Chapter 12: Problem 24
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) \(\lim _{x \rightarrow 0} f(x)\)
Short Answer
Step by step solution
Understand the function definition
Sketch the graph of the piecewise function
Find the left-hand limit as \( x \rightarrow 0^{-} \)
Find the right-hand limit as \( x \rightarrow 0^{+} \)
Evaluate the two-sided limit as \( x \rightarrow 0 \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limits
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 (\( \lim_{x \rightarrow 0^{+}} f(x) = 1 \)).
Graphing
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.
Continuous and Discontinuous Functions
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.