2\), it is a linear function segment given by \(y = x - 1\). At \(x=2\), there is a discontinuity because the left-hand limit \(y = 0\) and the right-hand limit \(y = 1\) differ.

Key Concepts

Understanding Limits

When dealing with piecewise functions, limits help us understand the behavior of the function as it approaches a specific point. Let's explore this concept using the given piecewise function.

• The left-hand limit, denoted as \( \lim_{x \rightarrow 2^-} f(x) \), examines what happens to the function as \( x \) approaches 2 from the left, where in this case the expression is \( 4-x^2 \).
• By substituting \( x=2 \) into \( 4-x^2 \), we find that \( \lim_{x \rightarrow 2^-} f(x) = 0 \).

Similarly, the right-hand limit, denoted as \( \lim_{x \rightarrow 2^+} f(x) \), examines the function as \( x \) approaches from the right, using the expression \( x-1 \).

• Substitute \( x = 2 \) into \( x-1 \) to get \( \lim_{x \rightarrow 2^+} f(x) = 1 \).
• Since these two limits do not match, the overall limit \( \lim_{x \rightarrow 2} f(x) \) does not exist.

Remember, for a limit to exist at a point in a piecewise function, the left-hand and right-hand limits must be equal.

Exploring Continuity

Continuity is a property that ensures a function has no breaks, jumps, or holes at a point. This means a function is continuous at a point if the limit from the left, the limit from the right, and the function's value at that point are all equal. However, our piecewise function shows a lack of continuity at \( x = 2 \).

• The left-hand limit at \( x=2 \) is 0.
• The right-hand limit at \( x=2 \) is 1.

Since the left-hand and right-hand limits are different, there is a discontinuity at this point. The function does not transition smoothly between the two pieces of the piecewise function. A clear indicator is the absence of a single value that represents \( f(2) \) such that the function is smooth without jumps. This discontinuity results from the differing limits on either side of \( x=2 \). Such behavior is typical in piecewise functions, especially at the boundaries where formula changes occur.

The Art of Graph Sketching

Graph sketching helps visualize the function's behavior and highlights features like continuity and limits. Let's break it down for the given piecewise function:

• For \( x \leq 2 \), \( f(x) = 4 - x^2 \) describes a downward-opening parabola. This segment reaches its maximum at \( x = 0 \) and ends at \( x = 2 \) with the value 0.
• For \( x > 2 \), \( f(x) = x - 1 \) illustrates a straight line with a slope of 1. It starts right after \( x = 2 \) at \( (2, 1) \).

To sketch:- Draw the parabola from \( x = -\infty \) to \( x = 2 \) ensuring it meets the point \( (2, 0) \).- Sketch the line starting from just above the point \( (2, 1) \) to keep the section continuous from \( x = 2 \) onward.Notice the gap at \( x=2 \), as the value shifts from 0 for \( x \leq 2 \) to 1 for \( x>2 \), reinforcing the discontinuity. Sketching piecewise functions helps students see complex relationships and transitions between function segments.