Problem 52 - Calculus: Early Transcendentals

Question

Let \( g(x) = \left\\{ \begin{array}{ll} x & \mbox{if \) x < 1 \(}\\\ 3 & \mbox{if \) x = 1 \(}\\\ 2 - x^2 & \mbox{if \) 1 < x \le 2 \(}\\\ x - 3 & \mbox{if \) x > 2 \(} \end{array} \right.\) (a) Evaluate each of the following, if it exists. (i) \( \displaystyle \lim_{x \to 1^-}g(x) \) (ii) \( \displaystyle \lim_{x \to 1}g(x) \) (iii) \( g(1) \) (iv) \( \displaystyle \lim_{x \to 2^-}g(x) \) (v) \( \displaystyle \lim_{x \to 2^+}g(x) \) (vi) \( \displaystyle \lim_{x \to 2}g(x) \) (b) Sketch the graph of \( g \).

Step-by-Step Solution

Answer

1: 1, 2: DNE, 3: 3, 4: -2, 5: -1, 6: DNE. The graph has distinct segments with discontinuities at 1 and 2.

Step 1: Evaluating the Left-Hand Limit at x = 1

To find \( \lim_{x \to 1^-} g(x) \), observe the behavior of \( g(x) \) as \( x \) approaches 1 from the left. Since \( x < 1 \), we use the expression \( g(x) = x \), so \( \lim_{x \to 1^-} g(x) = 1 \).

Step 2: Evaluating the Two-Sided Limit at x = 1

\( \lim_{x \to 1} g(x) \) involves checking both left-hand and right-hand limits. From Step 1, the left-hand limit is 1. To find the right-hand limit, we note that when \( x = 1 \), \( g(x) = 3 \) by definition. Therefore, \( \lim_{x \to 1} g(x) \) does not exist because the left-hand and right-hand limits are not equal.

Step 3: Evaluating g(1)

For \( g(1) \), we simply use the explicit value provided for \( x = 1 \) in the definition of the function. Thus, \( g(1) = 3 \).

Step 4: Evaluating the Left-Hand Limit at x = 2

To find \( \lim_{x \to 2^-} g(x) \), evaluate \( g(x) \) as \( x \) approaches 2 from the left (i.e., \( 1 < x < 2 \)). The relevant expression is \( g(x) = 2 - x^2 \). As \( x \to 2^- \), \( g(x) = 2 - 4 = -2 \). So, \( \lim_{x \to 2^-} g(x) = -2 \).

Step 5: Evaluating the Right-Hand Limit at x = 2

To find \( \lim_{x \to 2^+} g(x) \), observe \( g(x) \) for \( x > 2 \), where \( g(x) = x - 3 \). Thus, as \( x \to 2^+ \), \( g(x) = 2 - 3 = -1 \). Thus, \( \lim_{x \to 2^+} g(x) = -1 \).

Step 6: Evaluating the Two-Sided Limit at x = 2

\( \lim_{x \to 2} g(x) \) involves comparing the left-hand and right-hand limits (from Steps 4 and 5). Since \( \lim_{x \to 2^-} g(x) = -2 \) and \( \lim_{x \to 2^+} g(x) = -1 \), the limit does not exist as these limits are not equal.

Step 7: Sketching the Graph of g

To sketch the graph of \( g(x) \), start by plotting key points and intervals based on the given conditions. For \( x < 1 \), use the line \( y = x \). At \( x = 1 \), there is a point at \( y = 3 \) signifying a step or break. For \( 1 < x \leq 2 \), plot the curve \( y = 2 - x^2 \). Finally, for \( x > 2 \), draw the line \( y = x - 3 \). Maintain open or closed circles to reflect inclusivity at endpoints.

Key Concepts

Limits of Functions

Limits help us understand how a function behaves as it approaches a certain point. Whether the limit of a function exists depends on whether the function approaches the same value from both left and right sides at that point.

For piecewise functions like the one in the original problem, different expressions define the function in different intervals. This makes it crucial to verify the behavior from both directions. For instance, at \( x = 1 \), the left-hand limit \( \lim_{x \to 1^-}g(x) \) and the right-hand limit are checked separately.

• Left-hand limit: Uses the expression from a lesser value interval, which is \( g(x) = x \) leading to \( \lim_{x \to 1^-} g(x) = 1 \).
• Right-hand limit: Comes directly from the set value \( g(x) = 3 \), giving \( 3 \) as we approach 1 from the right.

Since the left and right limits do not match, \( \lim_{x \to 1} g(x) \) does not exist.

Graphing Piecewise Functions

Graphing piecewise functions can show potential discontinuities and behaviors of the function over different intervals. When sketching the graph of a piecewise function, each section of the function is represented in its defined interval.

To graph the function \( g(x) \):

• For \( x < 1 \), draw the line \( y = x \) until just before \( x = 1 \) without including it.
• At \( x = 1 \), plot a discrete point at \( y = 3 \).
• For \( 1 < x \leq 2 \), the expression \( y = 2 - x^2 \) forms a curve, which is plotted up to \( x = 2 \).
• For \( x > 2 \), use the line equation \( y = x - 3 \). Continue this line beyond \( x = 2 \).

Don't forget to indicate open or closed circles on the endpoints, showing where values are included or not.

Continuity

A function is continuous at a point if the limit of the function as it approaches from both the left and the right equals the function's value at that point. For piecewise functions, the potential for discontinuity at the boundaries of different function parts is high.

For our function \( g(x) \), consider these critical points:

• At \( x = 1 \), the left and right limits do not match, and \( g(1) = 3 \), implying a discontinuity due to the limit and function value discrepancy.
• The point \( x = 2 \) also reveals discontinuity since \( \lim_{x \to 2^-} g(x) \) and \( \lim_{x \to 2^+} g(x) \) differ, and the function isn't defined to match these limits at \( x = 2 \).

Continuity can be visualized while graphing, especially where pieces do not connect smoothly.

Calculation of Limits

The process of calculating limits in piecewise functions involves checking directionally from where \( x \) approaches a point. Step approach is crucial:

- **Left-Hand Limit Calculation**: Use the expression governing the interval when approaching from the left. For our function, \( \lim_{x \to 2^-} g(x) \) used \( 2 - x^2 \), resulting in \( -2 \).

- **Right-Hand Limit Calculation**: Use the expression for when approaching from the right. For instance, \( \lim_{x \to 2^+} g(x) \) used \( x - 3 \), resulting in \( -1 \).

• These illustrate the importance of different expressions contributing to possible different outcomes.

To fully determine \( \lim_{x \to c} g(x) \), confirm both approaches attain equal results. Unequal outcomes mean the limit doesn't exist.