Problem 24 - Precalculus: Mathematical for Calculus

Question

Graph the piecewise-defined function and use your graph to find the values of the limits, if they exist. $$f(x)=\left\\{\begin{array}{ll} 2 & \text { if } x<0 \\ x+1 & \text { if } x \geq 0 \end{array}\right.$$ (a) \(\lim _{x \rightarrow 0^{-}} f(x)\) (b) \(\lim _{x \rightarrow 0^{+}} f(x)\) (c) \(\lim _{x \rightarrow 0} f(x)\)

Step-by-Step Solution

Answer

(a) 2, (b) 1, (c) does not exist.

Step 1: Understand the function definition

The function is piecewise linear. For values of \( x < 0 \), \( f(x) = 2 \), which means the graph is a horizontal line at \( y = 2 \) for this domain. For values of \( x \geq 0 \), \( f(x) = x + 1 \), which represents a linear function starting at \( (0, 1) \) and having a slope of 1.

Step 2: Sketch the graph of the piecewise function

Plot two parts on a graph. For \( x < 0 \), draw a horizontal line at \( y = 2 \) extending to the left. For \( x \geq 0 \), draw a line starting at point \( (0,1) \) with a slope of 1 going to the right. The point at \( (0,1) \) is filled (solid dot) since \( f(0) \) is defined as \( 1 \).

Step 3: Find the left-hand limit as \( x \rightarrow 0^{-} \)

For \( x \rightarrow 0^{-} \), we look at \( f(x) \) values just to the left of 0, which come from the part of the function where \( f(x) = 2 \). Thus, \( \lim_{x \rightarrow 0^{-}} f(x) = 2 \).

Step 4: Find the right-hand limit as \( x \rightarrow 0^{+} \)

For \( x \rightarrow 0^{+} \), we consider \( f(x) = x + 1 \), which evaluates to 1 when \( x = 0 \). Hence, \( \lim_{x \rightarrow 0^{+}} f(x) = 1 \).

Step 5: Evaluate the two-sided limit as \( x \rightarrow 0 \)

To find \( \lim_{x \rightarrow 0} f(x) \), both one-sided limits must be equal. Since \( \lim_{x \rightarrow 0^{-}} f(x) = 2 eq \lim_{x \rightarrow 0^{+}} f(x) = 1 \), the two-sided limit \( \lim_{x \rightarrow 0} f(x) \) does not exist.

Key Concepts

Limits

When we talk about limits in the context of piecewise functions, we aim to figure out what value a function approaches as the input (or 'x-value') nears a particular number. In this exercise, we're focusing on the behavior of the function as x approaches 0 from two directions: the left and the right.
For the left-hand limit, written as \( \lim_{x \to 0^{-}} f(x) \), we examine the behavior of our piecewise function just before reaching 0. Since for \( x < 0 \), \( f(x) = 2 \), the left-hand limit is 2.

• The limit from the left (\( \lim_{x \rightarrow 0^{-}} f(x) = 2 \)).
• The limit from the right (\( \lim_{x \rightarrow 0^{+}} f(x) = 1 \)).

Notice that limits can vary depending on the direction of approach. When these one-sided limits aren't equal, the overall limit (\( \lim_{x \rightarrow 0} f(x) \)) doesn't exist at that point. In simpler terms, for the limit to exist, the graph must meet as you move from both sides towards the point.

Graphing

Graphing a piecewise function involves plotting different expressions depending on the x-values given by the function rules. In the case of our function, \( f(x) = 2 \) for \( x < 0 \) is just a straight horizontal line at \( y = 2 \). This gives us part of the graph for the negative x-values.
Then, for \( x \geq 0 \), the graph is described by \( f(x) = x + 1 \). This is a linear expression with a slope of 1, meaning that it steps up evenly as you move to the right on the x-axis, starting from the point (0, 1) where our function switches from the constant to the linear rule.
It’s crucial to accurately plot these on a coordinate graph:

• Draw a solid point at (0,1) because the expression \( x+1 \) is defined and included starting at \( x = 0 \).
• For \( x < 0 \), mark a horizontal line at y=2 extending left from just before \( x = 0 \).
• For \( x \geq 0 \), draw the line \( y = x + 1 \) starting firmly at (0,1) and rising up to the right.

Graphing this correctly helps us visually understand the limits and the behavior of the piecewise function near the switch at \( x=0 \).

Continuous and Discontinuous Functions

A function is considered continuous if you can draw it without lifting your pencil off the paper. This implies there are no sudden jumps or breaks in the graph. However, piecewise functions often introduce discontinuities, particularly at break-points where the definitions change.
In our given function, there is a notable discontinuity at \( x=0 \). On the left side of zero, the function holds steady at \( f(x) = 2 \), while on the right, it starts at the point (0,1) and increases with \( f(x) = x+1 \). These differing values mean the graph jumps vertically from \( y=2 \) for the left-hand limit to \( y=1 \) for the right-hand limit. This illustrates why the two-sided limit doesn’t exist at 0, due to this abrupt vertical leap.
For a function to be continuous at a point:

• The function must be defined at the point.
• The limit as x approaches the point from both sides must exist and be equal.
• The value of the function at the point must equal this limit.

Understanding continuity and discontinuity helps us see the fundamental nature of piecewise functions and anticipate how their graphs will behave at certain points.